Time evolution operator - Confusion

[Kalidor]

Hi everyone. I am given a somewhat common potential well $V(x)=0$ for $¦x¦<a$ and infinite elsewhere. I am told that at t = 0 my particle is in a state represented by the wavefunction

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

where A is a constant use for normalization. It turns easily out that this constant equals $\sqrt{\frac{1}{3a}}$. On the other hand, the eigenfunctions in this situations are very well known to be $\psi_n= \sqrt{\frac{1}{a}} \sin(\frac{n \pi x}{2a})$ for n even and $\psi_n= \sqrt{\frac{1}{a}} \cos(\frac{n \pi x}{2a})$ 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:

$e^{- \frac{i}{\hbar}(E_2-E_3)t} \psi(x,0)$

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
$e^{- \frac{i}{\hbar}E_2 t} \psi_2(x,0) - e^{- \frac{i}{\hbar}E_3 t} \psi_3(x,0)$?
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 $|2> + \sqrt{2} |3>$ is of course different from the sum of the kets $|2>$ and $|3>$ but then why isn't the right expression something like
$e^{- \frac{i}{\hbar}(E_2-\sqrt{2}E_3)t} \psi(x,0)$.

Am I somewhat clear?

 

[vanhees71]

Es you can easily check by pluging your solution of the Schrödinger eq. back into the equation, you are right and the prof. is wrong. The correct state is

$$|\psi,t \rangle>=A [\exp(-\mathrm{i} E_2 t) |2 \rangle+\sqrt{2} \exp(-\mathrm{i} E_3 t) |3 \rangle].$$

The norm should be 1, so you find $A=1/\sqrt(1+2)=1/\sqrt{3}$. Note that the $1/\sqrt{a}$ factor is already included in the norm of the state vectors, which are given in position representation.

I don't understand, how you come to the last forumula, which definitely is wrong.

 

[Kalidor]

Yeah, problem solved. The unexplainable factor came from a multiplication of exponentials. I must have been drunk when i first read the solution.
Thanks