- 743

- 4

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...

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...

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