# Additional understanding needed on proof involving Hermitian conjugates

1. Sep 26, 2011

### TheFerruccio

I encountered this part in Griffith's Introduction to Quantum Mechanics that I have been unable to figure out. It is probably obvious, but I am not seeing it. I probably need more practice with operators in order to have it fully understood.

Equation 2.64 in the second edition states:

$\int_{-\infty}^{\infty}{f^*}(a_{\pm}g)dx = \int_{-\infty}^{\infty}(a_{\mp}f)^*gdx$

It starts out by making the substitution where $a_\pm = \frac{1}{\sqrt{2\hbar m\omega}}(\mp\hbar\frac{d}{dx}+m\omega x)$...

$\int_{-\infty}^{\infty}f^*(a_\pm g)dx=\frac{1}{\sqrt{2\hbar m\omega}}\int_{-\infty}^{\infty}f^*(\mp\hbar\frac{d}{dx}+m\omega x)gdx$

From there, it states that the integrals must exist, which means that f(x) and g(x) must go to zero. This makes sense, since what is being done is normalizing, and a normalizable function must have values of 0 at the extrema.

What I don't understand is the next step, whereby it states that integration by parts takes $\int f^*(\frac{dg}{dx})dx$ to $-\int(\frac{df}{dx})^*gdx$

How did they arrive at this? I tried writing it out, but I think I am not dealing with the operators correctly. What are the steps to arrive at this conclusion? It says that it has to do with the fact that the function values at the extrema disappear, but I am not seeing it when I write it out. It just ends up getting messy. The book skips over these steps, but I always include the explicit steps in my notes.

2. Sep 26, 2011

### jfy4

Let $g'$ be the thing you integrate during integration by parts, and $f^{*}$ the thing you differentiate. Then remember exactly what you said earlier, that these functions (g, f) are zero at the bounds of integration...

3. Sep 26, 2011

### TheFerruccio

I think I am starting to see it.

I took $\int_{-\infty}^{\infty}f^*(\mp\hbar\frac{d}{dx}+m\omega x)gdx = \int_{-\infty}^{\infty}f^*\mp\hbar\frac{dg}{dx}dx+\int_{-\infty}^{\infty}f^*(m\omega x)dx$

That allows me to continue on to the final part of the proof. Thanks!

However, one thing I do not understand, is how the * moves about during the proof.

For instance, when I used the integrals, I was able to arrive at:

$\frac{1}{\sqrt{2\hbar m\omega}\int_{-\infty}^{\infty}(\pm\hbar g(\frac{df}{dx})^*+f^*m\omega xg)dx$

But, I do not see how I am to go from there to...

$$\frac{1}{\sqrt{2\hbar m\omega}\int_{-\infty}^{\infty}((\pm\hbar\frac{d}{dx}+m\omega x)f)^*gdx$$

Also, wow, my LaTeX is breaking for me. I don't see what I'm doing wrong that's causing the crazy errors. I'm getting lots of W's with lines through them, in boxes.

Anyway, in hopes that the LaTeX doesn't break again, here's another example of my confusion:

$\int_{-\infty}^{\infty}(a_\pm\psi_n)^*(a_\pm\psi_n)dx = \int_{-\infty}^{\infty}(a_\mp a_\pm\psi_n)^*\psi_n dx$

I am absolutely not seeing why the * orders that way. Why can't it be...

$\int_{-\infty}^{\infty}{\psi_n}^*a\mp a\pm\psi_n dx$ ?

Last edited: Sep 26, 2011
4. Sep 27, 2011

### Fredrik

Staff Emeritus
The two broken LaTeX formulas both contain an expression of the form
Code (Text):
\frac{A}{\sqrt{B}
That seems to be the only problem here. In case you're not aware of the 50 character bug, you also need to know that if you type 50 characters without a space, vBulletin will insert one that usually breaks the code. The workaround is to type more spaces.

I also recommend tex tags instead of itex when you want the math image to appear on a line of its own. (When you use tex tags, don't type any line breaks before and after. If you want a comma or a period at the end, put it before the closing tex tag).

5. Sep 27, 2011

### jfy4

Consider
$$\int_{-\infty}^{\infty}(f^* \frac{\partial g}{\partial x} + f^* x g)dx$$
Let $u=f^*$ and $v'=g'$. Then using integration by parts
$$uv-\int u'v=\int uv'$$
we can see that
$$\int_{-\infty}^{\infty}f^* \frac{\partial g}{\partial x} dx=f^* g |_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\frac{\partial f^{*}}{\partial x} g dx$$
but what are $f, g$ at infinity? The result follows.