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
\begin{equation}
{}[ A , B ] := AB - BA \, .
\end{equation}
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
\begin{equation}
A \,( B \, \phi_j ) = B \, ( A \, \phi_j) = B\, ( a_j \phi_j ) = a_j (B \, \phi_j)
\end{equation}
so \(B \, \phi_j\) is also an eigenfunction \(A\) with eigenvalue \(a_j\). But such wave function is unique up to normalisation and therefore
\begin{equation}
B \, \phi_j (x) = b_j \phi_j(x)
\end{equation}
for some \(b_j \in \mathbb{R}\). So \(\phi_j(x)\) is simultaneously an eigenfunction of \(B\).
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,
\begin{equation}
\Delta A \Delta B \geq \frac{1}{2} | \langle [A,B] \rangle| \, .
\end{equation}
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{align}
(\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{align}
where \(\psi_A := A \cdot \psi\). There is an identical statement for \(B\) and therefore we can write
\begin{equation}
(\Delta A)^2(\Delta B)^2 = \langle \psi_A,\psi_A\rangle \langle \psi_B,\psi_B\rangle \, .
\end{equation}
We can now use the Cauchy-Schwarz inequality,
\begin{equation}
\langle \psi_A , \psi_A \rangle \langle \psi_B , \psi_B \rangle \geq | \langle \psi_A,\psi_B\rangle |^2 \, .
\label{eq:cs}
\end{equation}
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{align}
\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{align}
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{align}
|\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{align}
This concludes the proof.
Position and Momentum. For position and momentum,
\begin{equation}
\Delta x \, \Delta p \geq \frac{\hbar}{2} \, ,
\end{equation}
which is Heisenberg's uncertainty principle.
Momentum and Energy. For momentum and energy,
\begin{equation}
\Delta p \, \Delta H \geq \frac{\hbar}{2} \langle V'(x)\rangle \, ,
\end{equation}
which vanishes automatically when \(V(x)\) is constant.
Position and Energy. For position and energy,
\begin{equation}
\Delta x \, \Delta H \geq \frac{\hbar}{2m} \langle p\rangle \, .
\end{equation}
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 [
2]).
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.
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