Finding meaning in the Phase of the wavefunction

[Irishdoug]

Homework Statement

Suppose $\psi_{0}$ is a properly normalised wavefunction with $<x_{\psi_0} >$ = $x_{0}$ and $<p_{\psi_0} >$ = $p_{0}$. Define a new wavefunction $\psi_{new}$(x) = $e^{{-iqx}/{\hbar}}$ $\psi_{0}$

Relevant Equations

N/A

Suppose $\psi_{0}$ is a properly normalised wavefunction with $<x_{\psi_0} >$ = $x_{0}$ and $<p_{\psi_0} >$ = $p_{0}$. Define a new wavefunction $\psi_{new}$(x) = $e^{{-iqx}/{\hbar}}$ $\psi_{0}$

What is the expectation value $<\psi_{new}>$ in the state given by $\psi_{new}$(x)?

The answer to the question is given here: https://ocw.mit.edu/courses/physics...pring-2013/assignments/MIT8_04S13_ps6_sol.pdf (It is $q + p_{0}$)

In doing the question myself I got as far as ($-i \hbar$) $(\frac{iq}{\hbar})$ $\int_{-\infty}^{+\infty} \psi_{0} e^{-iqx/\hbar} \psi_{0} e^{iqx/\hbar} dx$ + ($-i \hbar$) $\int_{-\infty}^{+\infty} \psi_{0} e^{-iqx/\hbar} \psi_{0}' e^{iqx/\hbar} dx$

Now, the 2nd integral on the right is just equal to $p_{0}$ . It is not clear to me how the first integral reduces to just q to give the answer of $q + p_{0}$ as we are left with $\int_{-\infty}^{+\infty} \psi_{0}^{2} e^{0} dx$

I did the integral and it is divergent. Does this mean it is treated as being 1 as it's unphysical and we are just left with q after the cancelations of i and $\hbar$?

 

[Irishdoug]

Ok I just realized after writng that all out that
$\int_{-\infty}^{+\infty} \psi_{0}^{2} dx$ is the probability density function which is = 1 by definition so I'm guessing that is why it reduces to 1?

 

[PeroK]

Ok I just realized after writng that all out that
$\int_{-\infty}^{+\infty} \psi_{0}^{2} dx$ is the probability density function which is = 1 by definition so I'm guessing that is why it reduces to 1?

You should have $\psi_0^* \psi_0$ in those integrals.

 

[Irishdoug]

Is it not the case that $\psi_{0}$ can be taken to be real as such $\psi^{*}_{0}$ = $\psi_{0}$?

 

[PeroK]

Is it not the case that $\psi_{0}$ can be taken to be real as such $\psi^{*}_{0}$ = $\psi_{0}$?

Not in general.

 

[Irishdoug]

Ok, I thought this was a general rule as such.

 

[PeroK]

Ok, I thought this was a general rule as such.

You can often choose a set of particular eigenfunctions to be real-valued. But, that's different from any set of eigenfunctions being real-valued.

 

[Irishdoug]

Am I correct in saying that the eigenvalue has to be real-valued though?

 

[PeroK]

Am I correct in saying that the eigenvalue has to be real-valued though?

Eigenvalues of a Hermitian operator are real.

 

[Irishdoug]

Ok, I need to be more careful with my language then. I thought in QM, in general, eigenvalues had to be real. I am assuming this presumption is incorrect?

 

[PeroK]

Ok, I need to be more careful with my language then. I thought in QM, in general, eigenvalues had to be real. I am assuming this presumption is incorrect?

The eigenvalues of operators in general are complex numbers. E.g. the eigenvalues of the lowering (annihilation) operator for the SHO can be any complex number.

The eigenvalues of Hermitian operators, which represent observables, are real.

In this case, however, all the problem says is that $\psi_0$ is a normalised wavefunction. You can't assume anything further.

 

[Irishdoug]

The eigenvalues of operators in general are complex numbers. E.g. the eigenvalues of the lowering (annihilation) operator for the SHO can be any complex number.

The eigenvalues of Hermitian operators, which represent observables, are real.

In this case, however, all the problem says is that $\psi_0$ is a normalised wavefunction. You can't assume anything further.

Ah that makes perfect sense now. Thankyou!