# Compute the energy for 2D wave function with discontinuous derivatives

#### entanglement witness

Problem Statement
Hi,

I am given the following set of trial wave functions depending on some parameter alpha.

$$\psi(x,y) = \begin{cases} A \left(1 - |xy|/a^2\right) e^{- \alpha} & |x| \leq a,\, |y| \leq a \\ A \left(1 - |x|/a\right) e^{- \alpha |y| / a} & |x| \leq a,\, |y| > a \\ A \left(1 - |y|/a\right) e^{- \alpha |x| / a} & |x| > a,\, |y| \leq a \\ 0 & \text{else} \\ \end{cases}$$

where a > 0 and asked to show that the expectation value of the free Hamiltonian reads:

$$<\psi | H | \psi> = \frac{3 \hbar^2}{m a^{2}} \left( \frac{a^{2} + 2 \alpha + 3}{6 + 11 \alpha} \right)$$
Relevant Equations
$$H = - \frac{\hbar^{2}}{2m} \left( \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} \right)$$
I have calculated the normalization constant, but I'm struggling with the discontinuities in the derivatives of the wave function. Due to the symmetry, it should suffice to consider the first two cases. The results should be (according to WolframAlpha):

$$\left( \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} \right) \left(1 - |xy|/a^2\right) e^{- \alpha} = -2 e^{-\alpha} (x^2 + y^2) \delta(x y)/a^2$$

and

$$\left( \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} \right) \left(1 - |x|/a\right) e^{- \alpha |y| / a} = (1 - |x|/a) \left( \frac{ \alpha^2 e^{-\alpha |y|/a}}{a^2} - \frac{2 \alpha \delta(y)}{a}\right) - \frac{2 \delta(x) e^{-\alpha |y|/a}}{a}$$

Then, again for symmetry reasons, I assume that I need to compute three integrals, namely:

$$-2 \int_{-a}^{a} \mathrm{d}x \int_{-a}^{a} \mathrm{d}y \left(1 - |xy|/a^2\right) e^{- 2 \alpha} x^2 \delta(xy)$$

(first equation, term with y^2 should give the same result)

$$\int_{0}^{a} \mathrm{d}x \int_{a}^{\infty} \mathrm{d}y \left(1 - x/a\right)^{2} e^{- 2 \alpha y / a} \frac{\alpha^{2}}{a^{2}}$$

(first term in the second equation)

$$- \frac{2}{a} \int_{-a}^{a} \mathrm{d}x \int_{a}^{\infty} \mathrm{d}y \left(1 - |x|/a\right) e^{- 2 \alpha y / a} \delta(x)$$

(third term in the second equation, the second should not contribute)

Could someone check if those integrals (including the limits) are correct? I get the wrong result when adding them up (assuming that the first integral contributes twice, the second eight times and the third four times)... I would just like to know if I'm completely on the wrong track or just made a mistake in the calculation.

Thanks a lot!

Related Advanced Physics Homework News on Phys.org

#### TSny

Homework Helper
Gold Member
$\nabla^2 \psi(x,y)$ produces four cases of delta-function expressions that will occur along certain lines in the x-y plane. These lines are indicated by the colored lines in the figure below:

By symmetry, each of the different colors contributes either 4 or 8 times. You should be able to see that the delta-function that occurs along the red line will not contribute when calculating $\iint \psi \nabla^2 \psi \,dx\,dy$. So, that leaves the three cases: blue, green, and orange. You can find the form of the delta function expression for each of these cases by looking at the discontinuity in either $\frac{\partial \psi}{\partial x}$ or $\frac{\partial \psi}{\partial y}$ when crossing perpendicularly one of the colored lines.

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