This lecture is motivated by the question:
Suppose I measure an observable \(A\), such as position, momentum or energy. How does that measurement affect subsequent measurements of another observable \(B\)?
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.
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:
Anti-symmetry: \([A,B] = -[B,A]\).
Linear: \([a_1A_1+a_2A_2 , B] = a_1[A_1,B]+a_2[A_2,B]\).
\([A,\cdot]\) 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 \(A \, \phi_j(x) = a_j \phi_j(x)\) 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 \(B\).
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 \[\begin{aligned} \hat x = x \qquad \hat p = - i \hbar \frac{\partial}{\partial x} \end{aligned}\] while \[\hat H = \frac{\hat p^2}{2m} + V(x) = \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \, .\]
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.
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.
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.
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 \(\phi_j(x)\).
What happens if we make another measurement of \(A\) immediately afterwards? As \(\phi_j(x)\) is an eigenfunction of \(A\), the measurement will again find \(a_j\) with probability \(1\). Correspondingly, the uncertainty of a normalised eigenfunction vanishes, \[\begin{aligned} (\Delta A)^2 & = \langle A^2 \rangle - \langle A\rangle^2 \\ & = \langle \phi_j, A^2 \phi_j\rangle - \langle\phi_j A,\phi_j\rangle^2 \\ & = a_j^2\langle \phi_j,\phi_j\rangle - (a_j\langle \phi_j,\phi_j\rangle)^2 \\ & = a_j^2-a_j^2=0\, . \end{aligned}\]
What happens if we make a measurement of \(B\) immediately afterwards?
If \([A,B] = 0\), \(\phi_j(x)\) 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 \(\Delta B=0\). In other words, we can simultaneously determine the values of both \(A\) and \(B\).
If \([A,B] \neq 0\), \(\phi_j(x)\) 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 \(\Delta B = 0\). But now \(\Delta A > 0\).
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.
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”:
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,
\([A,B]\) is anti-Hermitian \(\Rightarrow\) \(\langle [A,B] \rangle \in i \mathbb{R}\).
\(\{ A, B\}\) 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.
Position and Momentum. For position and momentum, \[\Delta x \, \Delta p \geq \frac{\hbar}{2} \, ,\] which is Heisenberg’s uncertainty principle.
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.
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.).
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.
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 \(\hat{B}\).
Show that measurements of position and measurements of energy of a particle in one dimension satisfy the uncertainty relation \[\label{e:dxdE} \Delta x \, \Delta E \geq \frac{\hbar}{2m} \big|\langle p\rangle \big|\,.\]
Solution ▶
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.
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.
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.