1. The normalized wave functions $\Psi_1$ and $\Psi_2$ correspond to the ground state and the first excited states of a particle in a potential. The operators $\hat{A}$ act on the wave function as: $$\hat{A}\Psi_1=\Psi_2\text{ and }\hat{A}\Psi_2=\Psi_1$$ The expectation value of the operator $\hat{A}$ for the state $\hat{A}=(3\Psi_1+4\Psi_2)/5$ is:
    (A) 0
    (B) -0.32
    (C) 0.75
    (D) 0.96

Not able to understand how to determine expectation value when ground state and first excited states are given.


closed as off-topic by Michael Seifert, ZeroTheHero, knzhou, stafusa, Kyle Kanos May 30 '18 at 10:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Michael Seifert, ZeroTheHero, knzhou, stafusa, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome to Physics.SE! Unfortunately, we have a policy against homework or "homework-like" questions here, so this question is likely to be closed as it stands. You might be able to avoid this if you provide more context about what you're already tried and what aspect of the problem you're confused about. $\endgroup$ – Michael Seifert May 29 '18 at 15:23
  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ – Kyle Kanos May 30 '18 at 9:57

I'm not sure why the state is called $\hat A$ since $\hat A$ is also an operator, but if you call the state $|\psi\rangle$ such that $|\psi\rangle=(3|\psi_1\rangle+4|\psi_2\rangle)/5$ then the expectation value of $\hat A$ is given by $$\langle\hat A\rangle=\langle\psi|\hat A|\psi\rangle$$ Since $\hat A$ is linear ( most quantum operators are linear) this can be expanded as follows: $$\tfrac{1}{5}\left(3\langle\psi_1|+4\langle\psi_2|\right)\hat A\tfrac{1}{5}\left(3|\psi_1\rangle+4|\psi_2\rangle\right)= \\\tfrac{1}{25}\left(9\langle\psi_1|\hat A|\psi_1\rangle+12\langle\psi_1|\hat A|\psi_2\rangle+12\langle\psi_2|\hat A|\psi_1\rangle+16\langle\psi_2|\hat A|\psi_2\rangle\right)$$ You can now replace $\hat A|\psi_1\rangle$ by $|\psi_2\rangle$ and $\hat A|\psi_2\rangle$ by $|\psi_1\rangle$.

What do you get as an expectation value now?

Hint: what are the values of $\langle\psi_1|\psi_1\rangle$, $\langle\psi_2|\psi_2\rangle$ and $\langle\psi_1|\psi_2\rangle$?

  • $\begingroup$ Yeah that’s what I didn’t understand why is operator given as a combination of psi1 and psi2. After solving as you suggested the answer comes to 24/25 which is 0.96. $\endgroup$ – Pratik. B May 29 '18 at 16:42

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