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Finding the Potential

  • Thread starter Reshma
  • Start date
743
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This one is from Liboff(p6.8)

Given the wavefunction:
[tex]\psi(x, t) = A exp[i(ax - bt)][/tex]
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:

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

[tex]\frac{\partial^2 \psi}{\partial x^2} = -a^2\psi[/tex]

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

Substituting the above values in this equation:

[tex]\hbar b \psi = \frac{\hbar^2 a^2}{2m}\psi + V(x)\psi[/tex]

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

Am I going right? Before I can proceed furthur...
 
Last edited:
2,035
2
Time independent Schrodinger's equation is:
[tex]i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi[/tex]
:surprised

The calculation seems correct, though.
 
743
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:surprised

The calculation seems correct, though.
Sorry, I realised my mistake but I got disconnected before I could correct it. I've corrected it now. :biggrin:
 
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Now I have to get an expression for the momentum.
[tex]p = \hbar k[/tex]
[tex]V(x) = {1\over 2}kx^2[/tex]

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

Gokul43201

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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?
 
743
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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?
Use the operator relation?
[tex]\hat p = -i\hbar\frac{\partial}{\partial x}[/tex]

[tex]\hat H = -\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2}[/tex]
 
3,761
8
Now I have to get an expression for the momentum.
[tex]p = \hbar k[/tex]
[tex]V(x) = {1\over 2}kx^2[/tex]
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
 
743
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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
Sorry! I made an erroneous assumption. :blushing: I think using the operator relation is the correct technique. Am I right?
 
3,761
8
Sorry! I made an erroneous assumption. :blushing: I think using the operator relation is the correct technique. Am I right?
yes

marlon
 

Gokul43201

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Use the operator relation?
[tex]\hat p = -i\hbar\frac{\partial}{\partial x}[/tex]
This is correct.

[tex]\hat H = -\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2}[/tex]
This isn't right. The RHS is simply the kinetic energy, not the total energy.
 
20
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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:
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Thanks for the responses Gokul, Marlon and Worzo! Things are making better sense now. :smile:
 

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