## Monday 28 April 2008

### Some more exciting math

I haven't written anything for a month and the thing that catches my mind is proving why the Hamiltonian operator is Hermitian, which is of course extremely exciting. What I liked about it, is the fact that the proof is not really more than the formula for partial integration.

A is Hermitian if

$<\Psi|A\Phi>=$

or equivalent

$<\Psi|A|\Phi>=<\Phi|A|\Psi>^*$

where <.|.> is a scalar product of a unitary vector space

As a warm up one can show that multiplying with any real constant λ is Hermitian. This is independent of the chosen scalar product and we can write (the last part is possible just because λ is real and therefore equal to its complex conjugate): [1]

$<\Psi|\lambda \Phi>=\lambda<\Psi|\Phi>=<\lambda^*\Psi|\Phi>=<\lambda\Psi|\Phi>$

Let's define the scalar product as:

$<\Psi|\Phi>=\int_{-\infty}^{\infty}\Psi(x)^* \Phi(x)dx$

Now comes the part that I mainly wanted to show, proving that 1/i d/dx is Hermitian with partial integration (please excuse that I am leaving out the lim's but I think it's easier to read this way)

$<\Psi|\frac{1}{i}\frac{d}{dx}\Phi>=\frac{1}{i}\int_{-\infty}^{\infty}\Psi(x)^* \frac{d}{dx}\Phi(x)dx=$

partial integration leads to

$=\frac{1}{i}\Psi(x)\Phi(x)|_{-\infty}^{\infty}-\frac{1}{i}\int_{-\infty}^{\infty}\frac{d}{dx}(\Psi(x))^* \Phi(x)dx=$

The first expression vanishes because the wave functions have to be square integrable and therefore disappear at infinity. In the second expression we can draw the operator into the complex conjugation and the complex conjugate of i is of course -i.

$=\int_{-\infty}^{\infty}(\frac{1}{i}\frac{d}{dx}\Psi(x))^* \Phi(x)dx=<\frac{1}{i}\frac{d}{dx}\Psi|\Phi>$

And that's what we wanted to show. Actually a pretty amazing result if one weren't used to it.

For showing that the Hamiltonian operator is Hermitian you have to prove a few more things, e.g. that the square of a Hermitian operator is Hermitian.

$<\Psi|A^2\Phi>==$

You can do similar things with summation or real scalar multiplication.

Finally you can enjoy yourself with multiple integrals and find out that there is no problem with those either:

$\int_{-\infty}^{\infty}\Psi(x_1,..,x_n)^*\frac{1}{i}\frac{\partial}{\partial x_1}\Phi(x_1,...,x_n)dx_1=\int_{-\infty}^{\infty}(\frac{1}{i}\frac{\partial}{\partial x_1}\Psi(x_1,...,x_n))^* \Phi(x_1,...,x_n)dx_1 \Rightarrow$
$\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}\Psi(x_1,..,x_n)^*\frac{1}{i}\frac{\partial}{\partial x_1}\Phi(x_1,...,x_n)dx_1)dx_2...dx_n=
\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}(\int_{-\infty}^{\infty}(\frac{1}{i}\frac{\partial}{\partial x_1}\Psi(x_1,...,x_n))^* \Phi(x_1,...,x_n)dx_1)dx_2...dx_n$

There is always only one variable involved in the operator and you can just add the other integrals on the outside. Of course this also holds for different xi than x1 because you can change the integration order.

[1] I am sorry I put a frownie face there.

Echiral said...

Not a constructive comment (sorry), but the proof is pretty; I like the integration by parts.

Actually, for the claim that the scalar multiple of a self-adjoint operator is self-adjoint ("You can do similar things with summation or scalar multiplication"), isn't it false for complex scalars? Or am I misunderstanding something?

(Using square brackets because I can't type the angle ones)
Given that [f, Ag] = [Af, g] for operator A and complex scalar &lambda,

[f, &lambda Ag] = &lambda [f, Ag]
= &lambda [Af, g]
= [&lambda *Af, g]
!= [&lambda Af, g]
if &lambda isn't real

Felix said...

yes, thanks it's only for real scalars. i changed it