# Hermitian operators

1. Oct 8, 2006

### holden

this question is two parts, both dealing with telling if combinations of hermitian operators are hermitian.

the first combination is PX + XP, where P stands for the momentum operator, (h bar /i)(d/x), and X is the "x operator", x. i have figured out that PX and XP are not hermitian by themselves, but i don't have any idea how to go about showing their linear combination is or isn't. so far i have tried calculating the expectation values for PX + XP and it's conjugate to see if they were the same, but i get the feeling this isn't a correct method.

the second combination is XPX.. i know if you have two operators multiplied, their product can only be hermitian if their commutator is zero.. but how do you do a commutator of a product of three operators?

any help is greatly appreciated! thanks so much.

2. Oct 8, 2006

### Ahmes

You should take a dagger to this new operator and show it is equal to the operator itself, example:

* means dagger!
(PX+XP)* = (PX)* + (XP)* = X*P* + P*X* ={X*=X}= XP + PX = (PX + XP)

3. Oct 8, 2006

oo thanks :)

4. Oct 8, 2006

### holden

think i got it - but how about the XPX combination? can you do the same technique of moving from the bra to the ket and vice versa, since the individual operators are hermitian?

5. Oct 8, 2006

### Ahmes

As you know (AB)* = B*A*, so the same reversal of the order will apply to three operators so (XPX)*=X*P*X*=XPX.
If you want a more rigorous proof:
(ABC)*=C*(AB)*=C*(B*A*)=C*B*A*

6. Oct 8, 2006

### holden

cool, that's what i was thinking too. thanks so much again.

7. Oct 9, 2006

### dextercioby

This is nonsense, since the X & P operators are unbounded on $\mathcal{H}=L^{2}(\mathbb{R})$.

A proof for the symmetry of the $XP+PX$ operator is quite difficult.

Daniel.

8. Oct 9, 2006

### Ahmes

Can you allaborate?
From what I know this was enough, proofs like this were shown to us in class.
What did you mean by unbounded?

9. Oct 9, 2006

### George Jones

Staff Emeritus
An linear operator A on a Hilbert is unbounded if for any positive real number c, there exists a Hilbert space element psi such that the Hilbert space element (A psi) is at least as large as the length of psi muliplied by c.

It turns out that for any state space, at least one of the operators X and P must be unbounded.

This type of stuff often is not mentioned in North American quantum theory courses because often it does not cause problems. Occasionally, however, it can lead one astray - see this thread that I started.

10. Oct 10, 2006

### Ahmes

Thanks! This thread looks very interesting. I'll read it all when I have time.