10 Commutators and Uncertainty Principle

10.1 The Commutator

Suppose we have two linear operators $A$ and $B$, such as position $x$, momentum $p$ or the Hamiltonian $H$. The commutator is defined by $$\begin{array}{}\text{(10.1)}& [A,B]:=AB-BA.\end{array}$$ It has the following properties:

• Anti-symmetry: $[A,B]=-[B,A]$.

• Linear: $[a_1{A}_1+a_2{A}_2,B]=a_1[A_1,B]+a_2[A_2,B]$.

• $[A,\cdot ]$ is a derivation: $[A,BC]=B[A,C]+[A,B]C$.

• Jacobi identity: $[A[B,C]]+[B,[C,A]]+[C,[A,B]]=0$.

The commutator plays an important role in quantum mechanics due to the following theorem:

Theorem: “Two commuting matrices are simultaneously diagonalizable." If $A$ and $B$ are Hermitian operators with $[A,B]=0$, it is possible to find an orthonormal basis of wave functions that are simultaneous eigenfunctions of $A$ and $B$.

Proof: To keep things simple, we will assume the spectrum of eigenvalues $\{a_j\}$ of $A$ is discrete and non-degenerate. This means that up to normalisation there is a unique solution to $A{\varphi }_j(x)=a_j{\varphi }_j(x)$ for each eigenvalue $a_j$. The normalisation can be chosen to make the basis orthonormal, $⟨{\varphi }_i,{\varphi }_j⟩={\delta }_{ij}$.

We want to prove that ${\varphi }_j(x)$ is simultaneously an eigenfunction of $B$. The commutator $[A,B]=0$ is equivalent to $AB=BA$. This means $$\begin{array}{}\text{(10.2)}& A(B{\varphi }_j)=B(A{\varphi }_j)=B(a_j{\varphi }_j)=a_j(B{\varphi }_j)\end{array}$$ so $B{\varphi }_j$ is also an eigenfunction $A$ with eigenvalue $a_j$. But such wave function is unique up to normalisation and therefore $$\begin{array}{}\text{(10.3)}& B{\varphi }_j(x)=b_j{\varphi }_j(x)\end{array}$$ for some $b_j\in \mathbb{R}$. So ${\varphi }_j(x)$ is simultaneously an eigenfunction of $B$.

10.2 Compatibility

This theorem motivates the following definition.

• Definition: Two observables represented by Hermitian operators $A$ and $B$ are called “compatible" if their commutator vanishes, $[A,B]=0$.

Let us first determine whether the observables we have encountered so far are compatible. Recall that the position and momentum operators are $$\begin{array}{}\text{(10.4)}& \begin{array}{r}\hat{x}=x\hat{p}=-i\hslash \frac{\partial }{\partial x}\end{array}\end{array}$$ while $$\begin{array}{}\text{(10.5)}& \hat{H}=\frac{{\hat{p}}^2}{2m}+V(x)=\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}+V(x).\end{array}$$

1. Position and Momentum. Let us compute the commutator of position and momentum acting on a generic wave function, $$\begin{array}{}\text{(10.6)}& [\hat{x},\hat{p}]\psi =x(-i\hslash \frac{\partial \psi }{\partial x})+i\hslash \frac{\partial }{\partial x}\left(x\psi \right)=i\hslash \psi .\end{array}$$ This is the “canonical commutation relation" $$\begin{array}{}\text{(10.7)}& [\hat{x},\hat{p}]=i\hslash ,\end{array}$$ which shows position and momentum are not compatible.

2. Momentum and Energy. A similar computation shows that $$\begin{array}{}\text{(10.8)}& [H,p]=[V(x),p]=i\hslash \frac{\partial V(x)}{\partial x}.\end{array}$$ So momentum $p$ and energy $H$ are compatible only if $V(x)$ is constant. We will return to this result later in the course.

3. Position and Energy. From the commutator of position and momentum, $$\begin{array}{}\text{(10.9)}& \begin{array}{r}[H,x]=\frac{1}{2m}[p^2,x]=\frac{1}{2m}(p[p,x]+[p,x]p)=-\frac{i\hslash }{m}p,\end{array}\end{array}$$ so position and energy are never compatible.

