10 Commutators and Uncertainty Principle

Discussing two effects of non-commuting operators: on the measurement process, and on the limitation of the wave function description of particles.

This lecture is motivated by the question:

To answer this question, we introduce the commutator \([A,B]\) of two Hermitian operators and explore its physical interpretation. We will prove a generalisation of Heisenberg’s uncertainty principle, which is a fundamental limitation on the precision that observables \(A\) and \(B\) can be determined simultaneously.

10.1 The Commutator

Two commuting operators share an (orthonormal) basis of eigenfunctions. This is extremely important, because eigenfunctions and eigenvalues are related to possible measurement outcomes.

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

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 \(A \, \phi_j(x) = a_j \phi_j(x)\) for each eigenvalue \(a_j\). The normalisation can be chosen to make the basis orthonormal, \(\langle \phi_i,\phi_j\rangle = \delta_{ij}\).

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

10.2 Compatibility

Quantum operators which originate from classical observables which have a vanishing Poisson bracket are called ‘‘compatible''.

This theorem motivates the following definition.

Let us first determine whether the observables we have encountered so far are compatible. Recall that the position and momentum operators are \[\begin{aligned} \hat x = x \qquad \hat p = - i \hbar \frac{\partial}{\partial x} \end{aligned}\] while \[\hat H = \frac{\hat p^2}{2m} + V(x) = \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \, .\]

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

  2. Momentum and Energy. A similar computation shows that \[{}[ H , p ] = [ V(x) , p ] = i \hbar \frac{\partial V(x)}{\partial x} \, .\] 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, \[\begin{aligned} {}\left[ H , x \right] = \frac{1}{2m}[p^2,x] = \frac{1}{2m} \left( p [p,x] + [p,x]p \right) = -\frac{i\hbar}{m} p \, , \end{aligned}\] 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 \(\phi_j(x)\).

If \(A\) and \(B\) do not commute, and we measure \(A\), immediately after that, the uncertainty of \(B\) has to be nonzero. This is a statement about subsequent measurements.

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

The generalised uncertainty principle relates the product of uncertainties of two operators to their commutator. It says something about the spread of two observables in the same state, not about subsequent measurements.

Theorem: For any square-normalisable wave function, \[\Delta A \Delta B \geq \frac{1}{2} | \langle [A,B] \rangle| \, .\]

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

With our assumption, the uncertainty in \(A\) can be expressed, \[\begin{aligned} (\Delta A)^2 & = \langle A^2\rangle = \langle \psi , A^2\psi\rangle = \langle A\psi ,A\psi\rangle = \langle \psi_A , \psi_A \rangle \, , \end{aligned}\] where \(\psi_A := A \cdot \psi\). There is an identical statement for \(B\) and therefore we can write \[(\Delta A)^2(\Delta B)^2 = \langle \psi_A,\psi_A\rangle \langle \psi_B,\psi_B\rangle \, .\] We can now use the Cauchy-Schwarz inequality, \[\langle \psi_A , \psi_A \rangle \langle \psi_B , \psi_B \rangle \geq | \langle \psi_A,\psi_B\rangle |^2 \, . \label{eq:cs}\] This result holds for any Hermitian inner product. It is analogous to the standard result \(| \mathbf{a}|^2 |\mathbf{b}|^2 \geq | \mathbf{a} \cdot \mathbf{b} |^2\) from real euclidean geometry, which follows from the formula \({\bf a} \cdot {\bf b} = |{\bf a}||{\bf b}| \cos(\theta)\) for the dot product.

The right-hand side of this inequality can be expressed as \[\begin{aligned} \langle \psi_A , \psi_B \rangle & = \langle AB \rangle \\ & = \frac{1}{2} \langle (AB-BA) \rangle + \frac{1}{2} \langle (AB+BA) \rangle \\ & = \frac{1}{2} \langle [A,B] \rangle + \frac{1}{2} \langle \{A,B\} \rangle \end{aligned}\] 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 \(\Rightarrow\) \(\langle [A,B] \rangle \in i \mathbb{R}\).

