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tom.stoer

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What led you to believe that it was?

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samalkhaiat

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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;

[tex]\langle x | \hat{V} | y \rangle = \delta (x-y) V(x)[/tex]

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

[tex]\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)[/tex]

where

[tex]H(x,-i\partial_{x}) = -\frac{\partial^{2}}{\partial x^{2}} + V(x)[/tex]

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

[tex]i\partial_{t} \Psi (x) = \int dy \langle x|\hat{H}|y\rangle \Psi(y)[/tex]

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;

[tex]\langle P|\hat{H}|\bar{P}\rangle = P^{2} \delta(P - \bar{P}) + V(P - \bar{P})[/tex]

where

[tex]V(p) = \int dx V(x) e^{iPx}[/tex]

and

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

regards

sam

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

[tex]\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)[/tex]

where

[tex]H(x,-i\partial_{x}) = -\frac{\partial^{2}}{\partial x^{2}} + V(x)[/tex]

[tex]\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)[/tex]

is true mathematically?

Thanks again.

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samalkhaiat

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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

[tex]\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)[/tex]

is true mathematically?

Thanks again.

In the X-representation, and for an arbitrary ket [itex]|\Psi\rangle[/itex], we have;

[tex]\langle x|\hat{X}|\Psi \rangle = x \langle x| \Psi \rangle \equiv x\Psi(x)[/tex]

[tex]\langle x|\hat{P}|\Psi\rangle = -i \partial_{x}\langle x|\Psi\rangle \equiv\ –i \partial_{x}\Psi(x)[/tex]

Now letting [itex]|\Psi\rangle[/itex] be the position eigenket [itex]|y\rangle[/itex], we get

[tex]\langle x |\hat{X}| y \rangle =x \langle x |y \rangle = x \delta (x-y)[/tex]

[tex]\langle x |\hat{P}| y \rangle = -i \partial_{x}\delta(x-y)[/tex]

Next, write

[tex]

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

[/tex]

and use the above relations to obtain

[tex]

\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

[/tex]

Again, let [itex] |\Psi \rangle = |y \rangle [/itex] to get

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

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

[tex] \langle x| \hat{H} |y \rangle = \langle x | \hat{P}^{2}| y\rangle + \langle x | \hat{V}| y \rangle [/tex]

you will get what you wanted.

Regards

sam

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