Time evolution operator - Confusion

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The discussion revolves around the time evolution of a quantum state represented by a wavefunction in a potential well. The initial wavefunction is a linear combination of two eigenstates, leading to confusion about the correct form of the time evolution operator. The professor's expression using the difference of energies in the exponent is questioned, with the expectation that the evolution should involve separate exponentials for each eigenstate. Ultimately, it is clarified that the correct time-dependent state includes both eigenstates evolving independently, resolving the initial misunderstanding. The conversation concludes with an acknowledgment of the error in interpreting the professor's solution.
Kalidor
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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?
 
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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.
 
Yeah, problem solved. The unexplainable factor came from a multiplication of exponentials. I must have been drunk when i first read the solution.
Thanks
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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