Question

Consider a 2D infinite well whose sides are of unequal length.

(a) Sketch the probability density as density of shading for the ground state.

(b) There are two likely choices for the next lowest energy. Sketch the probability density and explain how you know that this must be the next lowest energy. (Focus on the qualitative idea, avoiding unnecessary reference to calculations.)

Short answer

(a) The probability density for the ground state is,

(b) The probability density for the next lowest energy is,

Step-by-step solution

Define wave function and its energy.

The wavefunction for a particle trapped in a 2D-infinite well with unequal side lengths is,

${\mathit{\Psi }}_{\mathbf{n}\mathbf{x}\mathbf{,}\mathbf{n}\mathbf{y}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\mathit{A}\mathbf{sin}\left(\frac{n_x,\mathrm{\pi x}}{L_x}\right)\mathbf{sin}\left(\frac{n_y,\pi y}{L_y}\right)$ … (1)

Where n_x, n_y are integers and L_x, L_y are the length of the well in x and y directions, and m is the mass of the particle.

The energy of the wavefunction is given by,

${\mathit{E}}_{\mathbf{n}\mathbf{x}\mathbf{,}\mathbf{n}\mathbf{y}}\mathbf{=}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$ … (2)

The probability density of the shading for the ground state.

(a)

The idea here is that maximizing the probability density ${\left|\Psi (x,y)\right|}^2$gives us the position where the particle is most likely found. Accordingly, we can sketch the probability density.

For the ground state $(n_x,n_y)=(1,1)$, equation 1 becomes,

${\Psi }_{\left\{1,1\right\}}(x,y)=A\sin \left(\frac{\mathrm{\pi x}}{L_x}\right)\sin \left(\frac{\pi y}{L_y}\right)$

And the probability density has the form,

${\left|{\Psi }_{\left\{1,1\right\}}(x,y\right|}^2=A^2{\sin }^2\left(\frac{\mathrm{\pi x}}{L_x}\right){\sin }^2\left(\frac{\pi y}{L_y}\right)$ … (3)

The maximum of Equation 3 corresponds to the squared sines having their maximum value, which is.

localid="1662633062579" $$\begin{gathered} {\sin }^2\left(\frac{\mathrm{\pi x}}{L_x}\right)=1 \\ \frac{\mathrm{\pi x}}{L_x}=\frac{\mathrm{\pi }}{2} \\ x=\frac{L_x}{2} \\ {\sin }^2\left(\frac{\pi y}{L_y}\right)=1 \\ \frac{\pi y}{L_y}=\frac{\mathrm{\pi }}{2} \\ y=\frac{L_y}{2} \end{gathered}$$

Therefore, the particle is most likely found at localid="1662633067532" $\left(\frac{L_x}{2},\frac{L_y}{2}\right)$for the ground state (1,1). Correspondingly, localid="1662633071282" ${\left|\Psi (x,y)\right|}^2$can be represented as follows, in which the blackest part is at the center

The probability density of the shading for the next lowest level.

(b)

According to equation 2, the energy of the ground state (1,1) is,

$E_{\left(1,1\right)}=\left(\frac{1}{L_x^2}+\frac{1}{L_y^2}\right)\frac{{\sout{h}}^2{\mathrm{\pi }}^2}{2m}$

Then, the next lowest energy can be either (1,2) or (2,1) with the following energies,

$$\begin{gathered} E_{\left(1,2\right)}=\left(\frac{1}{L_x^2}+\frac{4}{L_y^2}\right)\frac{{\sout{h}}^2{\mathrm{\pi }}^2}{2m} \\ {E}_{\left(2,1\right)}=\left(\frac{4}{L_x^2}+\frac{1}{L_y^2}\right)\frac{{\sout{h}}^2{\mathrm{\pi }}^2}{2m} \end{gathered}$$

The energy choice is based on which is greater $L_x\mathrm{or}{L}_y$. Assume $L_x$is greater than $L_y$, then $\frac{1}{L_x^2}$is less than $\frac{1}{L_x^2}$.

Therefore, role="math" localid="1662631800237" $E_{(2,1)}$is less than role="math" localid="1662631810378" $E_{(1,2)}$, and the next lowest energy state is (2,1).

For this state, equation 1 becomes,

${\Psi }_{\left\{2,1\right\}}(x,y)=A\sin \left(\frac{2\mathrm{\pi x}}{L_x}\right)\sin \left(\frac{\mathrm{\pi }x}{L_y}\right)$

And the probability density has the form,

${\left|{\Psi }_{\left\{2,1\right\}}(x,y)\right|}^2=A^2{\sin }^2\left(\frac{2\mathrm{\pi x}}{L_x}\right){\sin }^2\left(\frac{\mathrm{\pi y}}{L_y}\right)$ … (4)

The maximum of Equation 4 corresponds to the squared sines having their maximum value, which is 1.

$$\begin{gathered} {\sin }^2\left(\frac{2\mathrm{\pi x}}{L_x}\right)=1 \\ \frac{2\mathrm{\pi x}}{L_x}=\frac{\mathrm{\pi }}{2},\frac{3L_x}{2} \\ x=\frac{L_x}{4},\frac{3L_x}{4} \\ {\sin }^2\left(\frac{\mathrm{\pi y}}{L_y}\right)=1 \\ \frac{\mathrm{\pi y}}{L_y}=\frac{\mathrm{\pi }}{2} \\ y=\frac{L_x}{2} \end{gathered}$$

Therefore, the particle is most likely found at $\left(\frac{L_x}{4},\frac{L_y}{2}\right)$and $\left(\frac{3L_x}{4},\frac{L_y}{2}\right)$for the state .

Correspondingly, ${\left|\Psi (x,y)\right|}^2$can be represented as follows: