3
$\begingroup$

Let $\hat{A}$ and $\hat{B}$ be Hermitian observables with spectra labeled by $a$ and $b$. Then we can write \begin{equation} \hat{A} = \sum_a a\, \hat{P}_a \end{equation} \begin{equation} \hat{B} = \sum_b b\, \hat{P}_b \end{equation}

where $\hat{P}_a$ and $\hat{P}_b$ are the eigenspace projection operators for $\hat{A}$ and $\hat{B}$.

Let $|\psi \rangle$ be a state and consider the expectation value \begin{align} \langle\psi|\hat{A} \hat{B}|\psi \rangle = \sum_{a,b} ab\,\langle \psi|\hat{P}_a \hat{P}_b |\psi\rangle. \end{align}

If $[\hat{A},\hat{B}] = 0$, then it is clear how to interpret $\langle\psi|\hat{A} \hat{B}|\psi \rangle$: when we perform a simultaneous measurement of $\hat{A}$ and $\hat{B}$, we project the state onto a simultaneous eigenspace and measure the corresponding eigenvalues $a,b$. The probability that the state will be projected onto the $a,b$ eigenspace is simply $\langle \psi|\hat{P}_a \hat{P}_b |\psi\rangle$. Then from the above equation, we see that $\langle\psi|\hat{A} \hat{B}|\psi \rangle$ represents the average value of the product of the measurements, taken over a large number of repeated experiments.

My question concerns the meaning of $\langle\psi|\hat{A} \hat{B}|\psi \rangle$ when $\hat{A}$ and $\hat{B}$ do not commute. My first thought is that it represents the average value of the product of the measurements when $\hat{B}$ is measured first, followed by a measurement of $\hat{A}$. But this quantity, it seems to me, should instead be computed as follows:

  1. First make a measurement of $\hat{B}$. This projects the state onto an eigenspace with eigenvalue $b$ with probability $\langle \psi|\hat{P}_b |\psi \rangle$. After making this measurement, our new state is now \begin{equation} |\psi'\rangle = \frac{1}{\sqrt{\langle\psi |\hat{P_b} |\psi \rangle}}\hat{P}_b|\psi\rangle \end{equation}

  2. We now perform a measurement of $\hat{A}$ on $|\psi'\rangle$. This projects onto an eigenspace with eigenvalue $a$ with probability $\langle \psi'|\hat{P}_a|\psi'\rangle$

  3. Hence, the probability of obtaining first an eigenvalue $b$ followed by an eigenvalue $a$ is

\begin{equation} \langle \psi'|\hat{P}_a|\psi'\rangle\,\langle \psi|\hat{P}_b|\psi\rangle = \langle \psi|\hat{P}_b \hat{P}_a \hat{P}_b|\psi\rangle \end{equation}

  1. And thus I would expect the average value of the product of the measurements to be \begin{equation} \sum_{a,b} ab\,\langle \psi|\hat{P}_b \hat{P}_a \hat{P}_b|\psi\rangle. \end{equation}

If this reasoning is correct, then how am I to interpret $\langle \psi | \hat{A} \hat{B} |\psi \rangle$ (since it is clearly not equal to the result in bullet point 4 when the projectors do not commute)? If the reasoning is incorrect, what is wrong with it?

$\endgroup$
2
$\begingroup$

Right, in general you're not going to see a straightforward equivalence there. We can use Dirac notation with $\hat P_b = |b\rangle\langle b|$ to see that $\langle \hat A \hat B \rangle = \sum_{a,b} a~b~\langle \psi | a \rangle~\langle a | b \rangle~\langle b | \psi \rangle$ and even inserting an identity matrix for $b$ (call it $b'$) gives: $$ \begin{align}\langle \hat A \hat B \rangle =& \sum_{a,b,b'} a~b~\langle \psi | b'\rangle~\langle b'|a \rangle~\langle a | b \rangle~\langle b | \psi \rangle\\=&\sum_{b,b'} b~\psi^*(b')~\psi(b) ~ \langle b'|\hat A| b\rangle\end{align}$$Indeed, you need to insert some sort of $\delta_{bb'}$ into this last sum to get the $b ~ \langle b|\hat A | b\rangle$ sense of "measure $B$ first, then $A$," which this expression doesn't have unless it's hidden in that $\hat A$ term.

There is a very simple reason why you do not see this straightforward equivalence. Let's work in a finite-dimensional Hilbert space $\psi \in \mathbb C^N.$ Then the matrix $\hat C = \hat A \hat B$ is really given by the Einstein sum $$C_{ik} = A_{ij} ~B_{jk}.$$This is Hermitian if and only if $C_{ik}^* = C_{ki}$ but the complex conjugate here is$$C_{ik}^* = A_{ij}^* ~B_{jk}^* = B_{kj}~A_{ji}$$and demanding that this is equal to $C_{ki}$ is therefore demanding that $B_{kj} A_{ji} = A_{kj} B_{ji}$ or therefore $[\hat A, \hat B] = 0.$

In other words, the product of two Hermitian matrices is only Hermitian if they commute. In general the expectation $\langle \hat A \hat B \rangle$ is going to be a complex number when they do not commute.

If you want something which is Hermitian (say you have a classical expression involving $\langle x~ p\rangle$ that you want to generalize into the quantum case) then you will probably do a symmetric product $\frac 12 (\hat A \hat B + \hat B \hat A)$, which is then again Hermitian if its constituent matrices are.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.