10.3 Compatibility and Measurement

To understand the physical significance of compatibility, suppose we measure $A$ and find the eigenvalue $a_j$. Then the wave function immediately after the measurement is the eigenfunction ${\varphi }_j(x)$.

• What happens if we make another measurement of $A$ immediately afterwards? As ${\varphi }_j(x)$ is an eigenfunction of $A$, the measurement will again find $a_j$ with probability $1$. Correspondingly, the uncertainty of a normalised eigenfunction vanishes, $$\begin{array}{}\text{(10.10)}& \begin{array}{rl}(\mathrm{\Delta }A{)}^2& =⟨A^2⟩-⟨A{⟩}^2\\ & =⟨{\varphi }_j,A^2{\varphi }_j⟩-⟨{\varphi }_{j}A,{\varphi }_j{⟩}^2\\ & =a_j^2⟨{\varphi }_j,{\varphi }_j⟩-(a_j⟨{\varphi }_j,{\varphi }_j⟩{)}^2\\ & =a_j^2-a_j^2=0.\end{array}\end{array}$$

• What happens if we make a measurement of $B$ immediately afterwards?

  • If $[A,B]=0$, ${\varphi }_j(x)$ is also an eigenfunction of $B$ with eigenvalue $b_j$. So a measurement of $B$ will find $b_j$ with probability $1$. Correspondingly, the same argument as above shows that $\mathrm{\Delta }B=0$. In other words, we can simultaneously determine the values of both $A$ and $B$.

  • If $[A,B]\ne 0$, ${\varphi }_j(x)$ is not an eigenfunction of $B$ and there are multiple potential outcomes. Suppose the measurement of $B$ yields a particular eigenvalue, say $b$. Then the wave function jumps to the corresponding eigenfunction of $B$ with $\mathrm{\Delta }B=0$. But now $\mathrm{\Delta }A>0$.

It is important to emphasise that the subsequent measurements discussed in this section are measurements without time between them. A generic eigenstate of an operator will not stay an eigenstate under time evolution, so a measurement of $A$, followed by a delay and then another measurement of $A$, will typically still lead to a non-zero spread. Only eigenstates of the Hamiltonian operator, or operators which commute with it, remain eigenstates under time evolution.

10.4 The Generalised Uncertainty Principle

If $[A,B]\ne 0$, we cannot necessarily find simultaneous eigenfunctions of $A$ and $B$ with both $\mathrm{\Delta }A=0$ and $\mathrm{\Delta }B=0$. In fact, there is a fundamental limitation in quantum mechanics on the how small we can simultaneously make the uncertainties $\mathrm{\Delta }A$ and $\mathrm{\Delta }B$. This is quantified by the “Generalised Uncertainty Principle”:

Theorem: For any square-normalisable wave function, $$\begin{array}{}\text{(10.11)}& \mathrm{\Delta }A\mathrm{\Delta }B\ge \frac{1}{2}|⟨[A,B]⟩|.\end{array}$$

Proof: We will assume here that $⟨A⟩=⟨B⟩=0$ for the wave function in question. This will simplify the argument without losing any of its essence. The translation to $⟨A⟩\ne 0$, $⟨B⟩\ne 0$ as an exercise for the interested reader.

With our assumption, the uncertainty in $A$ can be expressed, $$\begin{array}{}\text{(10.12)}& \begin{array}{rl}(\mathrm{\Delta }A{)}^2& =⟨A^2⟩=⟨\psi ,A^2\psi ⟩=⟨A\psi ,A\psi ⟩=⟨{\psi }_A,{\psi }_A⟩,\end{array}\end{array}$$ where ${\psi }_A:=A\cdot \psi$. There is an identical statement for $B$ and therefore we can write $$\begin{array}{}\text{(10.13)}& (\mathrm{\Delta }A{)}^2(\mathrm{\Delta }B{)}^2=⟨{\psi }_A,{\psi }_A⟩⟨{\psi }_B,{\psi }_B⟩.\end{array}$$ We can now use the Cauchy-Schwarz inequality, $$\begin{array}{}\text{(10.14)}& ⟨{\psi }_A,{\psi }_A⟩⟨{\psi }_B,{\psi }_B⟩\ge |⟨{\psi }_A,{\psi }_B⟩{|}^2.\end{array}$$ This result holds for any Hermitian inner product. It is analogous to the standard result $|\mathbf{a}{|}^2|\mathbf{b}{|}^2\ge |\mathbf{a}\cdot \mathbf{b}{|}^2$ from real euclidean geometry, which follows from the formula $\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos (\theta )$ for the dot product.

