Homework 9

Problems 6.2, 6.5, 6.6, 6.7, 6.11, 6.17

6.2

$$-\frac{{\hslash }^2}{2m}\left(\frac{d^2}{d{x}^2}+\frac{d^2}{d{y}^2}+\frac{d^2}{d{z}^2}\right)\psi (x,y,z)+\frac{1}{2}m{\omega }^2(x^2+y^2+z^2)=E\psi (x,y,z)$$

assuming eigenfunctions ${\psi }_n(x)$ with one-dimensional energy eigenvalues $E_n=(n+\frac{1}{2})\hslash \omega$ First excited state is given by $E_1=(\frac{3}{2})\hslash \omega$. The Hamiltonian is given by:

$$H=\frac{{\hat{p}}^2}{2m}+\frac{1}{2}m{\omega }^2(x^2+y^2+z^2)$$

Expanding this, we get:

$$H=\frac{{\hat{p}}^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2+\frac{1}{2}m{\omega }^2{y}^2+\frac{1}{2}m{\omega }^2{z}^2$$

Separating the Momentum Operator into its 1-D components, we get:

$$H=\frac{{\widehat{p_x}}^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2+\frac{{\widehat{p_y}}^2}{2m}+\frac{1}{2}m{\omega }^2{y}^2+\frac{{\widehat{p_z}}^2}{2m}+\frac{1}{2}m{\omega }^2{z}^2$$

And if we take

$$H_n=\frac{{\hat{p}}_n^2}{2m}+\frac{1}{2}m{\omega }^2{n}^2$$

for any variable $n$, then we can rewrite the Hamiltonian as the following:

$$H=H_x+H_y+H_z$$

which means that the energy level is the sum of

$$(n_x+n_y+n_z)$$

Plugging in the energy level into the energy equation we get:

$$E_n=(n_x+\frac{1}{2})\hslash {\omega }_x+(n_y+\frac{1}{2})\hslash {\omega }_y+(n_z+\frac{1}{2})\hslash {\omega }_z$$ $$E_n=\left(n_x+n_y+n_z+\frac{3}{2}\right)\hslash \omega$$

Looking at this function, there is a 6-fold degeneracy, because for x,y and z, one could be 2 and the rest 0, or there could be two 1’s and one zero. That’s 2 possibilities per coordinate, and there’s three coordinates yielding a 6-fold degeneracy.

6.5

Prove that the operator $L_{zop}=\frac{\hslash }{i}\frac{d}{d\varphi }$ is a Hermitian Operators. Wave function must be single valued.

To determine if an operator is Hermitian, we have to see if $L_{zop}^{†}=L_{zop}$ Functionally, this means proving that

$${\int }_{-\infty }^{\infty }{\phi }^{\ast }(L\Psi )d\varphi ={\int }_{-\infty }^{\infty }(L\phi {)}^{\ast }\Psi d\varphi$$ $${\int }_{-\infty }^{\infty }{\phi }^{\ast }\left(\frac{\hslash }{i}\frac{d}{d\varphi }\Psi \right)d\varphi ={\int }_{-\infty }^{\infty }(\frac{\hslash }{i}\frac{d}{d\varphi }\phi {)}^{\ast }\Psi d\varphi$$ $${\int }_{-\infty }^{\infty }{\phi }^{\ast }\frac{\hslash }{i}\frac{d}{d\varphi }\Psi d\varphi =-{\int }_{-\infty }^{\infty }{\left(\frac{d\varphi }{dx}\right)}^{\ast }\psi d\varphi +\frac{\hslash }{i}{\varphi }^{\ast }\psi {|}_{-\infty }^{\infty }={\int }_{-\infty }^{\infty }(\frac{\hslash }{i}\frac{d}{d\varphi }\phi {)}^{\ast }\Psi d\varphi$$

Therefore, $L_{zop}$ is Hermitian.

6.6

Suppose that

$$\Psi (\varphi )=\sqrt{\frac{1}{3\pi }}+\sqrt{\frac{1}{6\pi }}{e}^{i\varphi }$$

Show that $\Psi (x)$ is properly normalized. What are possible results of a measurement of $L_z$ for a particle with wave function $\Psi (\varphi )$, and what are the probabilities of these results.

$${\int }_0^{2\pi }\Psi (\varphi {)}^{\ast }\Psi (\varphi )d\varphi ={\int }_0^{2\pi }\left(\sqrt{\frac{1}{3\pi }}+\sqrt{\frac{1}{6\pi }}{e}^{-i\varphi }\right)\left(\sqrt{\frac{1}{3\pi }}+\sqrt{\frac{1}{6\pi }}{e}^{i\varphi }\right)d\varphi =$$ $${\int }_0^{2\pi }\frac{2\sqrt{2}\cos (\varphi )}{6\pi }+\frac{3}{6\pi }d\varphi ={\int }_0^{2\pi }\frac{2\sqrt{2}}{6\pi }\cos (\varphi )d\varphi +{\int }_0^{2\pi }\frac{3}{6\pi }d\varphi$$ $$\frac{\sqrt{2}}{3\pi }{\int }_0^{2\pi }\cos (\varphi )d\varphi +{\int }_0^{2\pi }\frac{1}{2\pi }d\varphi =\frac{\sqrt{2}}{3\pi }[\sin \varphi ]{|}_0^{2\pi }+\left[\frac{1}{2\pi }\varphi \right]{|}_0^{2\pi }$$

First term goes to 0, second term yields

$$\frac{1}{2\pi }2\pi -0=1$$

Therefore the wave-function is normalized.

