# Operators and Observables: From Hermitian Operators and Commutators to the Uncertainty Relation

> Why observables are represented by Hermitian operators, derived from the reality of expectation values; the meaning of the canonical commutation relation, a proof of Heisenberg's principle via Robertson's inequality, and measurement as projection.
> https://rikai.mugen-giken.com/en/physics/quantum-mechanics/operators-and-observables

## 0. Key points

- Physical quantities (observables) are represented by **Hermitian operators** on the state space. This is not an arbitrary convention imposed from above: it is forced upon us as soon as we demand that expectation values be real in every state.
- The values obtained in a measurement are the **eigenvalues** of the operator. What supports this correspondence is that the eigenvalues of a Hermitian operator are real and that eigenvectors belonging to distinct eigenvalues are orthogonal.
- Position and momentum satisfy the canonical commutation relation $[\hat X, \hat P] = i\hbar\hat I$. This relation can never be realised in finite dimensions, and that is why quantum mechanics requires an infinite-dimensional state space.
- For any two observables one has $\Delta A\,\Delta B \ge \frac{1}{2}\bigl|\langle [\hat A,\hat B]\rangle\bigr|$ (Robertson's inequality). Heisenberg's uncertainty principle is the special case, and it follows from the Cauchy–Schwarz inequality in a few lines.
- Two observables can possess simultaneously definite values only when they commute.
- Measurement **projects** the state onto an eigenspace (collapse of the wave function). This is a separate axiom, independent of the time evolution generated by the Schrödinger equation, and it is confirmed directly by sequential Stern–Gerlach measurements.

## 1. Motivation: how did observables become "matrices"?

In classical mechanics a physical quantity is a function on phase space. Fix a position $x$ and a momentum $p$, and the energy $H(x,p)$ and the angular momentum are read off as uniquely determined real numbers. To "measure" was to copy down a value already sitting there. This picture is brought into its finished form in [Hamiltonian mechanics](/physics/mechanics/hamiltonian-mechanics) (<Ref to="physics/mechanics/hamiltonian-mechanics#def-phase-space" text="phase space and canonical coordinates" />).

Atomic spectra, however, collided with this picture head-on. A hydrogen atom does not emit light at arbitrary continuous frequencies; it emits only a discrete set of spectral lines ([The birth of quantum mechanics](/en/physics/quantum-mechanics/birth-of-quantum-mechanics)). A discrete set of values does not arise naturally from a continuous function on phase space.

In 1925 Heisenberg adopted the policy of banishing from the theory every quantity not accessible to observation (such as the orbital radius of an electron) and rewriting mechanics using only the observable ones (transition frequencies and intensities). The dynamical variables then cease to be single numbers and become **arrays of quantities indexed by pairs of states**, $x_{nm}$. The product of two such arrays is a matrix product, and interchanging the order changes the value. Born and Jordan identified the source of this noncommutativity and found that position and momentum obey

$$
\hat P\hat X - \hat X\hat P = \frac{\hbar}{i}\hat I .
$$

This is one of the two protagonists of the present article: the canonical commutation relation.

On the other side, in [The Schrödinger equation and the wave function](/en/physics/quantum-mechanics/schrodinger-equation) we used the rule of "substituting"

$$
E \;\longrightarrow\; i\hbar\frac{\partial}{\partial t},\qquad
\boldsymbol{p} \;\longrightarrow\; -i\hbar\nabla
$$

for energy and momentum (<Ref to="physics/quantum-mechanics/schrodinger-equation#prop-momentum" text="derivation of the momentum operator" />). For the moment that substitution is no more than a device for writing down an equation. But Heisenberg's matrices and Schrödinger's differential operators are two faces of one and the same structure. That structure is this: **a physical quantity is a linear operator acting on the state space**.

Why a linear operator? There are two reasons. First, quantum states superpose, that is, the state space is a vector space; if the rule for extracting probabilities from a physical quantity is to be compatible with this linear structure, the quantity itself must be a linear object. Second, a linear operator comes equipped from the outset with a "set of values" that may be discrete, namely its **eigenvalues**. Eigenvalues answer precisely to the demand that discrete energy levels be explained.

This chapter answers three questions.

1. What properties must the operator corresponding to a physical quantity possess?
2. What does it mean physically that two operators fail to commute?
3. What kind of operation on a state is "performing a measurement"?

## 2. Preliminaries: the state space and Dirac notation

The state of a system is represented by a vector (of norm 1) in a complex Hilbert space $\mathcal H$. We write a vector as $|\psi\rangle$ (a ket) and the inner product as $\langle\phi|\psi\rangle$. The inner product is taken to be antilinear in the first argument and linear in the second; that is, for a complex number $c$,

$$
\langle c\phi|\psi\rangle = \bar{c}\,\langle\phi|\psi\rangle,\qquad
\langle\phi|c\psi\rangle = c\,\langle\phi|\psi\rangle,\qquad
\overline{\langle\phi|\psi\rangle} = \langle\psi|\phi\rangle .
$$

The norm is $\|\psi\| = \sqrt{\langle\psi|\psi\rangle}$. The general theory of inner product spaces is collected in [Inner product spaces and Gram–Schmidt orthogonalisation](/mathematics/linear-algebra/inner-product-spaces) (<Ref to="mathematics/linear-algebra/inner-product-spaces#def-inner-product" text="the definition of an inner product space" />).

For a single particle in one dimension, $\mathcal H = L^2(\mathbb{R})$, the space of functions satisfying $\int_{-\infty}^{\infty}|\psi(x)|^2\,dx < \infty$, with inner product

$$
\langle\phi|\psi\rangle = \int_{-\infty}^{\infty}\overline{\phi(x)}\,\psi(x)\,dx .
$$

A linear map $\hat A$ on $\mathcal H$ is called an operator, and $\langle\phi|\hat A|\psi\rangle := \langle\phi|\hat A\psi\rangle$ is called a matrix element. In finite dimensions, choosing an orthonormal basis $\{|e_k\rangle\}$, the numbers $A_{jk} = \langle e_j|\hat A|e_k\rangle$ are exactly the matrix entries.

<Aside type="note">
In what follows we freely interchange differentiation and integration and assume that boundary terms in integration by parts vanish (that $\psi\to 0$ as $|x|\to\infty$). Strictly speaking one must specify the domain of an operator; the relevant subtleties are collected in <Ref to="rem-domain" />. For the time being, think of the discussion as taking place on smooth functions decaying rapidly at infinity (the Schwartz space $\mathcal S(\mathbb{R})$).
</Aside>

## 3. Observables are Hermitian operators

<Definition id="def-adjoint" title="Adjoint operator">
Given an operator $\hat A$, an operator $\hat A^{*}$ satisfying

$$
\langle \hat A\phi\,|\,\psi\rangle = \langle \phi\,|\,\hat A^{*}\psi\rangle
$$

for all $|\phi\rangle, |\psi\rangle \in \mathcal H$ is called the **adjoint** of $\hat A$.
</Definition>

In finite dimensions with an orthonormal basis one has $(\hat A^{*})_{jk} = \overline{A_{kj}}$, so the adjoint corresponds to the conjugate transpose of a matrix. The definition gives $(\hat A\hat B)^{*} = \hat B^{*}\hat A^{*}$ and $(\hat A^{*})^{*} = \hat A$. The first of these can be read off from $\langle \hat A\hat B\phi|\psi\rangle = \langle \hat B\phi|\hat A^{*}\psi\rangle = \langle\phi|\hat B^{*}\hat A^{*}\psi\rangle$.

<Definition id="def-hermitian" title="Hermitian operator">
An operator satisfying $\hat A^{*} = \hat A$, that is, one for which

$$
\langle \hat A\phi\,|\,\psi\rangle = \langle \phi\,|\,\hat A\psi\rangle
$$

holds for all $|\phi\rangle,|\psi\rangle$, is called a **Hermitian** operator (a symmetric operator).
</Definition>

The basic convention of quantum mechanics is that "observables are represented by Hermitian operators", but the following proposition derives it from a more naive requirement. Define the expectation value of a physical quantity $\hat A$ in the state $|\psi\rangle$ (with $\|\psi\|=1$) by $\langle\hat A\rangle_\psi := \langle\psi|\hat A|\psi\rangle$. Since an expectation value is the average of measured values, it must be a real number.

<Proposition id="prop-observable-hermitian" title="Reality of expectation values and Hermiticity">
Let $\mathcal H$ be a complex inner product space and $\hat A$ a linear operator on $\mathcal H$. The following are equivalent.

1. $\langle\psi|\hat A|\psi\rangle \in \mathbb{R}$ for all $|\psi\rangle\in\mathcal H$.
2. $\hat A$ is Hermitian.
</Proposition>

<Proof of="prop-observable-hermitian">
**(2) implies (1).** If $\hat A$ is Hermitian, the conjugate symmetry $\overline{\langle\phi|\psi\rangle} = \langle\psi|\phi\rangle$ of the inner product gives

$$
\overline{\langle\psi|\hat A\psi\rangle} = \langle \hat A\psi|\psi\rangle = \langle\psi|\hat A\psi\rangle ,
$$

the last equality being <Ref to="def-hermitian" />. A number equal to its own complex conjugate is real, so (1) holds.

