Answer to Question #194195 in Quantum Mechanics for Aman Kumar

Question

2) Consider a 2D infinite potential well with the potential U(x, y) = 0 for 0 ≤ x ≤ a & 0 ≤ y ≤ B, and U(x, y) = ∞, otherwise. Solve the time-independent Schrodinger equation, and find the normalized wavefunction and the corresponding energies.

Answer 1

Let us now consider the Schrödinger Equation for an electron confined to a two dimensional box, $0\leq x\leq a$ and $0\leq y\leq b$ .  That is to say, within this rectangle the electron wavefunction behaves as a free particle $U(x,y)=0$ , but the walls are impenetrable so the wavefunction $\psi(x,y,t)=0$ at the walls

Extending the (time-independent) Schrödinger equation for a one-dimensional system

$\dfrac{−ℏ^2}{2m}\dfrac{d^2ψ(x)}{d x^2}+U(x)ψ(x)=Eψ(x)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$

to a two-dimensional system

$\dfrac{−ℏ^2}{2m}\bigg(\dfrac{∂^2ψ(x,y)}{∂x^2}+\dfrac{∂^2ψ(x,y)}{∂y^2}\bigg)+U(x,y)ψ(x,y)=Eψ(x,y)\space\space\space\space\space\space\space\space\space\space\space\space\space(2)$

Equation 2 can be simplified for the particle in a 2D box since we know that $U(x,y)=0$

is within the box and $U(x,y)=\infin$ is outside the box.

$−\dfrac{ℏ^2}{2m}\bigg(\dfrac{∂^2ψ(x,y)}{∂x^2}+∂^2ψ(x,y)∂y^2\bigg)=Eψ(x,y)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(3)$

Since the Hamiltonian (i.e. left side of Equation 3) is the sum of two terms with independent (separate) variables, we try a product wavefunction like in the Separation of Variables approach used to separate time-dependence from the spatial dependence previously. Within this approach we express the 2-D wavefunction as a product of two independent 1-D components

$ψ(x,y)=X(x)Y(y)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(4)$

This ansatz separates Equation 3 into two independent one-dimensional Schrödinger equations

$−\dfrac{ℏ^2}{2m}\bigg(\dfrac{∂^2X(x)}{∂x^2}\bigg)=ε_xX(x)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(5)$

$−\dfrac{ℏ^2}{2m}\bigg(\dfrac{∂^2Y(y)}{∂y^2}\bigg)=ε_yY(y)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(6)$

where the total energy of the particle is the sum of the energies from each one-dimensional Schrödinger equation

$E=ε_x+ε_y\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(7)$

The differential equations in Equations 5 and 6 are familiar as they were found for the particle in a 1-D box. They have the general solution

$X(x)=A_xsin(kx_x)+B_xcos(kx_x)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(8)$

$Y(y)=A_ysin(ky_y)+Bycos(ky_y)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(9)$

Applying Boundary Conditions

The general solutions in Equations 8 and 9 can be simplified to address boundary conditions dictated by the potential, i.e $ψ(0,y)=0$ and $ψ(x,0)=0 .$ Therefore, $B_x=0$ and $B_y=0.$

Thus, we can combine Equations 8, 9, and 4 to construct the wavefunction $ψ(x,y)$ for a particle in a 2D box of the form

$ψ(x,y)=Nsin\bigg(\sqrt{\dfrac{2mε_x}{ℏ^2}}x\bigg)sin\bigg(\sqrt{\dfrac{2mε_y}{ℏ^2}}y\bigg)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(10)$

We still need to satisfy the remaining boundary conditions $ψ(L,y)=0 ψ(L,y)=0\space and\space ψ(x,L)=0$

$ψ(x,y)=Nsin\bigg(\sqrt{\dfrac{2mε_x}{ℏ^2}}L\bigg)sin\bigg(\sqrt{\dfrac{2mε_y}{ℏ^2}}y\bigg)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(11)$

and

$ψ(x,y)=Nsin\bigg(\sqrt{\dfrac{2mε_x}{ℏ^2}}x\bigg)sin\bigg(\sqrt{\dfrac{2mε_y}{ℏ^2}}L\bigg)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(12)$

Equation 11 can be satisfied if

$sin\bigg(\sqrt\dfrac{2mε_x}{ℏ^2}L\bigg)=0\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(13)$

independent of the value of y, while equation 12 can be satisfied if

$sin\bigg(\sqrt\dfrac{2mε_y}{ℏ^2}L\bigg)=0\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(14)$

independent of the value of x. These are the same conditions that we encountered for the one-dimensional box, hence we already know the sin function in each case can be zero in many places. In fact, these two conditions are satisfied if

$\sqrt{\dfrac{2mε_x}{ℏ^2}}L=n_xπ\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(15)$

and

$\sqrt{\dfrac{2mε_y}{ℏ^2}}L=n_yπ\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(16)$

which yield the allowed values of $ε_x$ and $ε_y$ as

$ε_{n_x}=\dfrac{ℏ^2π^2}{2mL^2}n^2_x\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(17)$

and

$ε_{n_y}=\dfrac{ℏ^2π^2}{2mL^2}n^2_y\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(18)$

We need two different integers $n_x$ and $n_y$ because the conditions are completely independent and can be satisfied by any two different (or similar) values of these integers. The allowed values of the total energy are now given by

$E_{n_x,n_y}=\dfrac{ℏ^2π^2}{2mL^2}(n^2x+n^2y)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(19)$

Once the conditions on $ε_{n_x}$ and $ε_{n_y}$ are substituted into Equation 10, the wavefunctions become

$ψ_{n_x,n_y}(x,y)=Nsin\bigg(\dfrac{n_x\pi x}{L}\bigg)sin\bigg(\dfrac{n_y\pi y}{L}\bigg)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(20)$

The constant N is now determined by the normalization condition

$∫^L_0∫^L_0|ψ_{n_x,n_y}(x,y)|^2dxdy=1$

$N^2∫^L_0sin^2\bigg(\dfrac{n_xπx}{L}\bigg)dx∫^L_0sin^2\bigg(\dfrac{n_yπy}{L}\bigg)dy=1$

$N^2\dfrac{L}{2}⋅\dfrac{L}{2}=1$

$N=\dfrac{2}{L}$

so that the complete normalized 2D wavefunction is

$ψ_{n_x,n_y}(x,y)=\dfrac2Lsin\bigg(\dfrac{n_xπx}{L}\bigg)sin\bigg(\dfrac{n_yπy}{L}\bigg)$