91影视

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

${\mathit{唯}}_{\mathbf{n}\mathbf{x}\mathbf{,}\mathbf{n}\mathbf{y}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\mathit{A}\mathbf{sin}\left(\frac{n_x,\mathrm{蟺虫}}{L_x}\right)\mathbf{sin}\left(\frac{n_y,蟺y}{L_y}\right)$ 鈥� (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,

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

(a)

The idea here is that maximizing the probability density ${\left|唯(x,y)\right|}^2$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,

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

And the probability density has the form,

${\left|{唯}_{\left\{1,1\right\}}(x,y\right|}^2=A^2{\sin }^2\left(\frac{\mathrm{蟺虫}}{L_x}\right){\sin }^2\left(\frac{蟺y}{L_y}\right)$ 鈥� (3)

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

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

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" ${\left|唯(x,y)\right|}^2$can be represented as follows, in which the blackest part is at the center

(b)

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

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

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

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

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,

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

And the probability density has the form,

${\left|{唯}_{\left\{2,1\right\}}(x,y)\right|}^2=A^2{\sin }^2\left(\frac{2\mathrm{蟺虫}}{L_x}\right){\sin }^2\left(\frac{\mathrm{蟺测}}{L_y}\right)$ 鈥� (4)

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

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

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