**Required math: calculus **

**Required physics: none**

Hermite’s differential equation shows up during the solution of the Schrödinger equation for the harmonic oscillator. The differential equation can be written in the form

but an analysis of the series solution of this equation shows that the parameter has to have the form

for some integer , so we can rewrite the differential equation as

We know the solutions of this equation are polynomials in , and we got (from the series solution) a recursion formula for the coefficients of the polynomials, but a recursion formula can be difficult to work with, and it turns out that there is another form that can be used to work with these polynomials. This uses the idea of the *generating function*.

The idea is that we can write a function , where is the same as in the differential equation, and is a kind of dummy variable that allows us to do calculations (as we’ll see in a moment). Suppose we define this function as follows:

From the expansion of the exponential in a Taylor series, we can also write this as

At first (and probably second) glance, this formula seems to have little relation to Hermite polynomials, but let’s write out the first few terms of the series

In the second line, we regrouped the series so that terms with the same power of are grouped together. The term in the series contains terms involving to the and higher powers only, so if we want to isolate those terms for a particular power (say the power) of we need look at only the first terms of the series. What do we get if we look at terms involving each successive power of , starting with the zeroth power? As can be seen above, the term involving is multiplied by a polynomial in and by comparing these polynomials with those obtained by our earlier definition of the Hermite polynomials, we can see that each polynomial here is . That is

Obviously we haven’t *proved* this in general, but this function may also be taken as the definition of Hermite polynomials, as the other definition that we used earlier can be derived from it, as we’ll see at the end of this post.

The Hermite polynomials can be obtained from this generating function by taking derivatives, as follows. Since the derivative of is zero if , taking this derivative will eliminate all terms with The derivative of is the constant . For all higher powers where , the derivative will leave a term . So if we take the derivative of and then set we will isolate the single term involving :

This is the reason that is called a generating function: it provides a relatively simple way of generating all the Hermite polynomials.

Since we started by defining the generating function, we should prove that the polynomials that it generates really are solutions of Hermite’s differential equation. We can do this by taking derivatives of the generating function (but without the step of setting ). We take derivatives of 4 and 9 and then set them equal to each other.

Now we use the old trick of requiring these two results to be equal for all values of , which implies that the coefficients of each power of must be equal independently. That is

In the second line, we adjusted the summation index on the left so that the power of was . On the right, we dropped the term since anyway (since ). The two sums are now aligned, so we can say

By a similar process we can take the other derivative with respect to :

We have ignored the term in the last line, since the derivative of the first term in the series with respect to is zero. Aligning the powers of gives

This relation is valid for all even though the case is a bit fortuitous. With we get which is true, since and .

From these results we can show that the polynomials do in fact solve Hermite’s differential equation. We do this by showing that the results above allow us to reconstruct the equation. From the second result:

From the first result, , and so substituting these into the last line above, we get

which is Hermite’s equation. QED.

One final bit of business is to show that the generating function approach is equivalent to the other definition of Hermite polynomials, that is, that it is equivalent to saying

The generating function 4 can be written as

so taking the derivative, we get

since for any function , . Setting , we reclaim the original definition:

so the two definitions are equivalent.

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AaronBy golly, this is an excellent resource. This is well written. It is clear and concise. It has been extremely useful to me.

DUYENcan you explain to me how you can come up with (25) from (24). I do not know why there is 2Hm. Please it is very urgent so answer me as soon as possible. Thank you for help a lots

gwrowePost authorEquation (25) is just the derivative of (24) with respect to y. Use the product rule to get the first two terms on the RHS of (25).