Normalizing a Wave Function Combining Two Infinite-Well States

[Xkaliber]

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.

\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}

Show that \Psi(x,t) is properly normalized.


Answer:

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

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

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

\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

\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

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

\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

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 \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}}, which depend upon L. Can someone please tell me where I went wrong?

 

[Dr Transport]

Remember that for all wave functions whaich are part of a complete set

\int \psi_{n}(x)\psi_{m}(x) dx = \delta_{nm}

so you don't need to get into integrating all of the sine functions, and you have forgotten to take the exponential funcitons out of the intergral because they are not functions of x.

 

[Xkaliber]

Thanks for the reply. Could you please explain what \delta_{nm} is? I don't believe I have been introduced to that term.

Regarding the exponential functions, I did not take them out of the integrals above but did take them out on my paper which is how I obtained the \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}}
. Sorry for the confusion.

 

[quasar987]

\delta_{nm}=\left\{ \begin{array} {c} 1 \ \ \mbox{if} \ \ m=n & 0 \ \ \mbox{otherwise} \end{array}

 

[OlderDan]

Thanks for the reply. Could you please explain what \delta_{nm} is? I don't believe I have been introduced to that term.

Regarding the exponential functions, I did not take them out of the integrals above but did take them out on my paper which is how I obtained the \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}}
. Sorry for the confusion.

\delta_{nm} is the Kronecker Delta. It has the value 1 when the two indices are equal and zero otherwise. There has to be something wrong with your sine expressions. The argument of a trig function must be dimensionless.

The point Dr Transport was making is that you already know the wave functions are orthonormal. You should not have to work out the integrals again.

 

[Xkaliber]

I think I can see where you are going with this. However, I still do not understand why \int_{0}^{L} \psi_{n}(x)\psi_{m}(x) dx = 0 in this case. Can someone show (by doing the integral) or otherwise explain why this is so?

(Dan, is the L in the sine expression, in fact, dimensionless since the x/L fractions (before the integration took place) canceled out the units of length?)

 

[OlderDan]

I think I can see where you are going with this. However, I still do not understand why \int_{0}^{L} \psi_{n}(x)\psi_{m}(x) dx = 0 in this case. Can someone show (by doing the integral) or otherwise explain why this is so?

(Dan, is the L in the sine expression, in fact, dimensionless since the x/L fractions (before the integration took place) canceled out the units of length?)

Before you integrated, you had x/L in the trig functions and dx/L in the integral with limits proportional to L. Everything was dimensionless. When you looked up your integrals, you somehow lost track of things. You might want to change to a dimensionless variable before integrting. y = x/L would do nicely.

 

[quasar987]

I think I can see where you are going with this. However, I still do not understand why \int_{0}^{L} \psi_{n}(x)\psi_{m}(x) dx = 0 in this case. Can someone show (by doing the integral) or otherwise explain why this is so?

It is a postulate of QM that the (normalized )eigenfunctions of the hamiltonian form an orthonormal basis for the system. For regular vectors, orthonormality of a basis means that

\hat{e}_n\cdot \hat{e}_m=\delta_{nm}

right? Well for eigenfunctions we extend this definition to mean that

\int_{-\infty}^{+\infty} \psi_{n}(x)\psi_{m}(x) dx = \delta_{nm}

In the particular case of the infinite well eigenfunctions, you can verify that they are orthonormal by computing the intgegral.. for inxtance by use of the identity 2sinAsinB=(cos(A+B/2)+sin(A-B/2)) [or somethign like that].

 

[Xkaliber]

Thanks guys. I got the integral to equal zero using Dan's suggestion. It's satisfying to mathematically see how a value is obtained rather than just taking it for granted. I have a few other parts related to this question to answer so I may be back if I encounter any more roadblocks.