- #1

- 150

- 17

Even using the uncertainty principle, we might get it if we know the uncertainty in position but not the exact value as we might get only the lower bound of it.

so, if any other ways, please do share!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter VVS2000
- Start date

I said I have already found the expectation value using ehrenfest theorem. just asked for some other ways that is all.f

- #1

- 150

- 17

Even using the uncertainty principle, we might get it if we know the uncertainty in position but not the exact value as we might get only the lower bound of it.

so, if any other ways, please do share!

- #2

Gold Member

- 2,123

- 1,128

[tex]\int \psi^{*}(x) \frac{\partial}{i \hbar \partial x} \psi(x) dx=\int \phi^{*} (p) p\ \phi(p) dp[/tex]

where ##\psi(x)## and ##\phi(p)## are Fourier or inverse Fourier transform of the other.

- #3

- 150

- 17

yeah I mentioned the integral method in the OP

[tex]\int \psi^{*}(x) \frac{\partial}{i \hbar \partial x} \psi(x) dx=\int \phi^{*} (p) p\ \phi(p) dp[/tex]

where ##\psi(x)## and ##\phi(p)## are Fourier or inverse Fourier transform of the other.

anything other than this?

- #4

Gold Member

- 2,123

- 1,128

By symmetrical treatment of coordinate x and momentum p as is shown in Fourier transforms, your question applies similarly on coordinate x , i.e.," Are there ways to get expectation value <x> and <p> other thananything other than this?

[tex]<x>=\int \psi(x)^* x \psi(x) dx[/tex]

[tex]<p>=\int \phi(p)^* p \phi(p) dp[/tex]?"

I regard these formula as definitions of <x> and <p>. If you happen to find other ways, I am afraid you scarcely do, the values of which must coincide with the values given by them.

- #5

- 150

- 17

Yeah I know there are few methods other than thisBy symmetrical treatment of coordinate x and momentum p as is shown in Fourier transforms, your question applies similarly on coordinate x , i.e.," Are there ways to get expectation value <x> and <p> other than

[tex]<x>=\int \psi(x)^* x \psi(x) dx[/tex]

[tex]<p>=\int \phi(p)^* p \phi(p) dp[/tex]?"

I regard these formula as definitions of <x> and <p>. If you happen to find other ways, I am afraid you scarcely do, the values of which must coincide with the values given by them.

I only asked because I was solving a problem and solving the integral to find the expectation value was very tedious and long

Just wanted to know whether there are other ways of finding it, that is all

- #6

Mentor

- 43,052

- 20,527

What problem? Being more specific about the problem you are trying to solve might help.I was solving a problem

- #7

- 150

- 17

Expectation value of momentum of particle in a box and the wave function is time dependent as wellWhat problem? Being more specific about the problem you are trying to solve might help.

- #8

Gold Member

- 2,123

- 1,128

What is the initial condition of your case? As an easy case of symmetric momentum wave function

[tex]\phi(p,t)=\phi(-p,t)[/tex]

Obviously <p>=0.

[tex]\phi(p,t)=\phi(-p,t)[/tex]

Obviously <p>=0.

Last edited:

- #9

- 150

- 17

it is not symmetric, but as I said I have already found the expectation value using ehrenfest theorem. just asked for some other ways that is all

[tex]\phi(p,t)=\phi(-p,t)[/tex]

Obviously <p>=0.

Share:

- Replies
- 14

- Views
- 618

- Replies
- 4

- Views
- 647

- Replies
- 2

- Views
- 837

- Replies
- 1

- Views
- 620

- Replies
- 3

- Views
- 622

- Replies
- 3

- Views
- 798

- Replies
- 7

- Views
- 414

- Replies
- 1

- Views
- 929

- Replies
- 12

- Views
- 2K

- Replies
- 5

- Views
- 1K