# Group symmetry of Hamiltonian

1. Dec 17, 2013

### LagrangeEuler

$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2x^2$
Parity
$Px=-x$
end $e$ neutral are group of symmetry of Hamiltonian.
$PH=H$
$eH=H$
so I said it is group of symmetry because don't change Hamiltonian? And $e$ and $P$ form a group under multiplication. Is there any way to right some representation of this group? Thx for the answer.

2. Dec 17, 2013

### ChrisVer

what is $e$?
Your representation depends on the space it acts. For example the parity when acting on the 4 Lorentz vectors in Minkowski space can be represented by diag[1,-1,-1,-1].
On three dim space, it is still diag[-1,-1,-1] so I guess the parity for that is equivalent to $Z_{2}$ so that's why in the Harmonic oscillator you give, your Hamiltonian eigenstates to that of the parity, are the same and they are conserved (being either cosine or sinus )
I am not sure, but I also somehow think it corresponds to the reflections (although it's not the group itself).

So to write a representation, you have to define the vector space the representation acts on.

3. Dec 17, 2013

### LagrangeEuler

$e$ is neutral.
$e^2=e$
This iz $Z_2$ group.
Hamiltonian eigenstates are Hermite functions.
So could I say that $H$ is invariant under $Z_2$?

4. Dec 17, 2013

### ChrisVer

the identity ok.
Hermite funcs are funcs of fixed parity as well...
I'm not sure if you can, personally I'd prefer saying parity...

5. Dec 18, 2013

### LagrangeEuler

I'm just confused what I have now when I know that $\{e,P\}$ is symmetry group of Hamiltonian? I know that $PH=HP=H$. So $[P,H]=0$. Right?

6. Dec 19, 2013

### nonequilibrium

Note that PH (or HP) is not H! Your mistake is that you are acting P on (just) H, but the way you should interpret "PH" is as follows: for any function f, PH is defined as (PH)(f) := P(Hf). And note that for general functions f, P(Hf) won't be the same as Hf! Hence PH is not H.

What is true, however, is indeed [P,H] = 0. Why? Well, again that statement is equivalent to "for any f, [P,H]f = 0", and this is true, because
$PH f(x) = P \left[ \left( \alpha \frac{\mathrm d^2}{\mathrm d x^2} + \beta x^2 \right) f(x)\right] = \left( \alpha \frac{\mathrm d^2}{\mathrm d (-x)^2} + \beta (-x)^2 \right) f(-x) = \left( \alpha \frac{\mathrm d^2}{\mathrm d x^2} + \beta x^2 \right) f(-x) = \left( \alpha \frac{\mathrm d^2}{\mathrm d x^2} + \beta x^2 \right) P f(x) = HP f(x)$

The value in knowing [P,H] = 0, is that now you know the eigenstates of H are eigenstates of P. Why? Well suppose f is an eigenstate of H, i.e. Hf = Ef.
Then the claim is $Pf = \pm f$. To see this, note that $E Pf = P E f = P Hf = HP f$, but this means Pf is an eigenstate of H with eigenvalue E. If we assume there is only one for every energy, then Pf has to be proportional to f! (and it's clear any eigenvalue of P has to square to 1)

Last edited: Dec 19, 2013