The right-hand side of this inequality can be expressed as $$\begin{array}{}\text{(10.15)}& \begin{array}{rl}⟨{\psi }_A,{\psi }_B⟩& =⟨AB⟩\\ & =\frac{1}{2}⟨(AB-BA)⟩+\frac{1}{2}⟨(AB+BA)⟩\\ & =\frac{1}{2}⟨[A,B]⟩+\frac{1}{2}⟨\{A,B\}⟩\end{array}\end{array}$$ where $[A,B]:=AB-BA$ is the commutator and $\{A,B\}:=AB+BA$ is the “anti-commutator". It is straightforward to check that,

1. $[A,B]$ is anti-Hermitian $\Rightarrow$ $⟨[A,B]⟩\in i\mathbb{R}$.

2. $\{A,B\}$ is Hermitian $\Rightarrow$ $⟨\{A,B\}⟩\in \mathbb{R}$.

so the commutator and anti-commutator provide the imaginary and real parts of $⟨{\psi }_A,{\psi }_B⟩$. Recalling the formula $|z{|}^2=x^2+y^2$ for the modulus squared of a complex number $z=x+iy$, we have $$\begin{array}{}\text{(10.16)}& \begin{array}{rl}|⟨{\psi }_A,{\psi }_B⟩{|}^2& =\frac{1}{4}|⟨[A,B]⟩{|}^2+\frac{1}{4}|⟨\{A,B\}⟩{|}^2\\ & \ge \frac{1}{4}|⟨[A,B]⟩{|}^2.\end{array}\end{array}$$ This concludes the proof.

1. Position and Momentum. For position and momentum, $$\begin{array}{}\text{(10.17)}& \mathrm{\Delta }x\mathrm{\Delta }p\ge \frac{\hslash }{2},\end{array}$$ which is Heisenberg’s uncertainty principle.

2. Momentum and Energy. For momentum and energy, $$\begin{array}{}\text{(10.18)}& \mathrm{\Delta }p\mathrm{\Delta }H\ge \frac{\hslash }{2}⟨V^{\prime }(x)⟩,\end{array}$$ which vanishes automatically when $V(x)$ is constant.

3. Position and Energy. For position and energy, $$\begin{array}{}\text{(10.19)}& \mathrm{\Delta }x\mathrm{\Delta }H\ge \frac{\hslash }{2m}⟨p⟩.\end{array}$$ This implies that square-normalisable Hamiltonian eigenfunctions must have $⟨p⟩=0$. In fact, square-normalisable Hamiltonian eigenfunctions may always be chosen real, compatible with this statement.

Note that what is derived here is a statement which is different from the one in the previous section: the generalised uncertainty principle as derived above says nothing about subsequent measurements (see also Jacques Distler and Sonia Paban. Uncertainties in successive measurements. Physical Review A, Jun 2013. URL: http://dx.doi.org/10.1103/PhysRevA.87.062112, doi:10.1103/physreva.87.062112.).

It should be emphasised that this is a fundamental feature of quantum mechanics. Only in the classical limit, $\hslash \to 0$, can we simultaneously determine exactly the values of non-compatible observables such as position and momentum.

10.5 Problems

1. Generalised uncertainty principle:

   Derive the generalised uncertainty principle, $$\begin{array}{}\text{(10.20)}& (\mathrm{\Delta }A{)}^2(\mathrm{\Delta }B{)}^2\ge {(\frac{1}{2i}⟨[\hat{A},\hat{B}]⟩)}^2,\end{array}$$ where $(\mathrm{\Delta }\hat{A}{)}^2=⟨{\hat{A}}^2⟩-⟨\hat{A}{⟩}^2$ and similar for $\hat{B}$.

