Is the hamiltonian in the coordinate representation always diagonal?

[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?

 

[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.

 

[IcedEcliptic]

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?

What led you to believe that it was?

 

[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.

 

[samalkhaiat]

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?

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

 

[zedya]

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

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

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.

 

[samalkhaiat]

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.

In the X-representation, and for an arbitrary ket $|\Psi\rangle$, we have;

$$\langle x|\hat{X}|\Psi \rangle = x \langle x| \Psi \rangle \equiv x\Psi(x)$$

$$\langle x|\hat{P}|\Psi\rangle = -i \partial_{x}\langle x|\Psi\rangle \equiv\ –i \partial_{x}\Psi(x)$$

Now letting $|\Psi\rangle$ be the position eigenket $|y\rangle$, we get

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

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

Next, write

$$\langle x |\hat{P}^{2} | \Psi \rangle = \int \ dy \ \langle x |\hat{P} | y \rangle \langle y | \hat{P} | \Psi \rangle$$

and use the above relations to obtain

$$\langle x |\hat{P}^{2} | \Psi \rangle = (-i)^{2} \partial_{x} \int \ dy \delta(x-y) \partial_{y} \langle y | \psi \rangle = - \frac{\partial^{2}}{\partial x^{2}} \langle x | \Psi \rangle$$

Again, let $|\Psi \rangle = |y \rangle$ to get

$$\langle x | \hat{P}^{2} | y \rangle = -i \frac{\partial^{2}}{\partial x^{2}} \delta (x-y)$$

Now if you put the above in the Hamiltonian matrix ( 2m = 1),

$$\langle x| \hat{H} |y \rangle = \langle x | \hat{P}^{2}| y\rangle + \langle x | \hat{V}| y \rangle$$

you will get what you wanted.

Regards

sam