**(1) implies (2).** Put $\hat B := \hat A - \hat A^{*}$. By <Ref to="def-adjoint" /> we have $\langle\psi|\hat A^{*}\psi\rangle = \langle \hat A\psi|\psi\rangle = \overline{\langle\psi|\hat A\psi\rangle}$, so using hypothesis (1) (that $\langle\psi|\hat A\psi\rangle$ is real) we obtain

$$
\langle\psi|\hat B\psi\rangle = \langle\psi|\hat A\psi\rangle - \overline{\langle\psi|\hat A\psi\rangle} = 0
\qquad(\forall\,|\psi\rangle).
$$

Now take arbitrary $|\psi\rangle,|\phi\rangle$ and apply the identity above first to $|\psi\rangle+|\phi\rangle$. By linearity of $\hat B$ and sesquilinearity of the inner product,

$$
0 = \langle\psi|\hat B\psi\rangle + \langle\psi|\hat B\phi\rangle + \langle\phi|\hat B\psi\rangle + \langle\phi|\hat B\phi\rangle
= \langle\psi|\hat B\phi\rangle + \langle\phi|\hat B\psi\rangle .
$$

Next apply it to $|\psi\rangle + i|\phi\rangle$. Noting that antilinearity in the first argument gives $\langle i\phi| = -i\langle\phi|$, we get

$$
0 = \langle\psi|\hat B\psi\rangle + i\langle\psi|\hat B\phi\rangle - i\langle\phi|\hat B\psi\rangle + (-i)(i)\langle\phi|\hat B\phi\rangle
= i\bigl(\langle\psi|\hat B\phi\rangle - \langle\phi|\hat B\psi\rangle\bigr),
$$

that is, $\langle\psi|\hat B\phi\rangle = \langle\phi|\hat B\psi\rangle$. Adding this to the previous identity yields $2\langle\psi|\hat B\phi\rangle = 0$, so $\langle\psi|\hat B\phi\rangle = 0$ for all $|\psi\rangle,|\phi\rangle$. Choosing $|\psi\rangle = \hat B|\phi\rangle$ gives $\|\hat B\phi\|^2 = 0$, hence $\hat B = 0$, that is, $\hat A = \hat A^{*}$.
</Proof>

<Remark id="rem-complex-essential">
It is essential to this proof that the inner product space be **complex**. Over a real inner product space there are counterexamples. The rotation of $\mathbb{R}^2$ through $90^\circ$, $R = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$, satisfies $\langle \boldsymbol{v}, R\boldsymbol{v}\rangle = 0$ for every $\boldsymbol{v}$ (the rotated vector is orthogonal to the original), which is always real; yet $R^{\mathsf{T}} = -R \ne R$, so $R$ is not symmetric. The difference is that the step of substituting $|\psi\rangle + i|\phi\rangle$ is unavailable. It is in situations like this that the use of complex numbers in quantum mechanics pays off.
</Remark>

Not only expectation values but individual measured values can be read off from the operator. The grounds for this are the following theorem.

<Theorem id="thm-hermitian-spectrum" title="Eigenvalues and eigenvectors of a Hermitian operator">
Let $\hat A$ be a Hermitian operator on a complex inner product space $\mathcal H$.

1. Every eigenvalue of $\hat A$ is real; that is, if $\hat A|\psi\rangle = a|\psi\rangle$ with $|\psi\rangle \ne 0$, then $a\in\mathbb{R}$.
2. Eigenvectors belonging to distinct eigenvalues are orthogonal; that is, if $\hat A|\psi\rangle = a|\psi\rangle$, $\hat A|\phi\rangle = b|\phi\rangle$ and $a \ne b$, then $\langle\phi|\psi\rangle = 0$.
</Theorem>

<Proof of="thm-hermitian-spectrum">
**(1).** Let $\hat A|\psi\rangle = a|\psi\rangle$ with $|\psi\rangle\ne 0$. Pairing with $\langle\psi|$ on the left,

$$
\langle\psi|\hat A\psi\rangle = a\,\langle\psi|\psi\rangle .
$$

On the other hand <Ref to="def-hermitian" /> gives $\langle\psi|\hat A\psi\rangle = \langle \hat A\psi|\psi\rangle = \overline{a}\,\langle\psi|\psi\rangle$ (by antilinearity in the first argument). Subtracting the two identities gives $(a - \overline{a})\langle\psi|\psi\rangle = 0$. Since $|\psi\rangle\ne 0$ we have $\langle\psi|\psi\rangle = \|\psi\|^2 > 0$, hence $a = \overline{a}$, that is, $a$ is real.

**(2).** Compute the matrix element $\langle\phi|\hat A\psi\rangle$ in two ways. Using $\hat A|\psi\rangle = a|\psi\rangle$ directly gives $\langle\phi|\hat A\psi\rangle = a\langle\phi|\psi\rangle$. Moving $\hat A$ to the left by Hermiticity gives $\langle\phi|\hat A\psi\rangle = \langle \hat A\phi|\psi\rangle = \overline{b}\,\langle\phi|\psi\rangle$, and by (1) the number $b$ is real, so $\overline{b} = b$. Hence $(a-b)\langle\phi|\psi\rangle = 0$, and the hypothesis $a\ne b$ gives $\langle\phi|\psi\rangle = 0$.
</Proof>

Part (1) guarantees the obvious requirement that measured values be real, and part (2) says that states corresponding to distinct measured values are mutually distinguishable. Moreover, in finite dimensions (or in infinite dimensions under suitable conditions) the eigenvectors of a Hermitian operator form an orthonormal basis of $\mathcal H$. This is the [spectral theorem](/mathematics/linear-algebra/spectral-theorem) (<Ref to="mathematics/linear-algebra/spectral-theorem#thm-spectral" text="the spectral theorem for Hermitian matrices" />), and it underlies the measurement axiom (<Ref to="ax-measurement" />).

<Example id="ex-position-momentum-hamiltonian" title="Hermiticity of position, momentum and the Hamiltonian">
On $\mathcal H = L^2(\mathbb{R})$ set
$(\hat X\psi)(x) = x\,\psi(x)$ and $(\hat P\psi)(x) = -i\hbar\,\psi'(x)$.

**Position.** Since $x$ is real we have $\overline{x\phi(x)} = x\overline{\phi(x)}$, so

$$
\langle\phi|\hat X\psi\rangle = \int \overline{\phi(x)}\,x\psi(x)\,dx
= \int \overline{x\phi(x)}\,\psi(x)\,dx = \langle \hat X\phi|\psi\rangle .
$$

**Momentum.** Integrating by parts and using $\phi,\psi\to 0$ as $|x|\to\infty$, the boundary term drops out and

$$
\begin{aligned}
\langle\phi|\hat P\psi\rangle
&= \int \overline{\phi}\,(-i\hbar\psi')\,dx
= \Bigl[-i\hbar\,\overline{\phi}\,\psi\Bigr]_{-\infty}^{\infty} + i\hbar\int \overline{\phi}'\,\psi\,dx \\
&= \int \overline{(-i\hbar\phi')}\,\psi\,dx = \langle \hat P\phi|\psi\rangle .
\end{aligned}
$$

In the second line we used $\overline{-i\hbar\,\phi'} = +i\hbar\,\overline{\phi}'$. Without the imaginary unit (that is, for $d/dx$ alone) the signs would not match and the operator would not be Hermitian. The $-i$ in the momentum operator is needed for exactly this one point.

**Hamiltonian.** Let $\hat H = \hat P^2/(2m) + V(\hat X)$ with $V$ real-valued. Then $(\hat P^2)^{*} = \hat P^{*}\hat P^{*} = \hat P^2$ and $V(\hat X)^{*} = V(\hat X)$ (multiplication by a real-valued function), so $\hat H$ is Hermitian as well, since a real linear combination of Hermitian operators is Hermitian.
</Example>

<Remark id="rem-domain">
In infinite dimensions, "Hermitian (symmetric)" and "self-adjoint" are different notions. Strictly, an operator $\hat A$ carries a domain $D(\hat A)$, and the domain of $\hat A^{*}$ may satisfy $D(\hat A^{*}) \supseteq D(\hat A)$ strictly. When the two coincide the operator is called self-adjoint, and the spectral theorem, as well as unitarity of the time evolution $e^{-i\hat Ht/\hbar}$, requires this stronger condition.

