For an electron in an equal linear superposition of two eigenstates in a one-dimensional infinitely deep square potential well of width L, how do you isolate the time dependent part of the expectation value of the momentum?
I know when you have a superposition of two states and you add their two wave functions you end up with an additional part of the equation (namely the cos of the differences in energy over h-bar times t) that is now time dependent, and oscillates at a frequency of the differences in energy over h-bar.
And I know to calculate the expectation value you plug your specific operator into the middle of the integral (so for purposes of writing it out it looks as though it is sandwiched between the wave function in the integral).
I can see for the expectation value of the position how the time dependent part of the integral continues to just be the cos of the differences in energy over h-bar. I'm just struggling to see how to incorporate the momentum operator (-i h-bar gradient) into the integral, and then isolate the time dependent part of the integral.
Specifically, right now I am trying to solve the expectation value for momentum for an electron in the superposition of the first two eigenstates (so n=1 and n=2). But I do not see how the momentum operator affects the time dependent part of the equation…
Best Answer
Let's write the eigenstates for the infinite well as $$ \psi_n(x,t) = \phi_n(x)e^{-i\omega_n t} = |n\rangle e^{-i\omega_n t} , $$
where $\phi_n(x) = A \sin\left( n \pi x /L \right)$.
If your linear combination is $$ |\psi_0\rangle = \alpha |1\rangle e^{-i\omega_1 t}+ \beta |2\rangle e^{-i\omega_2 t}. $$ Then $$ \begin{eqnarray} \langle \psi_0 | \hat{p} | \psi_0 \rangle &=& \left( \alpha^* \langle 1| e^{i\omega_1 t}+ \beta^* \langle 2| e^{i\omega_2 t} \right)\hat{p} \left( \alpha |1\rangle e^{-i\omega_1 t}+ \beta |2\rangle e^{-i\omega_2 t} \right) \\ &=& |\alpha|^2 \langle 1 | \hat{p} | 1 \rangle + |\beta|^2 \langle 2 | \hat{p} | 2 \rangle + \alpha^*\beta \langle 1 | \hat{p} | 2 \rangle e^{-i\Delta\omega t} + \alpha\beta^* \langle 2 | \hat{p} | 1 \rangle e^{i\Delta\omega t} \end{eqnarray} $$ The $\hat{p}$ operator does not affect the time-depentent part. However, it will convert $\sin$ to $\cos$ making $\langle 1 | \hat{p} | 1 \rangle = \langle 2 | \hat{p} | 2 \rangle = 0$.
Now, the terms that remain are not zero and are complex conjugates of each other so, as with the expectation value of the position, it is proportional to $\cos\Delta\omega t$.