# Is the hamiltonian in the coordinate representation always diagonal?

1. May 17, 2010

### zedya

In the coordinate representation of a quantum mechanical system, is it always true that the Hamiltonian of the system is diagonal? If so, can someone explain to me why this is true?

2. May 17, 2010

### tom.stoer

No, the Hamiltonian of a qm system need not be diagonal in the coordinate rep. The Hamiltonian of a free particle is diagonal in p, not in x. The 1-dim. harmonic oscillator is neither diagonal in x nor in p.

3. May 17, 2010

### IcedEcliptic

What led you to believe that it was?

4. May 17, 2010

### zedya

I think someone said something of that sort a while back... and I made a note of it. It's been confusing me for several days. Because I was pretty sure that it made no sense. I wanted to check basically. Thanks for your replies.

5. May 19, 2010

### samalkhaiat

Let me explain to you what people mean when they say that.
In the X-representation, a potential is said to be local if it can be described by the following diagonal matrix;

$$\langle x | \hat{V} | y \rangle = \delta (x-y) V(x)$$

So, when the potential is local, the Hamiltonian is diagonal in the sense that it can be written as

$$\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)$$

where

$$H(x,-i\partial_{x}) = -\frac{\partial^{2}}{\partial x^{2}} + V(x)$$

This diagonal form of H is the one and only reason why Schrodinger equation is a differential equation in the X-representation;

$$i\partial_{t} \Psi (x) = \int dy \langle x|\hat{H}|y\rangle \Psi(y)$$

When V is not local, the Hamiltonian will not have a diagonal form and Schrodinger equation becomes an integral equation.

This also happens in the P-representation where the Hamiltonian has the following non-diagonal form;

$$\langle P|\hat{H}|\bar{P}\rangle = P^{2} \delta(P - \bar{P}) + V(P - \bar{P})$$

where

$$V(p) = \int dx V(x) e^{iPx}$$

and

$$i\partial_{t}\Psi(x) = P^{2} \Psi(x) + \int d \bar{P} V(P - \bar{P}) \Psi(\bar{P})$$

regards

sam

Last edited: May 19, 2010
6. May 24, 2010

### zedya

So, one more question (thanks a lot for this reply by the way, beautiful explanation)... how would you prove the above? I always get confused with notation stuff... how would you prove that

$$\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)$$

is true mathematically?

Thanks again.

7. May 26, 2010

### samalkhaiat

Last edited: May 27, 2010