2. Energy-position uncertainty relation:

   Show that measurements of position and measurements of energy of a particle in one dimension satisfy the uncertainty relation $$\begin{array}{}\text{(10.21)}& \mathrm{\Delta }x\mathrm{\Delta }E\ge \frac{\hslash }{2m}|⟨p⟩|.\end{array}$$

3. TUTORIAL 3
   Time-energy uncertainty relation (sort of) For an operator $\hat{Q}$ which does not depend on time explicitly, we have $$\begin{array}{}\text{(10.22)}& \frac{\mathrm{d}}{\mathrm{d}t}⟨\hat{Q}⟩=\frac{i}{\hslash }⟨[\hat{H},\hat{Q}]⟩.\end{array}$$ For small $\mathrm{\Delta }t$ and small standard-deviation $\mathrm{\Delta }Q$ we can write $$\begin{array}{}\text{(10.23)}& \mathrm{\Delta }Q\approx \left|{\displaystyle \frac{\mathrm{d}⟨\hat{Q}⟩}{\mathrm{d}t}}\right|\mathrm{\Delta }t.\end{array}$$ What is the meaning of $\mathrm{\Delta }t$ here? Use $\text{(10.23)}$ to derive the “time-energy uncertainty relation” $\mathrm{\Delta }H\mathrm{\Delta }t\ge \hslash /2$. What does this ‘uncertainty relation’ express?

   Solution ▶

   1. Writing $$\begin{array}{}\text{(10.24)}& a(x):=(\hat{A}-⟨\psi \hat{A}\psi ⟩)\psi (x)\end{array}$$ and ditto for $\hat{B}$ we can write $$\begin{array}{}\text{(10.25)}& (\mathrm{\Delta }A{)}^2(\mathrm{\Delta }B{)}^2=⟨a,a⟩⟨b,b⟩\ge |⟨a,b⟩{|}^2.\end{array}$$ Setting $z=⟨a,b⟩$ we then have $$\begin{array}{}\text{(10.26)}& |z{|}^2\ge \mathrm{I}\mathrm{m}(z{)}^2={(\frac{1}{2i}(z-z^{\ast }))}^2,\end{array}$$ For the particular $z$ we have $$\begin{array}{}\text{(10.27)}& ⟨a,b⟩=⟨\hat{A}\hat{B}⟩-⟨\hat{A}⟩⟨\hat{B}⟩\end{array}$$ and similar for the conjugate. Using that for the difference $z-z^{\ast }$ then gives the requested result.

   2. This requires computing $$\begin{array}{}\text{(10.28)}& =\frac{1}{2m}[\hat{x},{\hat{p}}^2]=\frac{i\hslash }{m}\hat{p}.\end{array}$$ and thus $$\begin{array}{}\text{(10.29)}& (\mathrm{\Delta }x{)}^2(\mathrm{\Delta }E{)}^2\ge {(\frac{\hslash }{2m}⟨\hat{p}⟩)}^2,\end{array}$$ from which the relation follows.

   3. Insertion of the Schrödinger equation into the uncertainty relation gives $$\begin{array}{}\text{(10.30)}& \mathrm{\Delta }H\mathrm{\Delta }Q\ge \frac{\hslash }{2}\left|\frac{\mathrm{d}⟨\hat{Q}⟩}{\mathrm{d}t}\right|.\end{array}$$ Then inserting the definition of $\mathrm{\Delta }t$ and rearranging factors produces the $E-t$ uncertainty relation.

      The symbol $\mathrm{\Delta }t$ is the time it takes the expectation value of the (arbitrary) operator $\hat{Q}$ to change by one standard deviation.

      The uncertainty relation expresses the fact that if all observables change rapidly ($\mathrm{\Delta }t$ small), then the uncertainty in the energy must be large, and if all observables change slowly, the uncertainty in the energy is small.

      See Griffiths for more detail and discussion.