# Finding the Potential

1. Dec 29, 2006

### Reshma

This one is from Liboff(p6.8)

Given the wavefunction:
$$\psi(x, t) = A exp[i(ax - bt)]$$
What is the Potential field V(x) in which the particle is moving?
If the momentum of the particle is measured, what value is found(in terms of a & b)?
If the energy is measured, what value is found?

My Work:

$$\psi(x, t) = A exp[i(ax - bt)]$$
I took the partial derivatives wrt to t and x:
$$\frac{\partial \psi}{\partial t} = -(ib)\psi$$

$$\frac{\partial^2 \psi}{\partial x^2} = -a^2\psi$$

Time dependent Schrodinger's equation is:
$$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi$$

Substituting the above values in this equation:

$$\hbar b \psi = \frac{\hbar^2 a^2}{2m}\psi + V(x)\psi$$

Dividing throughout by $\psi$ and rearranging, I get the potential field as:
$$V(x) = \hbar\left(b - \frac{\hbar a^2}{2m}\right)$$

Am I going right? Before I can proceed furthur...

Last edited: Dec 29, 2006
2. Dec 29, 2006

### neutrino

:surprised

The calculation seems correct, though.

3. Dec 29, 2006

### Reshma

Sorry, I realised my mistake but I got disconnected before I could correct it. I've corrected it now.

4. Dec 29, 2006

### Reshma

Now I have to get an expression for the momentum.
$$p = \hbar k$$
$$V(x) = {1\over 2}kx^2$$

Equating this to the other expression of V(x) would mean an x2 term in the momentum expression. I am confused over this.
$$k = {{2\hbar}\over x^2} \left(b - \frac{\hbar a^2}{2m}\right)$$

5. Dec 29, 2006

### Gokul43201

Staff Emeritus
Didn't you just find out that V(x) is a constant potential? Look at your answer for V(x) above. Why are you assuming that V(x) should now be a harmonic potential?

Given a wavefunction, how do you calculate the values of observables?

6. Dec 30, 2006

### Reshma

Use the operator relation?
$$\hat p = -i\hbar\frac{\partial}{\partial x}$$

$$\hat H = -\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2}$$

7. Dec 30, 2006

### marlon

Where does the above harmonic V(x) come from ?
Your original way of working is correct. I didn't check all the algebra though.

marlon

8. Dec 30, 2006

### Reshma

Sorry! I made an erroneous assumption. I think using the operator relation is the correct technique. Am I right?

9. Dec 30, 2006

### marlon

yes

marlon

10. Dec 30, 2006

### Gokul43201

Staff Emeritus
This is correct.

This isn't right. The RHS is simply the kinetic energy, not the total energy.

11. Dec 30, 2006

### Worzo

For the first part of working out the potential, you have to employ the TDSE, this is true.

However, you should recognise that this is the wavefunction for a free particle, in the form:

[psi] = Aexp[i(kx - wt)], where A = normalisation factor, p=[hbar]k, E = [hbar]w.

Hence in this example, p = [hbar]a, E = [hbar]b.

You may use the operator relations p[hat] = -i[hbar]d/dx, and H[hat] = i[hbar]d/dt to show this.

Notice also how V = [hbar]w - ([hbar]k)2/2m = [hbar]w - p2/2m
So V = E - KE
i.e. Potential Energy = Total Energy - Kinetic Energy

Which agrees with Total Energy = Kinetic Energy + Potential Energy

Last edited: Dec 30, 2006
12. Dec 31, 2006

### Reshma

Thanks for the responses Gokul, Marlon and Worzo! Things are making better sense now.