This is not pedantry. Consider $\hat P = -i\hbar\,d/dx$ on the $L^2$ space of the half-line $[0,\infty)$. It is symmetric (on smooth functions vanishing at the origin and at infinity), but it possesses **no self-adjoint extension at all**. The reason is that the solution $u = e^{-x}$ of $\hat P^{*}u = i\hbar u$ is square integrable while the solution $u = e^{x}$ of $\hat P^{*}u = -i\hbar u$ is not, so the deficiency indices are the asymmetric pair $(1,0)$. The naive observable "the momentum of a particle moving on a half-line" simply does not exist. On a finite interval $[0,L]$, by contrast, the deficiency indices are $(1,1)$, and there appears a family of self-adjoint extensions classified by the phase $\theta$ in $\psi(L) = e^{i\theta}\psi(0)$. See Chapter VIII of Reed–Simon for details.
</Remark>

## 4. The canonical commutation relation

<Definition id="def-commutator" title="Commutator">
For two operators $\hat A,\hat B$,

$$
[\hat A,\hat B] := \hat A\hat B - \hat B\hat A
$$

is called the **commutator**. When $[\hat A,\hat B] = 0$ we say that $\hat A$ and $\hat B$ **commute**.
</Definition>

<Lemma id="lem-commutator-rules" title="Algebraic properties of the commutator">
For any operators $\hat A,\hat B,\hat C$ and complex numbers $\alpha,\beta$ the following hold.

1. (Bilinearity) $[\alpha\hat A + \beta\hat B, \hat C] = \alpha[\hat A,\hat C] + \beta[\hat B,\hat C]$, and similarly in the second argument.
2. (Antisymmetry) $[\hat A,\hat B] = -[\hat B,\hat A]$.
3. (Leibniz rule) $[\hat A,\hat B\hat C] = [\hat A,\hat B]\hat C + \hat B[\hat A,\hat C]$ and $[\hat A\hat B,\hat C] = \hat A[\hat B,\hat C] + [\hat A,\hat C]\hat B$.
4. (Jacobi identity) $[\hat A,[\hat B,\hat C]] + [\hat B,[\hat C,\hat A]] + [\hat C,[\hat A,\hat B]] = 0$.
</Lemma>

<Proof of="lem-commutator-rules">
Parts (1) and (2) follow by writing out the definition. For (3), expanding the right-hand side gives

$$
(\hat A\hat B - \hat B\hat A)\hat C + \hat B(\hat A\hat C - \hat C\hat A)
= \hat A\hat B\hat C - \hat B\hat A\hat C + \hat B\hat A\hat C - \hat B\hat C\hat A
= \hat A\hat B\hat C - \hat B\hat C\hat A = [\hat A,\hat B\hat C]
$$

where the two middle terms cancel. Similarly $\hat A[\hat B,\hat C] + [\hat A,\hat C]\hat B = \hat A\hat B\hat C - \hat C\hat A\hat B = [\hat A\hat B,\hat C]$. For (4), expanding all three double commutators produces twelve terms; each of the six orderings of type $\hat A\hat B\hat C$ occurs twice, once with $+1$ and once with $-1$, and they cancel. For instance $\hat A\hat B\hat C$ appears as $+\hat A\hat B\hat C$ in the expansion of the first term $[\hat A,[\hat B,\hat C]]$ and as $-\hat A\hat B\hat C$ in the expansion of the third term $[\hat C,[\hat A,\hat B]]$.
</Proof>

Part (3) says that the commutator acts like a derivative. Indeed $[\hat A,\cdot\,]$ is a map obeying the Leibniz rule with respect to products, that is, a **derivation**. This point of view shows its power in the following computation.

<Theorem id="thm-ccr" title="Canonical commutation relation">
On $\mathcal H = L^2(\mathbb{R})$ let $(\hat X\psi)(x) = x\psi(x)$ and $(\hat P\psi)(x) = -i\hbar\psi'(x)$. Then for every differentiable $\psi$,

$$
[\hat X,\hat P]\,\psi = i\hbar\,\psi,
$$

that is, as an identity of operators, $[\hat X,\hat P] = i\hbar\hat I$.
</Theorem>

<Proof of="thm-ccr">
Apply the operators in both orders, following the definitions.

$$
(\hat X\hat P\psi)(x) = x\cdot\bigl(-i\hbar\psi'(x)\bigr) = -i\hbar\,x\psi'(x).
$$

In the other order $\hat P$ acts on the product $x\psi(x)$, so by the product rule

$$
(\hat P\hat X\psi)(x) = -i\hbar\,\frac{d}{dx}\bigl(x\psi(x)\bigr)
= -i\hbar\bigl(\psi(x) + x\psi'(x)\bigr).
$$

Taking the difference, the terms in $x\psi'$ cancel and

$$
([\hat X,\hat P]\psi)(x) = -i\hbar x\psi'(x) + i\hbar\psi(x) + i\hbar x\psi'(x) = i\hbar\,\psi(x) .
$$

Since $\psi$ was arbitrary, $[\hat X,\hat P] = i\hbar\hat I$. Note that the source of the noncommutativity is the extra term produced by the product rule.
</Proof>

This relation is a far stronger constraint than it looks.

<Corollary id="cor-no-finite-dim" title="The canonical commutation relation cannot be realised in finite dimensions">
Let $n \ge 1$. There exist no complex $n \times n$ matrices $A, B$ with $AB - BA = i\hbar I_n$ (where $\hbar \ne 0$).
</Corollary>

<Proof of="cor-no-finite-dim">
Use the trace. First, for any $n \times n$ matrices $A,B$,

$$
\operatorname{tr}(AB) = \sum_{j=1}^{n}\sum_{k=1}^{n} A_{jk}B_{kj}
= \sum_{k=1}^{n}\sum_{j=1}^{n} B_{kj}A_{jk} = \operatorname{tr}(BA)
$$

(the sums are finite, so their order may be interchanged freely). Hence, by linearity of the trace,

$$
\operatorname{tr}(AB - BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = 0 .
$$

On the other hand $\operatorname{tr}(i\hbar I_n) = i\hbar n \ne 0$ (since $\hbar\ne 0$ and $n\ge 1$). The traces of the two sides disagree, so no such $A,B$ exist.
</Proof>

In other words, the state space of a quantum system in which both position and momentum are defined must be **infinite-dimensional**. More strongly still, neither $\hat X$ nor $\hat P$ can be a bounded operator (<Ref to="thm-wintner" />). This is why the domain issues of <Ref to="rem-domain" /> cannot be avoided. Conversely, for degrees of freedom that close up in finite dimensions, such as spin, no relation of the canonical form appears ([Angular momentum and spin](/physics/quantum-mechanics/angular-momentum-and-spin)).

<Example id="ex-commutators" title="Computing commutators">
Part (3) of <Ref to="lem-commutator-rules" /> together with <Ref to="thm-ccr" /> suffices to compute essentially every commutator we need.

**(a) $[\hat X,\hat P^2]$.** Split $\hat P^2 = \hat P\cdot\hat P$ using the Leibniz rule:

$$
[\hat X,\hat P^2] = [\hat X,\hat P]\hat P + \hat P[\hat X,\hat P]
= i\hbar\hat P + \hat P\,i\hbar = 2i\hbar\hat P .
$$

**(b) $[\hat X^n,\hat P]$.** Iterating the Leibniz rule in the same way,
$[\hat X^n,\hat P] = \sum_{k=0}^{n-1}\hat X^{k}[\hat X,\hat P]\hat X^{n-1-k} = i\hbar\,n\hat X^{n-1}$.
The factors of $\hat X$ commute among themselves, so their order does not matter.

**(c) $[\hat P, V(\hat X)]$.** For a differentiable real-valued $V$ we compute directly in the position representation:

$$
\bigl([\hat P,V]\psi\bigr)(x) = -i\hbar\frac{d}{dx}\bigl(V(x)\psi(x)\bigr) - V(x)\bigl(-i\hbar\psi'(x)\bigr)
= -i\hbar\,V'(x)\psi(x).
$$

Hence $[\hat P, V(\hat X)] = -i\hbar\,V'(\hat X)$. Part (b) is an alternative derivation of the case $V(x)=x^n$ (the signs agree by antisymmetry, $[\hat X^n,\hat P] = -[\hat P,\hat X^n]$).
</Example>

<Remark id="rem-poisson">
In classical mechanics, functions $f,g$ on phase space have a Poisson bracket $\{f,g\} = \dfrac{\partial f}{\partial x}\dfrac{\partial g}{\partial p} - \dfrac{\partial f}{\partial p}\dfrac{\partial g}{\partial x}$, and $\{x,p\} = 1$ (see [Canonical transformations and Poisson brackets](/physics/mechanics/canonical-transformations), <Ref to="physics/mechanics/canonical-transformations#def-poisson" text="the definition of the Poisson bracket" />). Both the Poisson bracket and the commutator are bilinear and antisymmetric and satisfy the Leibniz rule and the Jacobi identity (<Ref to="lem-commutator-rules" />). They therefore carry the same algebraic structure, that of a Lie algebra.

Imposing **canonical quantisation**, that is, the correspondence

$$
\{f,g\} \;\longmapsto\; \frac{1}{i\hbar}[\hat f,\hat g],
$$

one obtains $[\hat X,\hat P] = i\hbar$ immediately from $\{x,p\}=1$. <Ref to="thm-ccr" /> is also a check that this prescription can be implemented consistently.

This correspondence cannot, however, be imposed on *all* observables at once. It is known that there is no quantisation map satisfying simultaneously linearity, $1\mapsto\hat I$, $x\mapsto\hat X$, $p\mapsto\hat P$, and the correspondence between Poisson brackets and commutators for all polynomials (the Groenewold–van Hove theorem); the obstruction appears at polynomials of degree three and higher. On the other hand, when $\hat X,\hat P$ are expressed in Weyl form (as a relation between the unitary groups $e^{ia\hat X}$ and $e^{ib\hat P}$), the irreducible representation on a separable Hilbert space is essentially unique, namely the Schrödinger representation on $L^2(\mathbb{R})$ (the Stone–von Neumann theorem). This is the mathematical content of the claim that Heisenberg's matrix mechanics and Schrödinger's wave mechanics are one and the same theory.
</Remark>

## 5. Uncertainty relations

<Definition id="def-uncertainty" title="Expectation value and uncertainty">
Let $\hat A$ be a Hermitian operator and $|\psi\rangle$ a state with $\|\psi\|=1$. We call

$$
\langle \hat A\rangle_\psi := \langle\psi|\hat A|\psi\rangle,\qquad
\Delta_\psi \hat A := \sqrt{\bigl\langle (\hat A - \langle\hat A\rangle_\psi)^2\bigr\rangle_\psi}
$$

the **expectation value** and the **uncertainty** (standard deviation) of $\hat A$, respectively. Below we drop the subscript $\psi$ when the state is clear from the context.
</Definition>

Let us check that the quantity under the square root is nonnegative. By <Ref to="prop-observable-hermitian" /> the number $\bar a := \langle\hat A\rangle_\psi$ is real, so $\hat A - \bar a\hat I$ is again Hermitian. Therefore

$$
\bigl\langle(\hat A-\bar a)^2\bigr\rangle_\psi
= \langle\psi|(\hat A-\bar a)(\hat A-\bar a)\psi\rangle
= \bigl\langle (\hat A-\bar a)\psi\,\bigl|\,(\hat A-\bar a)\psi\bigr\rangle
= \bigl\|(\hat A-\bar a)\psi\bigr\|^2 \ge 0
$$

(the second equality uses <Ref to="def-hermitian" />), and expanding shows that this equals $\langle\hat A^2\rangle - \langle\hat A\rangle^2$. This identity gives the following proposition at once.

<Proposition id="prop-sharp-value" title="Vanishing uncertainty occurs exactly in eigenstates">
Let $\hat A$ be a Hermitian operator and $|\psi\rangle$ a state with $\|\psi\|=1$. Then $\Delta_\psi\hat A = 0$ if and only if $|\psi\rangle$ is an eigenvector of $\hat A$ (with eigenvalue $\langle\hat A\rangle_\psi$).
</Proposition>

<Proof of="prop-sharp-value">
By the computation above, $(\Delta_\psi\hat A)^2 = \|(\hat A - \bar a)\psi\|^2$ with $\bar a = \langle\hat A\rangle_\psi$. A norm vanishes if and only if the vector vanishes, so $\Delta_\psi\hat A = 0 \iff (\hat A-\bar a)|\psi\rangle = 0 \iff \hat A|\psi\rangle = \bar a|\psi\rangle$. Since $\|\psi\|=1\ne0$, the vector $|\psi\rangle$ is an eigenvector. Conversely, if $\hat A|\psi\rangle = a|\psi\rangle$ then $\bar a = \langle\psi|a\psi\rangle = a$, and the same identity gives $\Delta_\psi\hat A = 0$.
</Proof>

This is the basic correspondence: "a state in which a physical quantity has a definite value" means "an eigenstate of that quantity". Can two quantities then have definite values **simultaneously**? The answer is supplied by the next theorem.

<Theorem id="thm-robertson" title="Robertson's uncertainty relation">
Let $\hat A,\hat B$ be Hermitian operators and $|\psi\rangle$ a state with $\|\psi\|=1$ lying in the domains of $\hat A\hat B$ and $\hat B\hat A$. Then

$$
\Delta_\psi \hat A \cdot \Delta_\psi \hat B \;\ge\; \frac{1}{2}\Bigl|\bigl\langle [\hat A,\hat B]\bigr\rangle_\psi\Bigr| .
$$
</Theorem>

<Proof of="thm-robertson">
Put $\bar a = \langle\hat A\rangle_\psi$ and $\bar b = \langle\hat B\rangle_\psi$ (both real by <Ref to="prop-observable-hermitian" />) and set

$$
\tilde A := \hat A - \bar a\hat I,\qquad \tilde B := \hat B - \bar b\hat I .
$$

Subtracting a real multiple of $\hat I$ preserves Hermiticity. Moreover $\hat I$ commutes with every operator, so by the bilinearity in <Ref to="lem-commutator-rules" />

$$
[\tilde A,\tilde B] = [\hat A,\hat B] .
$$

Set $|f\rangle := \tilde A|\psi\rangle$ and $|g\rangle := \tilde B|\psi\rangle$. The identity verified just after <Ref to="def-uncertainty" /> gives

$$
\|f\|^2 = (\Delta_\psi\hat A)^2,\qquad \|g\|^2 = (\Delta_\psi\hat B)^2 .
$$

**Step 1: the Cauchy–Schwarz inequality.** By the general inequality valid in any inner product space, <Ref to="mathematics/linear-algebra/inner-product-spaces#thm-cauchy-schwarz" text="the Cauchy–Schwarz inequality" />,

$$
\|f\|^2\,\|g\|^2 \;\ge\; \bigl|\langle f|g\rangle\bigr|^2 .
$$

**Step 2: keep only the imaginary part.** For a complex number $z$ we have $|z|^2 = (\operatorname{Re}z)^2 + (\operatorname{Im}z)^2 \ge (\operatorname{Im}z)^2$, hence

$$
\bigl|\langle f|g\rangle\bigr|^2 \;\ge\; \bigl(\operatorname{Im}\langle f|g\rangle\bigr)^2 .
$$

**Step 3: express the imaginary part by the commutator.** Hermiticity of $\tilde A$ (<Ref to="def-hermitian" />) gives

$$
\langle f|g\rangle = \langle \tilde A\psi|\tilde B\psi\rangle = \langle\psi|\tilde A\tilde B\psi\rangle,
\qquad
\overline{\langle f|g\rangle} = \langle g|f\rangle = \langle\psi|\tilde B\tilde A\psi\rangle .
$$

Therefore

$$
2i\operatorname{Im}\langle f|g\rangle = \langle f|g\rangle - \overline{\langle f|g\rangle}
= \bigl\langle [\tilde A,\tilde B]\bigr\rangle_\psi = \bigl\langle [\hat A,\hat B]\bigr\rangle_\psi .
$$

**Conclusion.** Combining the three steps,

$$
(\Delta_\psi\hat A)^2(\Delta_\psi\hat B)^2 \;\ge\; \bigl(\operatorname{Im}\langle f|g\rangle\bigr)^2
= \left|\frac{1}{2i}\bigl\langle[\hat A,\hat B]\bigr\rangle_\psi\right|^2
= \frac{1}{4}\Bigl|\bigl\langle[\hat A,\hat B]\bigr\rangle_\psi\Bigr|^2 .
$$

Taking square roots (the left-hand side is nonnegative) gives the assertion.
</Proof>

<Corollary id="cor-heisenberg" title="Heisenberg's uncertainty principle">
For every state $|\psi\rangle$ of a single particle in one dimension (with $\|\psi\|=1$ and lying in the domains of $\hat X\hat P$ and $\hat P\hat X$),

$$
\Delta_\psi \hat X \cdot \Delta_\psi \hat P \;\ge\; \frac{\hbar}{2}.
$$
</Corollary>

<Proof of="cor-heisenberg">
Take $\hat A = \hat X$ and $\hat B = \hat P$ in <Ref to="thm-robertson" />. By <Ref to="thm-ccr" /> we have $[\hat X,\hat P] = i\hbar\hat I$, so

$$
\bigl\langle[\hat X,\hat P]\bigr\rangle_\psi = i\hbar\,\langle\psi|\psi\rangle = i\hbar .
$$

Hence the right-hand side is $\frac{1}{2}|i\hbar| = \hbar/2$. Note that this value is a constant independent of the state.
</Proof>

The right-hand side is a state-independent constant only because the commutator has the special form of a constant multiple of $\hat I$. For general observables the right-hand side depends on the state and may even vanish.

<Example id="ex-gaussian-minimum" title="Equality is attained by Gaussian wave packets">
Let us determine when equality holds in <Ref to="cor-heisenberg" />. Equality in Step 1 of the proof (Cauchy–Schwarz) holds when $|g\rangle = \lambda|f\rangle$ for some $\lambda\in\mathbb{C}$, that is, when the two vectors are linearly dependent; equality in Step 2 holds when $\operatorname{Re}\langle f|g\rangle = 0$. Since $\langle f|g\rangle = \lambda\|f\|^2$, the latter means $\operatorname{Re}\lambda = 0$, that is, $\lambda = i\mu$ with $\mu\in\mathbb{R}$. The condition for equality is therefore

$$
(\hat P - \bar p)|\psi\rangle = i\mu\,(\hat X - \bar x)|\psi\rangle .
$$

To simplify the description take $\bar x = \bar p = 0$ (the general case follows by translation). In the position representation,

$$
-i\hbar\,\psi'(x) = i\mu\,x\,\psi(x)
\quad\Longleftrightarrow\quad
\frac{\psi'(x)}{\psi(x)} = -\frac{\mu}{\hbar}\,x .
$$

Integrating both sides gives $\log\psi = -\dfrac{\mu x^2}{2\hbar} + \text{const}$, that is,

$$
\psi(x) = C\exp\!\left(-\frac{\mu x^2}{2\hbar}\right).
$$

Square integrability requires $\mu > 0$. Setting $\sigma^2 := \hbar/(2\mu)$ and normalising,

$$
\psi(x) = \frac{1}{(2\pi\sigma^2)^{1/4}}\exp\!\left(-\frac{x^2}{4\sigma^2}\right),
\qquad
|\psi(x)|^2 = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\!\left(-\frac{x^2}{2\sigma^2}\right),
$$

so the probability density is a Gaussian of variance $\sigma^2$. From this, $\Delta\hat X = \sigma$.

