- #1

Xkaliber

- 59

- 0

Hi all,

I've got a tough problem that I need some guidance on.

Question: Consider a wave function that is a combination of two different infinite-well states, the nth and the mth.

[tex]\Psi(x,t)=\frac{1}{\sqrt{2}}\psi_n(x)e^{-i(\frac{E_n}{\hbar})t}+\frac{1}{\sqrt{2}}\psi_m(x)e^{-i(\frac{E_m}{\hbar})t}[/tex]

Show that [tex]\Psi(x,t)[/tex] is properly normalized.

Answer:

A wave function is normalized if [tex]\int_{0}^{L} |\Psi(x,t)|^2 dx = 1[/tex] where L is the length of the infinite well.

[tex]\int_{0}^{L} |\Psi(x,t)|^2 dx = 1[/tex]

[tex]\int_{0}^{L} \Psi^*(x,t)\Psi(x,t) dx = 1[/tex]

[tex]\int_{0}^{L} [\frac{1}{\sqrt{2}}\psi_n(x)e^{i(\frac{E_n}{\hbar})t}+\frac{1}{\sqrt{2}}\psi_m(x)e^{i(\frac{E_m}{\hbar})t}] [\frac{1}{\sqrt{2}}\psi_n(x)e^{-i(\frac{E_n}{\hbar})t}+\frac{1}{\sqrt{2}}\psi_m(x)e^{-i(\frac{E_m}{\hbar})t}]dx = 1[/tex]

[tex]\int_{0}^{L} \frac{1}{2}\psi^2_n(x)dx+\int_{0}^{L} \frac{1}{2}\psi_n(x)\psi_m(x)e^{[i(\frac{E_m}{\hbar})-i(\frac{E_n}{\hbar})]t}dx+\int_{0}^{L} \frac{1}{2}\psi_n(x)\psi_m(x)e^{[i(\frac{E_n}{\hbar})-i(\frac{E_m}{\hbar})]t}dx+\int_{0}^{L} \frac{1}{2}\psi^2_m(x)dx[/tex]

For an infinite well, we know that [tex]\psi_n(x)=\sqrt{\frac{2}{L}}sin\frac{n\pi x}{L}[/tex] and [tex]\psi_m(x)=\sqrt{\frac{2}{L}}sin\frac{m\pi x}{L}[/tex]. Substituting into the above integrals gives us

[tex]\frac{1}{2}\int_{0}^{L} \frac{2}{L}sin^2\frac{n\pi x}{L}dx+\frac{1}{2}\int_{0}^{L} \frac{2}{L}sin\frac{n\pi x}{L}sin\frac{m\pi x}{L}e^{[i(\frac{E_m}{\hbar})-i(\frac{E_n}{\hbar})]t}dx+\frac{1}{2}\int_{0}^{L}\frac{2}{L}sin\frac{n\pi x}{L}sin\frac{m\pi x}{L}e^{[i(\frac{E_n}{\hbar})-i(\frac{E_m}{\hbar})]t}dx+\frac{1}{2}\int_{0}^{L}\frac{2}{L}sin^2\frac{m\pi x}{L}dx[/tex]

The first and last integral terms are 1/2, which sum to one. Since all 4 terms must sum to one, that means the two middle integrals must sum to zero. I used an integral table to solve the two middle integrals. (

http://teachers.sduhsd.k12.ca.us/abrown/Classes/CalculusC/IntegralTablesStewart.pdf" #79) However, the result gives an answer involving [tex]\frac{sin(\frac{n\pi}{L}-\frac{m\pi}{L})}{\frac{n\pi}{L}-\frac{m\pi}{L}}-\frac{sin(\frac{n\pi}{L}+\frac{m\pi}{L})}{\frac{n\pi}{L}+\frac{m\pi}{L}}[/tex], which depend upon L. Can someone please tell me where I went wrong?

I've got a tough problem that I need some guidance on.

Question: Consider a wave function that is a combination of two different infinite-well states, the nth and the mth.

[tex]\Psi(x,t)=\frac{1}{\sqrt{2}}\psi_n(x)e^{-i(\frac{E_n}{\hbar})t}+\frac{1}{\sqrt{2}}\psi_m(x)e^{-i(\frac{E_m}{\hbar})t}[/tex]

Show that [tex]\Psi(x,t)[/tex] is properly normalized.

Answer:

A wave function is normalized if [tex]\int_{0}^{L} |\Psi(x,t)|^2 dx = 1[/tex] where L is the length of the infinite well.

[tex]\int_{0}^{L} |\Psi(x,t)|^2 dx = 1[/tex]

[tex]\int_{0}^{L} \Psi^*(x,t)\Psi(x,t) dx = 1[/tex]

[tex]\int_{0}^{L} [\frac{1}{\sqrt{2}}\psi_n(x)e^{i(\frac{E_n}{\hbar})t}+\frac{1}{\sqrt{2}}\psi_m(x)e^{i(\frac{E_m}{\hbar})t}] [\frac{1}{\sqrt{2}}\psi_n(x)e^{-i(\frac{E_n}{\hbar})t}+\frac{1}{\sqrt{2}}\psi_m(x)e^{-i(\frac{E_m}{\hbar})t}]dx = 1[/tex]

[tex]\int_{0}^{L} \frac{1}{2}\psi^2_n(x)dx+\int_{0}^{L} \frac{1}{2}\psi_n(x)\psi_m(x)e^{[i(\frac{E_m}{\hbar})-i(\frac{E_n}{\hbar})]t}dx+\int_{0}^{L} \frac{1}{2}\psi_n(x)\psi_m(x)e^{[i(\frac{E_n}{\hbar})-i(\frac{E_m}{\hbar})]t}dx+\int_{0}^{L} \frac{1}{2}\psi^2_m(x)dx[/tex]

For an infinite well, we know that [tex]\psi_n(x)=\sqrt{\frac{2}{L}}sin\frac{n\pi x}{L}[/tex] and [tex]\psi_m(x)=\sqrt{\frac{2}{L}}sin\frac{m\pi x}{L}[/tex]. Substituting into the above integrals gives us

[tex]\frac{1}{2}\int_{0}^{L} \frac{2}{L}sin^2\frac{n\pi x}{L}dx+\frac{1}{2}\int_{0}^{L} \frac{2}{L}sin\frac{n\pi x}{L}sin\frac{m\pi x}{L}e^{[i(\frac{E_m}{\hbar})-i(\frac{E_n}{\hbar})]t}dx+\frac{1}{2}\int_{0}^{L}\frac{2}{L}sin\frac{n\pi x}{L}sin\frac{m\pi x}{L}e^{[i(\frac{E_n}{\hbar})-i(\frac{E_m}{\hbar})]t}dx+\frac{1}{2}\int_{0}^{L}\frac{2}{L}sin^2\frac{m\pi x}{L}dx[/tex]

The first and last integral terms are 1/2, which sum to one. Since all 4 terms must sum to one, that means the two middle integrals must sum to zero. I used an integral table to solve the two middle integrals. (

http://teachers.sduhsd.k12.ca.us/abrown/Classes/CalculusC/IntegralTablesStewart.pdf" #79) However, the result gives an answer involving [tex]\frac{sin(\frac{n\pi}{L}-\frac{m\pi}{L})}{\frac{n\pi}{L}-\frac{m\pi}{L}}-\frac{sin(\frac{n\pi}{L}+\frac{m\pi}{L})}{\frac{n\pi}{L}+\frac{m\pi}{L}}[/tex], which depend upon L. Can someone please tell me where I went wrong?

Last edited by a moderator: