# Using generating function to normalize wave function

• thecourtholio
In summary, to prove the proper normalization of ##\psi_n## in Eq. 2.85, one must substitute generating functions in place of the Hermite polynomials in the normalization integral and equate the resulting Taylor series on both sides. As a by-product, it can be shown that ##\psi_m## and ##\psi_n## are orthogonal when m ≠ n. To evaluate the necessary integrals, one can consider the corresponding quantities using the generating functions, or use the Taylor expansions of the generating functions in terms of the Hermite polynomials.
thecourtholio

## Homework Statement

Prove that ##\psi_n## in Eq. 2.85 is properly normalized by substituting generating functions in place of the Hermite polynomials that appear in the normalization integral, then equating the resulting Taylor series that you obtain on the two sides of your equation. As a by-product, you should also find that ## \psi_m ## and ## \psi_n ## are orthogonal when m ≠ n.

## Homework Equations

Eq 2.85: $$\psi_n(x) = \bigg(\frac{m\omega}{\pi\hbar}\bigg)^{1/4} \frac{1}{\sqrt{2^nn!}}H_n(\xi) e^{-\xi^2/2}$$
where $$\xi \equiv \sqrt{\frac{m\omega}{\hbar}}x$$
and ## H_n(\xi) ## are the Hermite polynomials.

## The Attempt at a Solution

Ok so the main thing throwing me off here is where it says to equate the two taylor series obtained because I don't really see how I'm supposed to obtain a series on both sides.

First off, when I go to normalize I use
$$\int_{-\infty}^{\infty} |\psi_n|^2d\xi = 1$$
and equating all the constants in front of the ## H_n(\xi) ## term to ## A## the integral becomes
$$\int_{-\infty}^{\infty} A^2 H_n^2 e^{-\xi^2} d\xi = 1$$
Then, substituting the generating function (using the form ## e^{2\xi z-z^2} ##) in, this becomes
$$\int_{-\infty}^{\infty} A^2 e^{-(\xi-z)^2} = 1$$
But this is a simple gaussian integral right? So then its just
$$\int_{-\infty}^{\infty} A^2 e^{-(\xi-z)^2} = \sqrt{\pi}A^2$$

But I feel like all this is wrong because I don't see how a Taylor series comes into play. Am I doing the normalization all wrong or is there just something I'm not seeing?

Any advice is much appreciated!

I haven't worked out the problem, but I did notice a mistake in your attempt. The integral involves the Hermite polynomial squared, so you need to square the generating function to get a term that looks like that. Squaring the Taylor series the generating function is equal to will also produce the cross terms that you expect to vanish.

vela said:
I haven't worked out the problem, but I did notice a mistake in your attempt. The integral involves the Hermite polynomial squared, so you need to square the generating function to get a term that looks like that. Squaring the Taylor series the generating function is equal to will also produce the cross terms that you expect to vanish.
Oh thank you I hadn't caught that mistake!
So it should be ##\int_{-\infty}^{\infty} A^2 e^{4\xi z-z^2}e^{-\xi^2}d\xi## ? And should it be an integral of ##d\xi## or ##dx##?
But I'm still not sure how ## e^{4\xi z-2z^2} = \sum \frac{z^{2n}}{(n!)^2}H_n^2## gives me cross terms that vanish? What do you mean by that?

You need to be a little more careful here. Remember ##(a+b)^2 \ne a^2 + b^2##.
$$(e^{2\xi z- z^2})^2 = \left[\sum_{n=0}^\infty H_n(\xi) \frac{z^n}{n!}\right]^2 = \left[\sum_{m=0}^\infty H_m(\xi) \frac{z^m}{m!}\right]\left[\sum_{n=0}^\infty H_n(\xi) \frac{z^n}{n!}\right]$$

thecourtholio said:
$$\int_{-\infty}^{\infty} A^2 H_n^2 e^{-\xi^2} d\xi = 1$$
Then, substituting the generating function (using the form ## e^{2\xi z-z^2} ##) in, this becomes
$$\int_{-\infty}^{\infty} A^2 e^{-(\xi-z)^2} = 1$$
This step is not valid. ##H_n(\xi)## does not equal the generating function. So, you can't replace ##H_n(\xi)## by the generating function and still expect the integral to equal 1.

In normalizing the ## \psi_n(x)## and checking their orthonormality, you essentially need to find the integrals $$I_{nm} = \int_{-\infty}^{\infty} H_n(\xi) H_m(\xi) e^{-\xi^2} d\xi$$

The suggestion to obtain ##I_{nm}## is to consider the corresponding quantities $$G_{nm} = \int_{-\infty}^{\infty} g(\xi, y) g(\xi, z) e^{-\xi^2} d\xi$$ where the ##g##'s are the generating functions ##g(\xi, y) = e^{2\xi y-y^2} ## and ##g(\xi, z) = e^{2\xi z-z^2} ##.

See if you can evaluate ##G_{nm}## in two different ways:

(1) Using the expressions above for the generating functions

(2) Using the expansions of the generating functions, where the expansions are in terms of the Hermite polynomials.

Last edited:
vela said:
You need to be a little more careful here. Remember ##(a+b)^2 \ne a^2 + b^2##.
$$(e^{2\xi z- z^2})^2 = \left[\sum_{n=0}^\infty H_n(\xi) \frac{z^n}{n!}\right]^2 = \left[\sum_{m=0}^\infty H_m(\xi) \frac{z^m}{m!}\right]\left[\sum_{n=0}^\infty H_n(\xi) \frac{z^n}{n!}\right]$$

Oh yeah I see. I should've noticed that after seeing everyones comments about mod squared.

TSny said:
This step is not valid. ##H_n(\xi)## does not equal the generating function. So, you can't replace ##H_n(\xi)## by the generating function and still expect the integral to equal 1.

Yeah I was questioning that as I was typing it since ##e^{2\xi-z^2} = \sum \frac{z^n}{n!}H_n## and not just ##H_n## but I wasn't sure if it was possible to rearrange the summation.

TSny said:
In normalizing the ## \psi_n(x)## and checking their orthonormality, you essentially need to find the integrals $$I_{nm} = \int_{-\infty}^{\infty} H_n(\xi) H_m(\xi) e^{-\xi^2} d\xi$$

The suggestion to obtain ##I_{nm}## is to consider the corresponding quantities $$G_{nm} = \int_{-\infty}^{\infty} g(\xi, y) g(\xi, z) e^{-\xi^2} d\xi$$ where the ##g##'s are the generating functions ##g(\xi, y) = e^{2\xi y-y^2} ## and ##g(\xi, z) = e^{2\xi z-z^2} ##.

See if you can evaluate ##G_{nm}## in two different ways:

(1) Using the expressions above for the generating functions

(2) Using the Taylor expansions of the generating functions, where the expansions are in terms of the Hermite polynomials.

So combining your suggestion with vela's I need to evaluate something like ## \int_{-\infty}^{\infty} A^2 \bigg[ \sum\frac{z^m}{m!}H_m\bigg]\bigg[\sum \frac{z^n}{n!}H_n\bigg] e^{-\xi^2} d\xi ## ? Any tips on how to evaluate an integral like this?

thecourtholio said:
So combining your suggestion with vela's I need to evaluate something like ## \int_{-\infty}^{\infty} A^2 \bigg[ \sum\frac{z^m}{m!}H_m\bigg]\bigg[\sum \frac{z^n}{n!}H_n\bigg] e^{-\xi^2} d\xi ## ? Any tips on how to evaluate an integral like this?
You will not need to evaluate this expression directly. (However, I think that one of the sums should involve the variable ##y## instead of ##z## if you are going to follow the suggestion I gave in post #5. This yields the "second way" of evaluating ##G_{nm}## that I listed in post #5.)

The idea is to compare this result with the "first way" of evaluating ##G_{nm}## that I listed in post #5.

This "first way" should lead to a simple result for ##G_{nm}## in terms of ##y## and ##z##. You can then expand this result in a Taylor series expansion in powers of ##yz## and compare term-by-term with your double sum expression above (after writing one of your sums in terms of ##y## instead of ##z##).

TSny said:
You will not need to evaluate this expression directly. (However, I think that one of the sums should involve the variable ##y## instead of ##z## if you are going to follow the suggestion I gave in post #5. This yields the "second way" of evaluating ##G_{nm}## that I listed in post #5.)