Now the momentum side. Since $\psi$ is real, $\langle\hat P\rangle = \int\overline{\psi}(-i\hbar\psi')\,dx = -i\hbar\bigl[\psi^2/2\bigr]_{-\infty}^{\infty} = 0$. Next, integrating by parts (the boundary terms vanish),

$$
\langle\hat P^2\rangle = \int \overline{\psi}\,(-\hbar^2\psi'')\,dx = \hbar^2\int |\psi'(x)|^2\,dx .
$$

Since $\psi'(x) = -\dfrac{x}{2\sigma^2}\psi(x)$,

$$
\int|\psi'|^2\,dx = \frac{1}{4\sigma^4}\int x^2|\psi(x)|^2\,dx = \frac{1}{4\sigma^4}\cdot\sigma^2 = \frac{1}{4\sigma^2},
$$

where we used $\int x^2|\psi|^2dx = \sigma^2$ (the second moment of a Gaussian of variance $\sigma^2$). Hence $\langle\hat P^2\rangle = \hbar^2/(4\sigma^2)$ and $\Delta\hat P = \hbar/(2\sigma)$. Altogether

$$
\Delta\hat X\cdot\Delta\hat P = \sigma\cdot\frac{\hbar}{2\sigma} = \frac{\hbar}{2},
$$

so equality does indeed hold. Making $\sigma$ small sharpens the position, but the momentum spread grows like $1/\sigma$ and the product is unchanged.
</Example>

<Figure caption="Position distribution (left) and momentum distribution (right) of a Gaussian wave packet. The solid curve is a packet with small width σ, the dashed curve one with large σ; the packet that is sharp in position is broad in momentum. The product ΔX·ΔP equals ħ/2 for both.">
<svg viewBox="0 0 720 272" width="100%" role="img" aria-label="Two graphs showing that a packet with a sharp position distribution has a broad momentum distribution">
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    <text x="185" y="34" text-anchor="middle">Position probability distribution</text>
    <text x="545" y="34" text-anchor="middle">Momentum probability distribution</text>
    <text x="185" y="234" text-anchor="middle">position x</text>
    <text x="545" y="234" text-anchor="middle">momentum p</text>
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</Figure>

<Example id="ex-spin-uncertainty" title="The uncertainty relation for spin 1/2">
On a two-dimensional state space, write the spin operators as $\hat S_j = \frac{\hbar}{2}\sigma_j$, where

$$
\sigma_x = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\quad
\sigma_y = \begin{pmatrix}0 & -i\\ i & 0\end{pmatrix},\quad
\sigma_z = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}
$$

are the Pauli matrices. A direct computation gives $\sigma_x\sigma_y = i\sigma_z$ and $\sigma_y\sigma_x = -i\sigma_z$, so $[\hat S_x,\hat S_y] = i\hbar\hat S_z$. By <Ref to="thm-robertson" />,

$$
\Delta \hat S_x\cdot\Delta \hat S_y \ge \frac{\hbar}{2}\bigl|\langle \hat S_z\rangle\bigr| .
$$

Take the state $|{\uparrow_z}\rangle = \begin{pmatrix}1\\0\end{pmatrix}$; then $\langle\hat S_z\rangle = \hbar/2$, so the right-hand side is $\hbar^2/4$. Now compute the left-hand side. Since $\sigma_x|{\uparrow_z}\rangle = \begin{pmatrix}0\\1\end{pmatrix}$ we get $\langle\hat S_x\rangle = \frac{\hbar}{2}\langle{\uparrow_z}|\sigma_x|{\uparrow_z}\rangle = 0$, and from $\sigma_x^2 = I$ we get $\langle\hat S_x^2\rangle = \hbar^2/4$. Hence $\Delta\hat S_x = \hbar/2$, and likewise $\Delta\hat S_y = \hbar/2$, so the product is exactly $\hbar^2/4$: equality holds.

In a state with definite spin along $z$, the spin components along $x$ and $y$ are maximally uncertain. Combined with <Ref to="prop-sharp-value" />, the reason is seen to be that $|{\uparrow_z}\rangle$ is not an eigenstate of $\hat S_x$.
</Example>

<Remark id="rem-uncertainty-meaning">
The quantity $\Delta_\psi\hat A$ is **the spread of the measured values obtained by measuring $\hat A$ once on each of many systems all prepared in the same state $|\psi\rangle$**. It is not the "disturbance caused by measurement" when $\hat A$ and $\hat B$ are measured in succession on a single system. What Heisenberg discussed in his 1927 paper, using the thought experiment of the $\gamma$-ray microscope, was the latter (the relation between measurement error and disturbance), whereas <Ref to="thm-robertson" /> states the former (a spread carried by the state). These are two different claims; the correct universally valid inequality for measurement error and disturbance was formulated by Ozawa (2003). Note that <Ref to="thm-robertson" /> is determined by properties of the state alone and makes no reference whatsoever to a measuring apparatus.
</Remark>

## 6. Simultaneous measurement and the measurement axiom

<Theorem id="thm-simultaneous" title="Commuting Hermitian operators are simultaneously diagonalisable">
Let $\mathcal H$ be a finite-dimensional complex inner product space and $\hat A,\hat B$ Hermitian operators on $\mathcal H$. The following are equivalent.

1. $[\hat A,\hat B] = 0$.
2. There exists an orthonormal basis of $\mathcal H$ consisting of simultaneous eigenvectors of $\hat A$ and $\hat B$.
</Theorem>

<Proof of="thm-simultaneous">
**(2) implies (1).** Let $\{|e_k\rangle\}$ be such an orthonormal basis, with $\hat A|e_k\rangle = a_k|e_k\rangle$ and $\hat B|e_k\rangle = b_k|e_k\rangle$. Then

$$
\hat A\hat B|e_k\rangle = a_kb_k|e_k\rangle = \hat B\hat A|e_k\rangle
$$

for every $k$. Since $[\hat A,\hat B]$ is a linear operator sending every basis vector to $0$, it vanishes on all of $\mathcal H$.

**(1) implies (2).** By the [spectral theorem](/mathematics/linear-algebra/spectral-theorem), the eigenvalues of $\hat A$ are real and $\mathcal H$ decomposes as the orthogonal direct sum of the eigenspaces of $\hat A$, $\mathcal H = \bigoplus_{a} V_a$ with $V_a = \ker(\hat A - a\hat I)$.

We first show that $\hat B$ preserves each $V_a$. Let $|v\rangle\in V_a$. The hypothesis $\hat A\hat B = \hat B\hat A$ gives

$$
\hat A\bigl(\hat B|v\rangle\bigr) = \hat B\bigl(\hat A|v\rangle\bigr) = \hat B\,(a|v\rangle) = a\bigl(\hat B|v\rangle\bigr),
$$

so $\hat B|v\rangle$ is again an eigenvector with eigenvalue $a$ (or is $0$), that is, $\hat B|v\rangle \in V_a$.

Next consider the restriction $\hat B|_{V_a}$ of $\hat B$ to $V_a$. It is a linear operator on $V_a$, and it is Hermitian because $\langle\hat B\phi|\psi\rangle = \langle\phi|\hat B\psi\rangle$ for all $|\phi\rangle,|\psi\rangle \in V_a$ (a special case of an identity valid on all of $\mathcal H$). By the spectral theorem again, $V_a$ has an orthonormal basis of eigenvectors of $\hat B|_{V_a}$. These lie in $V_a$, hence are eigenvectors of $\hat A$ with eigenvalue $a$ as well, so they are simultaneous eigenvectors.

Finally, the spaces $V_a$ for distinct $a$ are orthogonal by part (2) of <Ref to="thm-hermitian-spectrum" />. Putting together the orthonormal bases constructed in each $V_a$ therefore yields an orthonormal basis of $\mathcal H$.
</Proof>

Read physically: states in which two observables have simultaneously definite values exist in sufficient abundance (enough to form a basis) precisely when the observables commute. A family of commuting observables whose simultaneous eigenstates specify a state uniquely is called a **complete set of commuting observables**. The typical example is the choice $\hat H,\hat{\boldsymbol{L}}^2,\hat L_z$ for the hydrogen atom ([Angular momentum and spin](/physics/quantum-mechanics/angular-momentum-and-spin)).

