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In quantum mechanics, the description of the infinite square well is given with the potential energy defined as $$V(x) = \begin{cases} 0 & \text{if } 0 \leq x \leq a,\\ \infty & \text{otherwise} . \end{cases} $$

Given that a particle in the infinite square well has its initial wave function as a mixture of the first two stationary states: $$\Psi(x,0) = \frac{1}{\sqrt{2}}[\psi_{1}(x) + \psi_{2}(x)]$$

Given this, as I understand the expectation value of the Hamiltonian for each stationary state is as follows:

$$<H_{1}> = \int_{0}^{a}\psi_{1}^{*}\hat{H_{1}}\psi_{1}dx = E_{1}\int_{0}^{a}|\psi_{1}|^{2}dx = E_{1}$$ and

$$<H_{2}> = \int_{0}^{a}\psi_{2}^{*}\hat{H_{2}}\psi_{2}dx = E_{2}\int_{0}^{a}|\psi_{2}|^{2}dx = E_{2}$$

Questions: 1. What I don't understand is why is the expectation value of the hamiltonian of the whole system is the average of the two, as follows $$<H> = \frac{1}{2}(E_{1} + E_{2})$$ and

  1. Why is it also concluded that the probability of getting either $E_{1}$ or $E_{2}$ is equal $(P_{1} = P_{2} = \frac{1}{2})$?

Thanks.

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  • $\begingroup$ What is $H_1$ and $H_2$ ? I mean the Hamiltonian is for the system, not for the states. $\endgroup$
    – user35952
    Dec 17, 2015 at 13:13
  • $\begingroup$ @user35952 The Hamiltonian for the separate stationary states. So if we had $\Psi(x,0) := \frac{1}{\sqrt{2}}\psi_{1}$ or $\Psi(x,0) := \frac{1}{\sqrt{2}}\psi_{2}$ then we would get the expectation value for the Hamiltonian as $<H_{1}>$ and $<H_{2}>$ respectively as I defined them. $\endgroup$
    – user101311
    Dec 17, 2015 at 13:16
  • $\begingroup$ That is actually more like $\left<H\right>_{1}$ and $\left<H\right>_{2}$ rather than what you said !! The Hamiltonian is the same. So E = $\left<H\right>_{\psi}$ = $\left<H\right>_{1}$ + $\left<H\right>_{2}$ $\endgroup$
    – user35952
    Dec 17, 2015 at 13:21
  • $\begingroup$ Am sorry, its rather $\left<H\right>_{\psi} = \frac{1}{2}( \left<H\right>_{1} + \left<H\right>_{2} )$ $\endgroup$
    – user35952
    Dec 17, 2015 at 13:30

1 Answer 1

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It seems like you have all the equations you need but are not combining them correctly.

$\left\langle H \right\rangle = \left\langle \Psi \left | H \right | \Psi\right\rangle = \left\langle \frac{1}{\sqrt2}\left ( \psi_1 + \psi_2 \right ) \left | H \right | \frac{1}{\sqrt2}\left ( \psi_1 + \psi_2 \right )\right\rangle = \frac{1}{2} \left\langle \psi_1 + \psi_2 \left | H \right | \psi_1 + \psi_2 \right\rangle = \frac{1}{2}\left ( \left\langle \psi_1 \left | H \right | \psi_1\right\rangle + \left\langle \psi_2 \left | H \right | \psi_2\right\rangle \right ) = \frac{1}{2}\left( E_1 + E_2 \right )$

Note that this reduction is possible because the cross terms, i.e.

$ \left\langle \psi_1 \left | H \right | \psi_2\right\rangle = E_2 \left\langle \psi_1 | \psi_2\right\rangle = 0 $ since $ \left | \psi1 \right\rangle$ and $\left | \psi2 \right\rangle$ are assumed to be orthogonal states by your description of the problem.

Your second question concerns the probability of outcomes. This information is encoded exclusively in the state you specified: $\left|\Psi\right\rangle = \frac{1}{\sqrt2}\left [ \left|\psi_1\right\rangle + \left|\psi_2\right\rangle \right ]$

The probability of finding the particle in $\left|\psi_1\right\rangle$ is given by $\left | \left \langle \psi_1 | \Psi \right \rangle \right |^2 = \left | \left \langle \psi_1 | \frac{1}{\sqrt2} \left[ \psi_1 + \psi_2 \right]\right \rangle \right |^2 = \left| \frac{1}{\sqrt2}\left[ \left\langle\psi_1 | \psi_1 \right \rangle + \left\langle\psi_1 | \psi_2 \right \rangle\right ]\right|^2 = \left| \frac{1}{\sqrt2}\right|^2 = \frac{1}{2}$ since, again, the states $ \left | \psi1 \right\rangle$ and $\left | \psi2 \right\rangle$ are orthogonal.

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