2
$\begingroup$

Expectation value of an operator $A$ is defined as: $\langle A\rangle=\langle\psi|A| \psi\rangle=\sum_{n} a_{n} \mathcal{P}_{a_{n}}$ where $a_n$ are the eigenvalues of $A$ and $\mathcal{P}_{a_{n}}$ is it's probability of occurance.

If $C=A+B$ having eigenstates and eigenvalues $(\left|c_{i}\right\rangle$, $c_n)$, such that none of $\left|c_{i}\right\rangle$ is an eigenstate of $A$ or $B$, its expectation value will be $\langle C\rangle=\langle\psi|A+B| \psi\rangle=\langle A\rangle+\langle B\rangle= \sum_{n} a_{n} \mathcal{P}_{a_{n}} + \sum_{n} b_{n} \mathcal{P}_{b_{n}}=\sum_{n} c_{n} \mathcal{P}_{c_{n}}$

Is there another way to see how

$\sum_{n} a_{n} \mathcal{P}_{a_{n}} + \sum_{n} b_{n} \mathcal{P}_{b_{n}}=\sum_{n} c_{n} \mathcal{P}_{c_{n}}$

I'm interested in it because it appears to me that the eigenstates, eigenvalues and probabilities of $C$ i.e $\left|c_{i}\right\rangle$, $\mathcal{P}_{c_{n}}$, $c_n$ can have any value irrespective of $(\left|a_{i}\right\rangle$, $\mathcal{P}_{a_{n}}$, $a_n)$ and $(\left|b_{i}\right\rangle$, $\mathcal{P}_{b_{n}}$, $b_n)$ so it's surprising how they are connected as $\sum_{n} a_{n} \mathcal{P}_{a_{n}} + \sum_{n} b_{n} \mathcal{P}_{b_{n}}=\sum_{n} c_{n} \mathcal{P}_{c_{n}}$

$\endgroup$
8
  • 3
    $\begingroup$ I'm not sure what you want as an answer to this question - you already know how to show that this statement is true (and in fact, this is how expectation values of all random variables behave, not just in QM), and you haven't explained to us at all why it "appears" to you that the eigenstates and values of $C$ can "have any value irrespective" of those of $A$ and $B$. I don't understand this at all - $C$ is the sum of $A$ and $B$, why do you refuse to believe that its properties should be related to those of its summands?! $\endgroup$
    – ACuriousMind
    Feb 2, 2022 at 13:01
  • $\begingroup$ A state can be an eigenstate of $C$ without being an eigenstate of either of $A$ or $B$. As a simple example, consider $$ \begin{bmatrix}1&0\\0&1\end{bmatrix} = \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix} $$ Then $\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$ is an eigenvector of $C$ but not of $A$ and not of $B$. $\endgroup$
    – Kashmiri
    Feb 2, 2022 at 13:26
  • $\begingroup$ Found the above example here:math.stackexchange.com/questions/3475896/… $\endgroup$
    – Kashmiri
    Feb 2, 2022 at 13:27
  • 1
    $\begingroup$ I don't see how that's a response to my comment. I didn't claim that the eigenvectors of $C$ had to be the same as those of $A$ or $B$, I just said that I don't understand your claim that they can have "any value irrespective[...]". The equation you're asking about in your question also doesn't imply that the eigenvectors of $C$ should be the same as those of $A$ or $B$ - perhaps you think it does? If you do, you should explain why you think so. $\endgroup$
    – ACuriousMind
    Feb 2, 2022 at 14:03
  • 1
    $\begingroup$ @Quantum Mechanic , yes I agree now. I was just confused. It's first introduction to qm and I'm self studying so I just slipped. $\endgroup$
    – Kashmiri
    Feb 2, 2022 at 16:10

1 Answer 1

1
$\begingroup$

The notation is somewhat misleading here, since probabilities $\mathcal{P}_{a_n}$, $\mathcal{P}_{b_n}$ and $\mathcal{P}_{c_n}$ actually refer to different eigenbases. Thus, if operator $A$ has eigenstates $|a_n\rangle$, i.e., $A|a_n\rangle=a_n|a_n\rangle$, then $$\mathcal{P}_{a_n}^A=|\langle a_n|\psi\rangle|^2,$$ and similarly $$\mathcal{P}_{b_n}^B=|\langle b_n|\psi\rangle|^2, \mathcal{P}_{c_n}^C=|\langle c_n|\psi\rangle|^2,$$ so that the relation takes form $$ \sum_na_n\mathcal{P}_{a_n}^A+ \sum_nb_n\mathcal{P}_{b_n}^B= \sum_nc_n\mathcal{P}_{c_n}^C. $$ Note that the summation over $n$ is misleading as well, as the different space are not necessarily indexed in the same way and not necessarily mapped one-to-one among each other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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