Hamiltonian in position basis

Question

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.

Answer 1

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.

Answer 2

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 $H$.

Answer 3

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.

Answer 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 $$ \langle x|P^2|x' \rangle = -\hbar^2 \frac{d^2}{dx^2} $$

Answer 5

Wave functions of position states are Dirac delta functions: $$| x' \rangle \leftrightarrow \varphi_{x'}(\xi) = \delta(\xi - x')$$

If we apply the Hamiltonian to the wave function, we obtain $$\hat{H} \varphi_{x'}(\xi) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial \xi^2} \varphi_{x'}(\xi) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial \xi^2} \delta(\xi - x')$$

Finally, we take the inner product with $\langle x |$ and apply the sifting property of delta functions:

$$\langle x | \hat{H} | x' \rangle = \int \varphi_x^*(\xi) \hat{H} \varphi_{x'}(\xi) \, d \xi = -\frac{\hbar^2}{2 m} \int \delta(\xi - x) \frac{\partial^2}{\partial \xi^2} \delta(\xi - x') \, d \xi = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \delta(x - x')$$

Answer 6

$$<x|H|x'>=<x|-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}|x'>$$ $$=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}<x|x'>$$ $$=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\delta(x-x').$$