  2. \(\{ A, B\}\) is Hermitian \(\Rightarrow\) \(\langle \{A,B\} \rangle \in \mathbb{R}\).

so the commutator and anti-commutator provide the imaginary and real parts of \(\langle \psi_A, \psi_B\rangle\). Recalling the formula \(|z|^2 = x^2 + y^2\) for the modulus squared of a complex number \(z = x + i y\), we have \[\begin{aligned} |\langle \psi_A , \psi_B \rangle|^2 & = \frac{1}{4} | \langle [A,B] \rangle|^2 + \frac{1}{4} | \langle \{A ,B\} \rangle|^2 \\ & \geq \frac{1}{4} | \langle [A,B] \rangle|^2 \, . \label{eq:second-step} \end{aligned}\] This concludes the proof.

  1. Position and Momentum. For position and momentum, \[\Delta x \, \Delta p \geq \frac{\hbar}{2} \, ,\] which is Heisenberg’s uncertainty principle.

  2. Momentum and Energy. For momentum and energy, \[\Delta p \, \Delta H \geq \frac{\hbar}{2} \langle V'(x)\rangle \, ,\] which vanishes automatically when \(V(x)\) is constant.

  3. Position and Energy. For position and energy, \[\Delta x \, \Delta H \geq \frac{\hbar}{2m} \langle p\rangle \, .\] This implies that square-normalisable Hamiltonian eigenfunctions must have \(\langle p \rangle = 0\). 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.).

The uncertainty relation always contains \(\hbar\) on the right hand side, and is thus fundamentally ‘‘quantum''.

It should be emphasised that this is a fundamental feature of quantum mechanics. Only in the classical limit, \(\hbar \to 0\), 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, \[(\Delta A)^2\, (\Delta B)^2 \geq \left( \frac{1}{2i} \langle [\hat{A}, \hat{B}]\rangle\right)^2\,,\] where \((\Delta \hat{A})^2 = \langle \hat{A}^2\rangle - \langle \hat{A}\rangle^2\) 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 \[\label{e:dxdE} \Delta x \, \Delta E \geq \frac{\hbar}{2m} \big|\langle p\rangle \big|\,.\]

  3. TUTORIAL 3
    Time-energy uncertainty relation (sort of) For an operator \(\hat{Q}\) which does not depend on time explicitly, we have \[\frac{{\rm d}}{{\rm d}t} \langle \hat{Q}\rangle = \frac{i}{\hbar} \langle [\hat{H}, \hat{Q}]\rangle\,.\] For small \(\Delta t\) and small standard-deviation \(\Delta Q\) we can write \[\label{e:Deltat} \Delta Q \approx \left|\displaystyle\frac{{\rm d}\langle\hat{Q}\rangle}{{\rm d}t}\right| \Delta t \,.\] What is the meaning of \(\Delta t\) here? Use \(\eqref{e:Deltat}\) to derive the “time-energy uncertainty relation” \(\Delta H \Delta t \geq \hbar/2\). What does this ‘uncertainty relation’ express?

    Solution

    1. Writing \[a(x) := (\hat A - \langle\psi \hat{A} \psi\rangle) \psi(x)\] and ditto for \(\hat{B}\) we can write \[(\Delta A)^2 (\Delta B)^2 = \langle a, a\rangle \langle b, b\rangle \geq |\langle a, b\rangle|^2\,.\] Setting \(z=\langle a, b\rangle\) we then have \[|z|^2 \geq {\rm Im}(z)^2 = \left(\frac{1}{2i}(z-z^*)\right)^2\,,\] For the particular \(z\) we have \[\langle a, b\rangle = \langle \hat{A}\hat{B}\rangle - \langle \hat{A}\rangle \langle \hat{B}\rangle\] and similar for the conjugate. Using that for the difference \(z-z^*\) then gives the requested result.

    2. This requires computing \[= \frac{1}{2m} [\hat{x}, \hat{p}^2] = \frac{i\hbar}{m} \hat{p}\,.\] and thus \[(\Delta x)^2 (\Delta E)^2 \geq \left( \frac{\hbar}{2m} \langle \hat{p}\rangle \right)^2\,,\] from which the relation follows.

    3. Insertion of the Schrödinger equation into the uncertainty relation gives \[\Delta H\, \Delta Q \geq \frac{\hbar}{2} \left| \frac{{\rm d}\langle \hat{Q}\rangle}{{\rm d}t} \right|\,.\] Then inserting the definition of \(\Delta t\) and rearranging factors produces the \(E-t\) uncertainty relation.

      The symbol \(\Delta t\) 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 (\(\Delta t\) 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.