How to calculate a wave function in time t

[Haorong Wu]

Homework Statement

Suppose $\psi _p \left ( x \right ) =\frac 1 {\sqrt {2 \pi \hbar}} e^{ip_0 x/ \hbar}$, and $H=\frac {p^2} {2m}$. The energy of the state is $\frac {p_0^2} {2m}$. if $\psi \left ( x,0 \right ) = \psi _p \left ( x \right )$, then calculate $\psi \left ( x , t \right)$.

Relevant Equations

$\psi \left ( x, t \right ) = e^{-iEt/\hbar} \psi \left ( x,0 \right )$

I use the equation
$\psi \left ( x, t \right ) = e^{-iEt/\hbar} \psi \left ( x,0 \right )$ to calculate $\psi \left ( x , t \right)$, and the result is $\psi \left ( x , t \right) = \frac 1 {\sqrt {2 \pi \hbar}} exp \left [ \frac {ip_0 x} {\hbar} - \frac {i p^2 t} {2m \hbar} \right ]$.

However, it is wrong, and in the solution, first, the wave function is converted to $\phi \left ( p \right )=\frac 1 {\sqrt {2 \pi \hbar}} \int e^{ip_0 x /\hbar} e^{-ipx/\hbar} dx = \sqrt {2 \pi \hbar} \delta \left ( p-p_0 \right )$.
Then, $\psi \left ( x , t \right) = \frac 1 {\sqrt {2 \pi \hbar}} \int \sqrt {2 \pi \hbar} e^{ipx/\hbar} \delta \left ( p-p_0 \right ) dp \cdot e^{-i E t /\hbar} = e^{i p_0 x /\hbar - iEt /\hbar}$.

There is a difference of a factor of $\frac 1 {\sqrt {2 \pi \hbar}}$, and I am confusing why it has to be converted to $\phi \left ( p \right )$ first?

Thanks!

 

[DrClaude]

Homework Equations: $\psi \left ( x, t \right ) = e^{-iEt/\hbar} \psi \left ( x,0 \right )$

This is incorrect. The actual equation (for a time-independent Hamiltonian) is
$$\psi \left ( x, t \right ) = e^{-i \hat{H} t/\hbar} \psi \left ( x,0 \right )$$
It is only in the case where the exponential of the operator is acting on an eigenstate of that operator that you can replace the operator in the exponential by the corresponding eigenvalue.

This is why you have to go to momentum space, where $\hat{H} \phi(p) = \frac{\hat{p}^2}{2m} \phi(p) = \frac{p^2}{2m}\phi(p)$ (the last equality is the one where the momentum operator becomes "multiply by $p$").

 

[Haorong Wu]

This is incorrect. The actual equation (for a time-independent Hamiltonian) is
$$\psi \left ( x, t \right ) = e^{-i \hat{H} t/\hbar} \psi \left ( x,0 \right )$$
It is only in the case where the exponential of the operator is acting on an eigenstate of that operator that you can replace the operator in the exponential by the corresponding eigenvalue.

This is why you have to go to momentum space, where $\hat{H} \phi(p) = \frac{\hat{p}^2}{2m} \phi(p) = \frac{p^2}{2m}\phi(p)$ (the last equality is the one where the momentum operator becomes "multiply by $p$").

Thanks, DrClaude.

Meanwhile, $H \left | \psi _p \right > = \frac {- \hbar ^2 } {2m} \frac {d} {dx} \frac 1 {\sqrt {2 \pi \hbar}} e^{ip_0 x/ \hbar} = \frac {- \hbar ^2 } {2m} {\left ( \frac {ip_0} \hbar \right )}^2 \frac 1 {\sqrt {2 \pi \hbar}} e^{ip_0 x/ \hbar}=\frac {- p_0 ^2 } {2m} \frac 1 {\sqrt {2 \pi \hbar}} e^{ip_0 x/ \hbar}=\frac {- p_0 ^2 } {2m} \left | \psi _p \right >$, then $\left | \psi _p \right >$ is a eigenstate of $H$. So why $\psi \left ( x, t \right ) = e^{-iEt/\hbar} \psi \left ( x,0 \right )$ does not apply here?

 

[DrClaude]

Thanks, DrClaude.

Meanwhile, $H \left | \psi _p \right > = \frac {- \hbar ^2 } {2m} \frac {d} {dx} \frac 1 {\sqrt {2 \pi \hbar}} e^{ip_0 x/ \hbar} = \frac {- \hbar ^2 } {2m} {\left ( \frac {ip_0} \hbar \right )}^2 \frac 1 {\sqrt {2 \pi \hbar}} e^{ip_0 x/ \hbar}=\frac {- p_0 ^2 } {2m} \frac 1 {\sqrt {2 \pi \hbar}} e^{ip_0 x/ \hbar}=\frac {- p_0 ^2 } {2m} \left | \psi _p \right >$, then $\left | \psi _p \right >$ is a eigenstate of $H$. So why $\psi \left ( x, t \right ) = e^{-iEt/\hbar} \psi \left ( x,0 \right )$ does not apply here?

Sorry, I hadn't paid attention to the particular wave function you were working with. (There is a typo in what you wrote: it should be $d^2/dx^2$.) I was answering the second part of

There is a difference of a factor of $\frac 1 {\sqrt {2 \pi \hbar}}$, and I am confusing why it has to be converted to $\phi \left ( p \right )$ first?

As for the first part, there is a typo in the answer. Norm is conserved and the normalization factor remains unchanged (notice that the only thing that happens with the time evolution is that the wave function changes its complex phase).