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  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.

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

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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 $\langle\psi_1|\psi_2\rangle$?

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  • $\begingroup$ 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

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