# How to calculate a wave function in time t

• Haorong Wu
In summary: 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?The 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.
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!

Delta2
Haorong Wu said:
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##").

DEvens and Delta2
DrClaude said:
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?

Haorong Wu said:
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
Haorong Wu said:
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).

Haorong Wu

## 1. What is a wave function?

A wave function is a mathematical description of the behavior of a quantum system, which includes information about the position, momentum, and energy of a particle at a given time.

## 2. How do you calculate a wave function in time t?

To calculate a wave function in time t, you need to use the Schrödinger equation, which is a mathematical equation that describes how a quantum system evolves over time. This equation takes into account the potential energy of the system and the mass of the particle to determine the wave function at a specific time t.

## 3. What factors affect the wave function in time t?

The wave function in time t is affected by the potential energy of the system, the mass of the particle, and any external forces acting on the particle. Additionally, the initial conditions of the system, such as the initial position and momentum of the particle, also play a role in determining the wave function.

## 4. Can the wave function change over time?

Yes, the wave function can change over time due to the probabilistic nature of quantum mechanics. As a particle moves through space, its position and momentum change, which affects the wave function. Additionally, external forces acting on the particle can also alter the wave function.

## 5. How is the wave function used in quantum mechanics?

The wave function is a fundamental concept in quantum mechanics and is used to predict the behavior of particles at the atomic and subatomic level. It is used to calculate the probability of finding a particle in a certain position or with a certain momentum, and to make predictions about the future behavior of a quantum system.

• Advanced Physics Homework Help
Replies
10
Views
705
• Advanced Physics Homework Help
Replies
3
Views
1K
• Advanced Physics Homework Help
Replies
10
Views
933
• Advanced Physics Homework Help
Replies
3
Views
1K
• Advanced Physics Homework Help
Replies
2
Views
989
• Advanced Physics Homework Help
Replies
1
Views
2K
• Advanced Physics Homework Help
Replies
3
Views
909
• Advanced Physics Homework Help
Replies
9
Views
1K
• Advanced Physics Homework Help
Replies
2
Views
1K
• Advanced Physics Homework Help
Replies
24
Views
1K