Stationary states of free particle

  • Thread starter Mindscrape
  • Start date
  • #1
1,860
0
The problem is to obtain the stationary states for a free particle in three dimensions by separating the variables in Schrödinger's equation.

So take

[tex]\psi(\mathbf{r},t) = \psi_1(x) \psi_2(y) \psi_3(z) \phi(t)[/tex]

and substitute it into the time-dependent Schrödinger equation. For the stationary states set U=0 and obtain

[tex] \frac{-\hbar^2}{2m} \nabla^2 \psi(\mathbf{r},t) = i\hbar \frac{\partial \psi(\mathbf{r},t)}{\partial t}[/tex]

Then divide by the wavefunction, and I get

[tex]i\hbar \frac{\partial \phi(t)}{\partial t} = \frac{-\hbar^2}{2m} \left( \frac{\partial^2 \psi_1}{\partial x^2} + \frac{\partial^2 \psi_2}{\partial y^2} + \frac{\partial^2 \psi_3}{\partial z^2} )\right [/tex]

I know that each one of the unknown functions must make a separate equation, but I don't know what to solve for without energy. For the time-independent equation they will all essentially be infinite square wells, but I don't know what to do with the time dependency.
 

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,764
293
The problem is to obtain the stationary states for a free particle in three dimensions by separating the variables in Schrödinger's equation.

So take

[tex]\psi(\mathbf{r},t) = \psi_1(x) \psi_2(y) \psi_3(z) \phi(t)[/tex]

and substitute it into the time-dependent Schrödinger equation. For the stationary states set U=0 and obtain

[tex] \frac{-\hbar^2}{2m} \nabla^2 \psi(\mathbf{r},t) = i\hbar \frac{\partial \psi(\mathbf{r},t)}{\partial t}[/tex]

Then divide by the wavefunction, and I get

[tex]i\hbar \frac{\partial \phi(t)}{\partial t} = \frac{-\hbar^2}{2m} \left( \frac{\partial^2 \psi_1}{\partial x^2} + \frac{\partial^2 \psi_2}{\partial y^2} + \frac{\partial^2 \psi_3}{\partial z^2} )\right [/tex]

I know that each one of the unknown functions must make a separate equation, but I don't know what to solve for without energy. For the time-independent equation they will all essentially be infinite square wells, but I don't know what to do with the time dependency.

Oops. A correction. I had no noticed that you had put in the wavefunction and then divided by it.

Ok, so at first you should get



[tex]i\hbar \psi_1 \psi_2 \psi_3 \frac{\partial \phi(t)}{\partial t} = \frac{-\hbar^2}{2m} \left( \phi \psi_1 \psi_3 \frac{\partial^2 \psi_1}{\partial x^2} +\phi \psi_1 \psi_3 \frac{\partial^2 \psi_2}{\partial y^2} + \phi \psi_1 \psi_2 \frac{\partial^2 \psi_3}{\partial z^2} )\right [/tex]

Then the next step is to divide everything by [itex] \phi \psi_1 \psi_2 \psi_3 [/itex] and then you should get (instead of what you wrote):

[tex]i\hbar {1 \over
\phi} \frac{\partial \phi(t)}{\partial t} = \frac{-\hbar^2}{2m} \left( {1 \over \psi_1} \frac{\partial^2 \psi_1}{\partial x^2} +{ 1 \over \psi_2} \frac{\partial^2 \psi_2}{\partial y^2} + {1 \over \psi_3} \frac{\partial^2 \psi_3}{\partial z^2} )\right [/tex]
and then to use the usual argument of separation of variables to show that [itex] \phi(t) [/itex] obeys
[tex] i \hbar {\partial \phi(t) \over \partial t} = E \phi(t) [/itex] where E is the constant of separation. So [itex] \phi(t) = A e^{-iE t / \hbar} [/itex].

Then you go on to separate the equations in x, y and z. You end up with three separate 1-dimensional Schrodinger equations ).

Hope this helps.

Patrick
 
Last edited:

Related Threads on Stationary states of free particle

  • Last Post
Replies
1
Views
2K
Replies
0
Views
892
Replies
4
Views
2K
  • Last Post
Replies
3
Views
185
Replies
4
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
3K
  • Last Post
Replies
1
Views
840
Top