Hi everyone. I am given a somewhat common potential well [itex] V(x)=0 [/itex] for [itex] ¦x¦<a [/itex] and infinite elsewhere. I am told that at t = 0 my particle is in a state represented by the wavefunction(adsbygoogle = window.adsbygoogle || []).push({});

[itex] \psi(x,0)= A(\sin{(\frac{\pi x}{a})}+ \sqrt{2} \cos{(\frac{3 \pi x}{2 a})}) [/itex]

where A is a constant use for normalization. It turns easily out that this constant equals [itex] \sqrt{\frac{1}{3a}} [/itex]. On the other hand, the eigenfunctions in this situations are very well known to be [itex]\psi_n= \sqrt{\frac{1}{a}} \sin(\frac{n \pi x}{2a}) [/itex] for n even and [itex]\psi_n= \sqrt{\frac{1}{a}} \cos(\frac{n \pi x}{2a}) [/itex] for n odd. So given wave function can easily be written as a linear combination of the second and third eigenstates.

Now come the problems. At some point of the exercise I am asked to calculate the expected value of some operator but at some time t. This is how the professor writes the wavefunction at the generic time t:

[itex] e^{- \frac{i}{\hbar}(E_2-E_3)t} \psi(x,0) [/itex]

Now my question is: since the given state is a combination of 2 eigenstates, why in the world does he write the temporal evolution this way, with (E_2-E_3) in the exponent? Shouldn't it be something like

[itex] e^{- \frac{i}{\hbar}E_2 t} \psi_2(x,0) - e^{- \frac{i}{\hbar}E_3 t} \psi_3(x,0) [/itex]?

Maybe some details are wrong but what I mean is I expect to find the sum of 2 exponentials whereas there is the exponential of a sum.

My idea is that (simplifying the notation) the ket [itex] |2> + \sqrt{2} |3> [/itex] is of course different from the sum of the kets [itex] |2> [/itex] and [itex] |3> [/itex] but then why isn't the right expression something like

[itex] e^{- \frac{i}{\hbar}(E_2-\sqrt{2}E_3)t} \psi(x,0) [/itex].

Am I somewhat clear?

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# Time evolution operator - Confusion

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