With this in hand, we state the axiom governing measurement.

<Axiom id="ax-measurement" title="Measurement axiom">
Let the observable $\hat A$ be a Hermitian operator with spectral decomposition

$$
\hat A = \sum_{a} a\,\hat P_a,\qquad
\hat P_a\hat P_{a'} = \delta_{aa'}\hat P_a,\qquad
\sum_a \hat P_a = \hat I
$$

(where $\hat P_a$ is the orthogonal projection onto the eigenspace of the eigenvalue $a$). When $\hat A$ is measured on a state $|\psi\rangle$ with $\|\psi\|=1$:

1. **The value obtained** is one of the eigenvalues $a$.
2. **Born rule**: the probability of obtaining the value $a$ is $p(a) = \|\hat P_a|\psi\rangle\|^2 = \langle\psi|\hat P_a|\psi\rangle$.
3. **Projection postulate (collapse of the wave function)**: immediately after the value $a$ is obtained, the state is $|\psi'\rangle = \hat P_a|\psi\rangle \,/\, \bigl\|\hat P_a|\psi\rangle\bigr\|$.
</Axiom>

The third item is a discontinuous, non-unitary change, utterly unlike the smooth unitary evolution generated by the Schrödinger equation. Here lies the heart of the interpretational problems of quantum mechanics.

<Proposition id="prop-born-consistency" title="Consistency of the Born rule with expectation value and uncertainty">
In the setting of <Ref to="ax-measurement" />, if $\|\psi\| = 1$ then the following hold.

1. $\displaystyle\sum_a p(a) = 1$.
2. The mean of the measured values is $\displaystyle\sum_a a\,p(a) = \langle\hat A\rangle_\psi$.
3. The variance of the measured values is $\displaystyle\sum_a \bigl(a - \langle\hat A\rangle_\psi\bigr)^2 p(a) = (\Delta_\psi\hat A)^2$.
</Proposition>

<Proof of="prop-born-consistency">
(1): $\sum_a p(a) = \sum_a\langle\psi|\hat P_a|\psi\rangle = \langle\psi|\bigl(\sum_a\hat P_a\bigr)|\psi\rangle = \langle\psi|\hat I|\psi\rangle = \|\psi\|^2 = 1$, using completeness of the projections, $\sum_a\hat P_a = \hat I$. Since $p(a) = \|\hat P_a\psi\|^2 \ge 0$, the function $p$ is indeed a probability distribution.

(2): interchanging the sum and the inner product in the same way,

$$
\sum_a a\,p(a) = \Bigl\langle\psi\Bigl|\Bigl(\sum_a a\hat P_a\Bigr)\Bigr|\psi\Bigr\rangle = \langle\psi|\hat A|\psi\rangle = \langle\hat A\rangle_\psi .
$$

(3): put $\bar a := \langle\hat A\rangle_\psi$. From the spectral decomposition and the property $\hat P_a\hat P_{a'} = \delta_{aa'}\hat P_a$ of the projections,

$$
(\hat A - \bar a\hat I)^2 = \Bigl(\sum_a (a-\bar a)\hat P_a\Bigr)^2 = \sum_{a}(a-\bar a)^2\hat P_a
$$

(the cross terms with $a\ne a'$ vanish because $\hat P_a\hat P_{a'} = 0$). Taking the expectation value of both sides in $|\psi\rangle$,

$$
(\Delta_\psi\hat A)^2 = \sum_a (a-\bar a)^2\langle\psi|\hat P_a|\psi\rangle = \sum_a (a-\bar a)^2 p(a) .
$$
</Proof>

Part (2) justifies calling $\langle\psi|\hat A|\psi\rangle$ an expectation value, and part (3) justifies regarding $\Delta_\psi\hat A$ as the spread of the measured values themselves. This is where the definition in <Ref to="def-uncertainty" /> is confirmed to be more than a play on symbols.

<Figure caption="The operations prescribed by the measurement axiom. The value is selected probabilistically, and immediately after the selection the state is projected onto the corresponding eigenspace.">
<Mermaid code={`flowchart TD
  S["state ψ before measurement"] --> M["measure the observable A"]
  M -->|"probability p(a1)"| R1["obtain the value a1 / state becomes P1 ψ normalised"]
  M -->|"probability p(a2)"| R2["obtain the value a2 / state becomes P2 ψ normalised"]
  R1 --> C1["an immediate remeasurement gives a1 with probability 1"]
  R2 --> C2["an immediate remeasurement gives a2 with probability 1"]`} />
</Figure>

The bottom row of the figure is a direct consequence of the projection postulate. Indeed, applying $\hat P_a$ to the collapsed state $|\psi'\rangle = \hat P_a|\psi\rangle/\|\hat P_a\psi\|$ gives $\hat P_a|\psi'\rangle = |\psi'\rangle$ because $\hat P_a^2 = \hat P_a$, so the Born rule yields $p'(a) = \|\psi'\|^2 = 1$. The reproducibility of measurement — that measuring the same quantity again at once returns the same value — is built into the axiom.

<Example id="ex-sequential-stern-gerlach" title="Sequential Stern–Gerlach measurements">
Send a beam of silver atoms through an apparatus with a magnetic field gradient along $z$ (call it SG$z$), which measures $\hat S_z$; the beam splits into two, corresponding to $\pm\hbar/2$. Extracting only the $+\hbar/2$ branch, the projection postulate says the state is $|{\uparrow_z}\rangle$.

Pass this beam next through SG$x$ (a measurement of $\hat S_x$). The eigenstates of $\hat S_x$ are $|{\uparrow_x}\rangle = \frac{1}{\sqrt2}(|{\uparrow_z}\rangle + |{\downarrow_z}\rangle)$ and $|{\downarrow_x}\rangle = \frac{1}{\sqrt2}(|{\uparrow_z}\rangle - |{\downarrow_z}\rangle)$. Solving in the other direction,

$$
|{\uparrow_z}\rangle = \frac{1}{\sqrt2}\bigl(|{\uparrow_x}\rangle + |{\downarrow_x}\rangle\bigr),
$$

so by the Born rule the values $\pm\hbar/2$ each occur with probability $|1/\sqrt2|^2 = 1/2$. Extracting only the $+\hbar/2$ branch collapses the state to $|{\uparrow_x}\rangle$.

Finally, pass the beam through SG$z$ once more. If the property "the $z$ component points up" had been preserved, everything would emerge in the $+\hbar/2$ channel. But since $|{\uparrow_x}\rangle = \frac{1}{\sqrt2}(|{\uparrow_z}\rangle + |{\downarrow_z}\rangle)$, the beam in fact **splits 50–50 again**. The measurement of $\hat S_x$ has erased the information about $\hat S_z$.

This can be understood as a consequence of <Ref to="thm-simultaneous" />. Since $[\hat S_x,\hat S_z] = -i\hbar\hat S_y \ne 0$, the operators $\hat S_x$ and $\hat S_z$ have no simultaneous eigenstates. Hence there is no state in which "$\hat S_z$ points up and $\hat S_x$ points right", and fixing one of them makes the other indefinite.
</Example>

<Remark id="rem-collapse">
The projection postulate does not specify when or where the collapse occurs. The apparatus is in principle a quantum system too, so the system and apparatus together ought to evolve according to the Schrödinger equation, in which no collapse appears. This tension is the measurement problem.

The standard modern treatment considers the process by which the interference terms of a superposition are rapidly lost through interaction of the system with a macroscopic environment (decoherence). Decoherence explains, within the framework of unitary evolution, why the classical alternatives are singled out; it does not answer why only one of them is realised. The many-worlds interpretation, Bohmian mechanics and spontaneous collapse theories represent different attitudes towards this remaining part. For the purposes of computation, <Ref to="ax-measurement" /> may be used as it stands, and it agrees with experiment.
</Remark>

## 7. Time evolution and commutators: Ehrenfest's theorem

Commutators govern not only measurement but time evolution as well.

