- #1

Vuldoraq

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## Homework Statement

Consider a particle in a potential ;

[tex]V(x)=0\ for\ 0<=x<=L[/tex] and

[tex]V(x)=\infty\ otherwise[/tex]

. Its wave function is a linear superposition of the lowest two stationary states and given by

[tex]\psi(x,t)=\frac{1}{2}*(\sqrt{3} \psi_{1}(x,t)-\psi_{2}(x,t)[/tex])

(a) Use the results fromquestion 2 and ﬁnd ψ(x,t). Then calculate [tex]|\psi(x,t)|^{2}[/tex].

Make sure that your result for [tex]|\psi(x,t)|^{2}[/tex] is manifestly real (no “i” left in there). What is the frequency of the oscillation in [tex]|\psi(x,t)|^{2}[/tex]?

(b) Calculate the expectation value of the Hamilton operator ˆH. Compare it with the energy eigenvalues

E1 and E2.

(c) What is the probability for the particle to be in state ψ1 and what is the probability for it to be in the state ψ2?

## Homework Equations

Time dependent wave function for the nth stationary state in this potential well in this case is;

[tex]\psi_{n}(x,t)=\sqrt{\frac{2}{L}}\sin(\frac{n\pix}{L})\exp{\frac{-iE_{n}t}{h/2\pi}\ n=1,2,3,...[/tex]

and

[tex]E_{n}=\frac{h^{2}n^{2}}{8mL^{2}}[/tex]

## The Attempt at a Solution

Hi, I can get the solution to the first part of (a), I think,

[tex]\psi(x,t)=\frac{1}{2}(\sqrt{\frac{6}{L}}\sin(\frac{\pi\x}{L})\exp{\frac{-iE_{1}t}{h/2\pi}-\sqrt{\frac{2}{L}}\sin(\frac{2\pi\x}{L})\exp{\frac{-iE_{2}t}{h/2\pi})[/tex]

Where

[tex]E_{1}=\frac{h^{2}}{8mL^{2}}[/tex] and

[tex]E_{2}=\frac{h^{2}}{2mL^{2}}[/tex]

But after that I have no idea of where to go. When I try and calculate the product of the conjugate times non-conjugate I end up with complex terms that I can't simplify, so my result isn't real. Also I have no idea how you find the frequancy of oscillation in [tex]|\psi(x,t)|^{2}[/tex], or how to calculate the expectation of the Hamiltonian operator. Any help would be a godsend to me, I feel completely lost now in QM.

Thanks in advanc

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