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Let $ H = \frac{-h^2}{2m}\frac{\partial^2 }{\partial x^2}$. I want to find the matrix elements of $H$ in position basis. It is written like this:

$\langle x \mid H \mid x' \rangle = \frac{-h^2}{2m}\frac{\partial^2}{\partial x^2} \delta(x -x')$.

How do we get this? are we allowed to do $\langle x | \frac{\partial^2}{\partial x^2} \mid x' \rangle = \frac{\partial^2}{\partial x^2} \langle x \mid x' \rangle$? Why? It seems some thing similar is done above.

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That's exactly what happens. Except since it's a continuous variable, the delta function appears as $\delta (x - x`)$ and not $\delta_{x,x`}$ –  Kitchi Nov 25 '12 at 11:17
@Kitchi But how can we take the differential operator out? –  DurgaDatta Nov 25 '12 at 12:29
The partial derivative is taken with $x$, but you're acting it on $x`$, which is independent of $x$. So you can take it out, and the result will be your answer. –  Kitchi Nov 25 '12 at 12:36
@Kitchi How would we argue if the LHS was complex conjugate i.e <x'|H|x>, of the original one? –  DurgaDatta Nov 25 '12 at 12:52

4 Answers 4

Indeed, the operator H is the sum of the kinetic energy operator and the potential energy operator. The kinetic energy operator is $$ T =\frac{P^2}{2m} $$ other part, the operator $P$ in the x basis is $$ \langle x|P^2|x' \rangle = -\hbar^2 \frac{d^2}{dx^2} $$

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+1. If you do not already know the Hamlitonian's representation in the position basis, this is the way to do it. Break down the kinetic energy operator into momentum, and then, if you must, break down the position eigenstates into the momentum basis. When you have nothing left but $\langle x | p \rangle$ and such terms, you do have to know that, at least, but they're Fourier duals, so you should know that by now. –  Muphrid Dec 26 '12 at 20:07

You're given $$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$$ This is an operator, so it acts on functions of x $$H\psi(x) = -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}$$

The LHS is just the inner product of $\langle x|$ with the new state $H|\psi \rangle$, and on the RHS, $\psi(x)$ is just the inner product of $\langle x|$ with $|\psi \rangle$, so $$ \langle x|H|\psi \rangle = -\frac{\hbar^2}{2m}\frac{d^2\langle x|\psi \rangle}{dx^2}$$ Subsitute the position eigenstate $|x' \rangle$ for $|\psi \rangle$ and the result follows.

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If the matrix elements of your Hamiltonian are given by $ \langle x|H|x`\rangle $.

This can be written as $\langle x|Hx`\rangle$, so if you take a complex conjugate, you end up with $\langle x`H|x\rangle$, since $H$ is hermitian. Therefore, the expression you have obtained is valid in both cases.

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Warning: I'm still not 100% convinced that this is the right method. If I can confirm that this is right/wrong, I'll post back here. Till then, please treat this with caution. –  Kitchi Nov 25 '12 at 14:55

The key here is that you can think of $\frac{\partial^2}{\partial x^2}$ as an operator acting on the $x$ variable - which, happily, does not occur in $| x^\prime \rangle$. Hence it is possible to move the operator out of the bracket (otherwise, there would be a pesky multiplication rule kicking in).

A different approach would be to calculate the eigenvalues of $H$ in the position basis rather than the matrix elements, since $H$ is diagonal in this basis, this works and the two results are equal. We then have:

$$\langle x | H | x^\prime \rangle = E_x \Psi(x) \delta(x-x^\prime) = E_x \langle x | x^\prime \rangle = H \langle x | x^\prime \rangle$$ where:

  • the first equality is due to the diagonality of $H$ in this basis.
  • the second equality is true as we can represent the wave function of a particle by $\Psi(r) = \langle r | x \rangle$.
  • the third equality is the eigenvalue equation for $H$.
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H is diagonal in the basis of energy eigenstates. How do we know that H is also diagonal in position basis? –  DurgaDatta Nov 25 '12 at 12:31
Good point, sorry, I was under the assumption that $H$, as it only depends on $\partial_x^2$, was also automatically diagonal in the position eigenbases (and the fact that a $\delta(x-x^\prime)$ appears supports said assumption). However, I currently don't have a conclusive argument as to why that would be necessarily true. –  Claudius Nov 25 '12 at 12:37

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