The idea is to compare this result with the "first way" of evaluating ##G_{nm}## that I listed in post #5.

This "first way" should lead to a simple result for ##G_{nm}## in terms of ##y## and ##z##. You can then expand this result in a Taylor series expansion in powers of ##yz## and compare term-by-term with your double sum expression above (after writing one of your sums in terms of ##y## instead of ##z##).

Ok I think I see what you are saying. So I evaluate ## \int e^{2\xi y-y^2}e^{2\xi z-z^2} e^{-\xi^2} d\xi ## and compare it to ##\int \bigg[\sum \frac{y^m}{m!}H_m\bigg]\bigg[\sum \frac{z^n}{n!}H_n\bigg] e^{-\xi^2}d\xi ## something like that?

thecourtholio said:
Ok I think I see what you are saying. So I evaluate ## \int e^{2\xi y-y^2}e^{2\xi z-z^2} e^{-\xi^2} d\xi ## and compare it to ##\int \bigg[\sum \frac{y^m}{m!}H_m\bigg]\bigg[\sum \frac{z^n}{n!}H_n\bigg] e^{-\xi^2}d\xi ## something like that?
Yes, that should work.

TSny said:
Yes, that should work.
Ok I got $$\int e^{2\xi y-y^2}e^{2\xi z-z^2}e^{-\xi^2}d\xi = \sqrt{\pi}e^{2yz} = \sqrt{\pi} \sum \frac{y^lz^l}{l!}$$
Is this the correct series? Cause in comparison to $$\int \frac{y^m}{m!}H_m \frac{z^n}{n!}H_n e^{-\xi^2} d\xi$$ the series and gaussian would cancel for m=n=l leaving just ##\int H_n H_m## which equals 1 since they're orthogonal. Is that right?

thecourtholio said:
Ok I got $$\int e^{2\xi y-y^2}e^{2\xi z-z^2}e^{-\xi^2}d\xi = \sqrt{\pi}e^{2yz} = \sqrt{\pi} \sum \frac{y^lz^l}{l!}$$ Is this the correct series?
Looks good except the right hand side is missing 2 raised to some power.
Cause in comparison to $$\int \frac{y^m}{m!}H_m \frac{z^n}{n!}H_n e^{-\xi^2} d\xi$$ the series and gaussian would cancel for m=n=l leaving just ##\int H_n H_m## which equals 1 since they're orthogonal. Is that right?
You want to use this analysis to prove that you have orthogonality and to deduce the value of ## \int H_n^2 e^{-\xi^2} d\xi ## . Since ##y## and ##z## are arbitrary numbers, terms of the same power in ##y## and ##z## in the comparison must match.

TSny said:
Looks good except the right hand side is missing 2 raised to some power.
You want to use this analysis to prove that you have orthogonality and to deduce the value of ## \int H_n^2 e^{-\xi^2} d\xi ## . Since ##y## and ##z## are arbitrary numbers, terms of the same power in ##y## and ##z## in the comparison must match.

Ok I think I understand it all now. Thank you so much for y'all's help!

## 1. What is a generating function?

A generating function is a mathematical tool used to represent a sequence or series of functions in a compact form. It allows us to manipulate and analyze the properties of these functions in a simpler way.

## 2. How does a generating function help in normalizing a wave function?

A generating function can be used to calculate the coefficients of a wave function, which represent the probability amplitudes of different states of a quantum system. By normalizing the coefficients, we can ensure that the total probability of all possible states is equal to 1, as required by the principles of quantum mechanics.

## 3. Can a generating function be used for any type of wave function?

Yes, a generating function can be used for any type of wave function, as long as it is a finite series or sequence of functions. It is a versatile tool that can be applied to various mathematical and physical systems.

## 4. Are there any limitations to using a generating function for normalization?

One limitation of using a generating function for normalization is that it may not work for some complex or non-linear systems. In these cases, other techniques may need to be used to normalize the wave function.

## 5. Are there any applications of using a generating function to normalize wave functions?

Yes, there are many applications of using a generating function to normalize wave functions. It is commonly used in quantum mechanics to calculate the probabilities of different states in a system. It is also used in signal processing, probability theory, and other areas of mathematics and physics.

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