<Theorem id="thm-ehrenfest" title="Ehrenfest's theorem">
Let $\hat H$ be a (time-independent) Hamiltonian and $|\psi(t)\rangle$ a solution of the Schrödinger equation $i\hbar\,\partial_t|\psi(t)\rangle = \hat H|\psi(t)\rangle$ with $\|\psi(t)\|=1$. Let $\hat A$ be a Hermitian operator with no explicit time dependence, and suppose the required interchanges of differentiation and inner product are permitted. Then

$$
\frac{d}{dt}\langle\hat A\rangle_{\psi(t)} = \frac{i}{\hbar}\bigl\langle[\hat H,\hat A]\bigr\rangle_{\psi(t)} .
$$
</Theorem>

<Proof of="thm-ehrenfest">
The Schrödinger equation gives $|\dot\psi\rangle = -\frac{i}{\hbar}\hat H|\psi\rangle$. By the product rule,

$$
\frac{d}{dt}\langle\psi|\hat A|\psi\rangle = \langle\dot\psi|\hat A\psi\rangle + \langle\psi|\hat A\dot\psi\rangle .
$$

For the first term, antilinearity in the first argument (the coefficient gets conjugated) gives

$$
\langle\dot\psi|\hat A\psi\rangle = \overline{\left(-\frac{i}{\hbar}\right)}\langle \hat H\psi|\hat A\psi\rangle
= \frac{i}{\hbar}\langle \hat H\psi|\hat A\psi\rangle
= \frac{i}{\hbar}\langle\psi|\hat H\hat A\psi\rangle,
$$

where the last equality uses the Hermiticity of $\hat H$ (<Ref to="def-hermitian" />). The second term is linear in the second argument, so

$$
\langle\psi|\hat A\dot\psi\rangle = -\frac{i}{\hbar}\langle\psi|\hat A\hat H\psi\rangle .
$$

Adding the two,

$$
\frac{d}{dt}\langle\hat A\rangle = \frac{i}{\hbar}\langle\psi|(\hat H\hat A - \hat A\hat H)|\psi\rangle = \frac{i}{\hbar}\bigl\langle[\hat H,\hat A]\bigr\rangle .
$$
</Proof>

<Corollary id="cor-ehrenfest-newton" title="Newton's equations satisfied by the expectation values">
For $\hat H = \dfrac{\hat P^2}{2m} + V(\hat X)$ with $V$ a differentiable real-valued function,

$$
\frac{d}{dt}\langle\hat X\rangle = \frac{\langle\hat P\rangle}{m},
\qquad
\frac{d}{dt}\langle\hat P\rangle = -\bigl\langle V'(\hat X)\bigr\rangle .
$$
</Corollary>

<Proof of="cor-ehrenfest-newton">
First compute $[\hat H,\hat X]$. The term $V(\hat X)$ commutes with $\hat X$ and contributes nothing, so by part (a) of <Ref to="ex-commutators" /> and antisymmetry,

$$
[\hat H,\hat X] = \frac{1}{2m}[\hat P^2,\hat X] = -\frac{1}{2m}[\hat X,\hat P^2] = -\frac{2i\hbar\hat P}{2m} = -\frac{i\hbar}{m}\hat P .
$$

Substituting into <Ref to="thm-ehrenfest" />,

$$
\frac{d}{dt}\langle\hat X\rangle = \frac{i}{\hbar}\left(-\frac{i\hbar}{m}\right)\langle\hat P\rangle = \frac{\langle\hat P\rangle}{m}.
$$

Next $[\hat H,\hat P]$. Since $\hat P^2$ commutes with $\hat P$, part (c) of <Ref to="ex-commutators" /> gives

$$
[\hat H,\hat P] = [V(\hat X),\hat P] = -[\hat P,V(\hat X)] = i\hbar\,V'(\hat X),
$$

hence $\dfrac{d}{dt}\langle\hat P\rangle = \dfrac{i}{\hbar}\cdot i\hbar\,\langle V'(\hat X)\rangle = -\langle V'(\hat X)\rangle$.
</Proof>

Formally, the expectation values obey Newton's equations of motion ([Foundations of Newtonian mechanics](/en/physics/mechanics/newtonian-mechanics), <Ref to="physics/mechanics/newtonian-mechanics#ax-second-law" text="the second law" />). The right-hand side, however, is $\langle V'(\hat X)\rangle$ and not $V'(\langle\hat X\rangle)$. The two agree when $V$ is a polynomial of degree at most two (a free particle, a uniform force, a harmonic oscillator), and in that case the centre of the wave packet follows the classical trajectory exactly. Otherwise, the broader the wave packet, the larger the deviation from classical mechanics.

There is one further important consequence. If $[\hat H,\hat A] = 0$, then <Ref to="thm-ehrenfest" /> shows that $\langle\hat A\rangle$ is constant in time. That is, **an observable commuting with the Hamiltonian is a conserved quantity**. The classical mechanism by which symmetries generate conservation laws ([Symmetries and conservation laws](/physics/mechanics/noethers-theorem), <Ref to="physics/mechanics/noethers-theorem#thm-noether" text="Noether's theorem" />) appears in quantum mechanics in the form "the generator of a symmetry transformation commutes with $\hat H$". Taking $\hat A = \hat H$ gives conservation of energy, $d\langle\hat H\rangle/dt = 0$, immediately.

## 8. Exercises

<Exercise id="exr-commutator-basics" difficulty="Easy">
Using only <Ref to="lem-commutator-rules" /> and <Ref to="thm-ccr" />, show the following.

(1) $[\hat X,\hat P^3] = 3i\hbar\hat P^2$

(2) $[\hat X^2,\hat P^2] = 2i\hbar(\hat X\hat P + \hat P\hat X)$

<Solution>
(1) Apply the Leibniz rule $[\hat A,\hat B\hat C] = [\hat A,\hat B]\hat C + \hat B[\hat A,\hat C]$ to $\hat P^3 = \hat P\cdot\hat P^2$ and substitute part (a) of <Ref to="ex-commutators" />:

$$
[\hat X,\hat P^3] = [\hat X,\hat P]\hat P^2 + \hat P[\hat X,\hat P^2]
= i\hbar\hat P^2 + \hat P\cdot 2i\hbar\hat P = 3i\hbar\hat P^2 .
$$

(2) This time apply the Leibniz rule on the left:

$$
[\hat X^2,\hat P^2] = \hat X[\hat X,\hat P^2] + [\hat X,\hat P^2]\hat X
= \hat X\cdot 2i\hbar\hat P + 2i\hbar\hat P\cdot\hat X = 2i\hbar(\hat X\hat P + \hat P\hat X).
$$

Since $\hat X\hat P \ne \hat P\hat X$, one must not combine these two terms into $4i\hbar\hat X\hat P$.
</Solution>
</Exercise>

<Exercise id="exr-hermitian-products" difficulty="Standard">
Let $\hat A,\hat B$ be Hermitian operators.

(1) Show that the product $\hat A\hat B$ is Hermitian if and only if $[\hat A,\hat B]=0$.

(2) Show that $\hat A\hat B + \hat B\hat A$ and $i[\hat A,\hat B]$ are both Hermitian.

<Solution>
(1) By the property $(\hat A\hat B)^{*} = \hat B^{*}\hat A^{*}$ stated just after <Ref to="def-adjoint" />, together with $\hat A^{*}=\hat A$ and $\hat B^{*}=\hat B$, we get $(\hat A\hat B)^{*} = \hat B\hat A$. Hence "$\hat A\hat B$ is Hermitian" is equivalent to $\hat A\hat B = \hat B\hat A$, which by <Ref to="def-commutator" /> is equivalent to $[\hat A,\hat B] = 0$.

(2) The adjoint is antilinear ($(c\hat C)^{*} = \bar c\,\hat C^{*}$) and preserves sums, so

$$
(\hat A\hat B + \hat B\hat A)^{*} = \hat B\hat A + \hat A\hat B = \hat A\hat B + \hat B\hat A,
$$

$$
\bigl(i[\hat A,\hat B]\bigr)^{*} = -i\,(\hat A\hat B - \hat B\hat A)^{*} = -i(\hat B\hat A - \hat A\hat B) = i[\hat A,\hat B].
$$

The second of these is the standard way of manufacturing a new observable out of noncommuting ones. That $\hat S_z$ is Hermitian, given $[\hat S_x,\hat S_y]=i\hbar\hat S_z$, is an instance of this general rule.
</Solution>
</Exercise>

<Exercise id="exr-schrodinger-relation" difficulty="Standard">
Show that if, in the proof of <Ref to="thm-robertson" />, one retains the information in the real part instead of using the "keep only the imaginary part" inequality of Step 2, one obtains the stronger relation

$$
(\Delta_\psi\hat A)^2(\Delta_\psi\hat B)^2 \ge
\left(\frac{1}{2}\bigl\langle \{\hat A,\hat B\}\bigr\rangle - \langle\hat A\rangle\langle\hat B\rangle\right)^2
+ \left(\frac{1}{2i}\bigl\langle[\hat A,\hat B]\bigr\rangle\right)^2
$$

(the Schrödinger uncertainty relation). Here $\{\hat A,\hat B\} := \hat A\hat B + \hat B\hat A$ is the anticommutator.

<Solution>
Keep the notation of the proof. With $|f\rangle = \tilde A|\psi\rangle$, $|g\rangle = \tilde B|\psi\rangle$, $\tilde A = \hat A - \bar a$ and $\tilde B = \hat B - \bar b$, the Cauchy–Schwarz inequality reads

$$
(\Delta_\psi\hat A)^2(\Delta_\psi\hat B)^2 = \|f\|^2\|g\|^2 \ge |\langle f|g\rangle|^2
= \bigl(\operatorname{Re}\langle f|g\rangle\bigr)^2 + \bigl(\operatorname{Im}\langle f|g\rangle\bigr)^2 .
$$

The imaginary part is exactly as in Step 3 of the proof: $\operatorname{Im}\langle f|g\rangle = \frac{1}{2i}\langle[\hat A,\hat B]\rangle$. For the real part,

$$
2\operatorname{Re}\langle f|g\rangle = \langle f|g\rangle + \overline{\langle f|g\rangle}
= \langle\psi|(\tilde A\tilde B + \tilde B\tilde A)|\psi\rangle = \bigl\langle\{\tilde A,\tilde B\}\bigr\rangle .
$$

Expanding the anticommutator, and using that $\bar a,\bar b$ are real constants,

$$
\{\tilde A,\tilde B\} = \{\hat A,\hat B\} - 2\bar b\hat A - 2\bar a\hat B + 2\bar a\bar b\hat I,
$$

so taking expectation values gives $\langle\{\tilde A,\tilde B\}\rangle = \langle\{\hat A,\hat B\}\rangle - 2\bar a\bar b$. Therefore

$$
\operatorname{Re}\langle f|g\rangle = \frac{1}{2}\bigl\langle\{\hat A,\hat B\}\bigr\rangle - \langle\hat A\rangle\langle\hat B\rangle ,
$$

and substituting into the Cauchy–Schwarz inequality above yields the assertion. Discarding the first term returns <Ref to="thm-robertson" />.
</Solution>
</Exercise>

<Exercise id="exr-infinite-well" difficulty="Hard">
For the ground state

$$
\psi_1(x) = \sqrt{\frac{2}{L}}\,\sin\frac{\pi x}{L}\qquad (0\le x\le L)
$$

of the infinite square well of width $L$ (the wave function vanishes outside $0 \le x \le L$), compute $\Delta\hat X$ and $\Delta\hat P$ and compare with <Ref to="cor-heisenberg" />. You may use
$\displaystyle\int_0^L x^2\sin^2\frac{\pi x}{L}\,dx = L^3\left(\frac{1}{6} - \frac{1}{4\pi^2}\right)$ if needed.

<Solution>
**Position.** Since $|\psi_1|^2$ is symmetric about $x = L/2$, we have $\langle\hat X\rangle = L/2$. The second moment follows from the given integral:

$$
\langle\hat X^2\rangle = \frac{2}{L}\cdot L^3\left(\frac16 - \frac{1}{4\pi^2}\right) = L^2\left(\frac13 - \frac{1}{2\pi^2}\right).
$$

Hence

$$
(\Delta\hat X)^2 = L^2\left(\frac13 - \frac{1}{2\pi^2}\right) - \frac{L^2}{4} = L^2\left(\frac{1}{12} - \frac{1}{2\pi^2}\right).
$$

Numerically, $1/12 = 0.08333$ and $1/(2\pi^2) = 0.05066$, so $(\Delta\hat X)^2 = 0.03267\,L^2$ and $\Delta\hat X = 0.1808\,L$.

**Momentum.** Since $\psi_1$ is real and vanishes at both ends,

$$
\langle\hat P\rangle = \int_0^L \psi_1(-i\hbar\psi_1')\,dx = -\frac{i\hbar}{2}\bigl[\psi_1^2\bigr]_0^L = 0 .
$$

Moreover $\psi_1'' = -(\pi/L)^2\psi_1$, so $\hat P^2\psi_1 = -\hbar^2\psi_1'' = \dfrac{\hbar^2\pi^2}{L^2}\psi_1$; that is, $\psi_1$ is an eigenstate of $\hat P^2$ and

$$
\langle\hat P^2\rangle = \frac{\hbar^2\pi^2}{L^2},\qquad \Delta\hat P = \frac{\hbar\pi}{L}.
$$

**The product.**

$$
\Delta\hat X\cdot\Delta\hat P = 0.1808\,L\cdot\frac{\hbar\pi}{L} = 0.5679\,\hbar \approx 1.14\times\frac{\hbar}{2}.
$$

This is indeed larger than $\hbar/2$, consistently with <Ref to="cor-heisenberg" />. Equality fails because, as we saw in <Ref to="ex-gaussian-minimum" />, only Gaussian wave functions realise it, and a sine is not one.

Note also that, as stated in <Ref to="rem-domain" />, defining $\hat P$ self-adjointly on a finite interval requires a choice of boundary condition. Here we used the vanishing of $\psi_1$ at both ends to drop the boundary terms in the integration by parts, and within that setting the computation is legitimate.
</Solution>
</Exercise>

## References

- J. J. Sakurai, J. Napolitano, *Modern Quantum Mechanics*, 3rd ed., Cambridge University Press, 2020 — Chapter 1 (Fundamental Concepts) traces the path from the Stern–Gerlach experiment through the operator formalism, commutation relations and uncertainty relations.
- P. A. M. Dirac, *The Principles of Quantum Mechanics*, 4th ed., Oxford University Press, 1958 — the original source for bra-ket notation and the theory of observables; the correspondence between Poisson brackets and commutators is discussed there as well.
- Akira Shimizu, *Shinpan Ryōshiron no Kiso — Sono Honshitsu no Yasashii Rikai no Tame ni*, Saiensu-sha, 2004 (in Japanese) — treats the correspondence between physical quantities and Hermitian operators, and the measurement axiom, carefully and with all hypotheses made explicit.
- M. Reed, B. Simon, *Methods of Modern Mathematical Physics I: Functional Analysis*, revised ed., Academic Press, 1980 — Chapter VIII, "Unbounded operators", contains the distinction between symmetric and self-adjoint operators, deficiency indices, and the theory of self-adjoint extensions.
- W. Heisenberg, "Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik", *Zeitschrift für Physik* **43** (1927), 172–198 — the original paper on the uncertainty principle.
- H. P. Robertson, "The Uncertainty Principle", *Physical Review* **34** (1929), 163–164 — the original paper for <Ref to="thm-robertson" />.
- M. Ozawa, "Universally valid reformulation of the Heisenberg uncertainty principle on noise and disturbance in measurement", *Physical Review A* **67** (2003), 042105 — the inequality for measurement error and disturbance ([arXiv:quant-ph/0207121](https://arxiv.org/abs/quant-ph/0207121)).

## Appendix: The canonical commutation relation cannot be realised by bounded operators

<Ref to="cor-no-finite-dim" /> asserted that finite dimensions do not suffice, but in fact a stronger conclusion holds. An operator is called bounded if its norm $\|\hat A\| = \sup_{\|\psi\|=1}\|\hat A\psi\|$ is finite. On the set of bounded operators the norm of a product satisfies $\|\hat A\hat B\| \le \|\hat A\|\,\|\hat B\|$.

<Theorem id="thm-wintner" title="Wintner–Wielandt theorem">
Let $\hbar \ne 0$. There exist no **bounded** operators $\hat A,\hat B$ on a Hilbert space $\mathcal H \ne \{0\}$ with $[\hat A,\hat B] = i\hbar\hat I$.
</Theorem>

<Proof of="thm-wintner">
Suppose such bounded operators $\hat A,\hat B$ existed.

**Step 1.** We show by induction on $n$ that

$$
[\hat A^n,\hat B] = i\hbar\,n\,\hat A^{n-1}
$$

for all $n\ge 1$. The case $n=1$ is the hypothesis itself. Assuming it for $n$ and applying part (3) of <Ref to="lem-commutator-rules" /> (in the form $[\hat A\hat B,\hat C] = \hat A[\hat B,\hat C] + [\hat A,\hat C]\hat B$) to $\hat A^{n+1} = \hat A^{n}\cdot\hat A$,

$$
[\hat A^{n+1},\hat B] = \hat A^{n}[\hat A,\hat B] + [\hat A^{n},\hat B]\hat A
= \hat A^{n}\,i\hbar + i\hbar n\hat A^{n-1}\hat A = i\hbar(n+1)\hat A^{n}.
$$

**Step 2: $\hat A^n \ne 0$ for all $n\ge 0$.** If $\hat A^{n} = 0$ for some $n\ge 1$, the left-hand side in Step 1 vanishes, so $i\hbar n\hat A^{n-1} = 0$, and since $\hbar\ne 0$ and $n\ne 0$ we get $\hat A^{n-1}=0$. Repeating this gives $\hat A^{0} = \hat I = 0$, contradicting $\mathcal H\ne\{0\}$.

**Step 3: estimating the norms.** Take norms of both sides of Step 1. On the left, using the triangle inequality and the estimate for products,

$$
\hbar\,n\,\|\hat A^{n-1}\| = \bigl\|[\hat A^{n},\hat B]\bigr\|
\le \|\hat A^{n}\hat B\| + \|\hat B\hat A^{n}\|
\le 2\|\hat A^{n}\|\,\|\hat B\|
\le 2\|\hat A\|\,\|\hat A^{n-1}\|\,\|\hat B\| .
$$

By Step 2 we have $\|\hat A^{n-1}\| \ne 0$, so dividing both sides by it gives

$$
\hbar\,n \le 2\|\hat A\|\,\|\hat B\|
$$

for every $n\ge 1$. The right-hand side is a finite value independent of $n$, so taking $n$ large enough produces a contradiction.
</Proof>

Consequently at least one of position and momentum must be unbounded. In fact both are: for $\hat X$, taking normalised functions supported in regions where $|x|$ is arbitrarily large makes $\|\hat X\psi\|$ arbitrarily large. An unbounded operator cannot be defined on the whole space, and the discussion of domains in <Ref to="rem-domain" /> becomes unavoidable. The mathematical complications are imposed by the canonical commutation relation itself; they are not of a kind that can be sidestepped.