To find the possible measurement results of $L_z$, we can apply the operator to the wave function:

$$L_z=\frac{\hslash }{i}\frac{d}{d\varphi }(\sqrt{\frac{1}{3\pi }}+\sqrt{\frac{1}{6\pi }}{e}^{i\varphi })$$ $$(\hslash \varphi \sqrt{\frac{1}{6\pi }}{e}^{i\varphi })=\alpha (\Psi )$$

There exists no constant $\alpha$ to make this happen, therefore the wave function is not an eigenfunction. This means it has to be a superposition of the normalized eigenfunctions of $L_z$. Looking at the form of the given wave function, we can determine that there are two possible states, either $m_1=0$ or $1$, yielding two wave functions based on the normalized eigenfunction of the $L_z$ operator ${\psi }_m=\frac{1}{\sqrt{2\pi }}{e}^{im\varphi }$

$${\psi }_1=\sqrt{\frac{1}{3}}{e}^{i\varphi }$$

and

$${\psi }_0=\sqrt{\frac{2}{3}}$$

Applying the rotational momentum operator to both of these wave functions yields our momentum eigenvalues:

$$L_z({\psi }_1)=\alpha {\psi }_1\&L_z({\psi }_0)=\alpha {\psi }_0$$

Where $\alpha =\hslash ,0$ respectively. These have respective probability amplitudes given by the square of their coefficients, $\frac{1}{3}$ and $\frac{2}{3}$ respectively.

6.7

Verify that $Y_{2,2}(\theta ,\varphi )=\sqrt{\frac{15}{32\pi }}{\sin }^2\theta e^{2i\varphi }$ is an eigenfunction of ${\mathbf{L}}^2$ and $L_z$ with appropriate eigenvalues

The operator $L_z$ is given by the following: $\frac{\hslash }{i}\frac{d}{d\varphi }$ Finding the eigenvalues:

$$L_z{Y}_{2,2}=\lambda Y_{2,2}$$

where eigenvalues are denoted by $\lambda$

$$2\hslash \sqrt{\frac{15}{32\pi }}{\sin }^2\theta e^{2i\varphi }=\lambda \sqrt{\frac{15}{32\pi }}{\sin }^2\theta e^{2i\varphi }$$

Thus yielding a value of $2\hslash$ for the eigenvalue and proving that this function is an eigenfunction of $L_z$

Now for the ${\mathbf{L}}^2$ operator:

$${\mathbf{L}}^2=-{\hslash }^2\left[\frac{1}{\sin \theta }\frac{d}{d\theta }\left(\sin \theta \frac{d}{d\theta }\right)+\frac{1}{{\sin }^2\theta }\frac{d^2}{d{\varphi }^2}\right]$$ $${\mathbf{L}}^2{Y}_{2,2}=\lambda Y_{2,2}$$ $${\mathbf{L}}^2\sqrt{\frac{15}{32\pi }}{\sin }^2\theta e^{2i\varphi }=\lambda \sqrt{\frac{15}{32\pi }}{\sin }^2\theta e^{2i\varphi }$$ $$-{\hslash }^2\left[\frac{1}{\sin \theta }\frac{d}{d\theta }\left(\sin \theta \frac{d\Theta }{d\theta }\right)-\frac{4}{{\sin }^2\theta }\Theta \right]=\lambda {\hslash }^2\Theta$$

Where

$$Y(\theta ,\varphi )=\Theta (\theta )e^{i{m}_1\varphi }$$

and

$$\Theta (\theta )=\sqrt{\frac{15}{32\pi }}{\sin }^2\theta$$ $$-\hslash \left[\frac{1}{\sin \theta }\frac{d}{d\theta }\left(\sin \theta \frac{\sqrt{15}\cos \left(\theta \right)\sin \left(\theta \right)}{2^{\frac{3}{2}}\sqrt{\pi }}\right)-\frac{4}{{\sin }^2\theta }\sqrt{\frac{15}{32\pi }}{\sin }^2\theta \right]$$ $$-\hslash \left[\frac{1}{\sin \theta }-\frac{\sqrt{15}\sin \left(\theta \right)\left({\sin }^2\left(\theta \right)-2{\cos }^2\left(\theta \right)\right)}{2^{\frac{3}{2}}\sqrt{\pi }}-\frac{4}{{\sin }^2\theta }\sqrt{\frac{15}{32\pi }}{\sin }^2\theta \right]$$

…

$$2(2+1){\hslash }^2\sqrt{\frac{15}{32\pi }}{\sin }^2\theta e^{2i\varphi }$$

For a spherical harmonic $Y_{l,m_1}$ we can assume it satisfies the following:

$${\mathbf{L}}^2{Y}_{l,m_1}(\theta ,\varphi )=l(l+1){\hslash }^2{Y}_{l,m_1}(\theta ,\varphi )$$

Therefore the eigenvalue is given by $6\hslash$ and it is an eigenfunction.

6.11

Normalized angular wave function for rigid rotator given by:

$$\psi (\theta ,\varphi )=\sqrt{\frac{3}{8\pi }}(-\sin \theta \cos \varphi +i\cos \theta )$$

Show that this wave function is an eigenfunction of:

$$L_y=\frac{\hslash }{i}\left(\cos \varphi \frac{d}{d\theta }-\cot \theta \sin \varphi \frac{d}{d\varphi }\right)$$

What is the eigenvalue?

$$\frac{d\psi }{d\theta }=\sqrt{\frac{3}{8\pi }}(-\cos \theta \cos \varphi -i\sin \theta )$$ $$\frac{d\psi }{d\varphi }=\sqrt{\frac{3}{8\pi }}(\sin \theta \sin \varphi )$$ $$L_{yop}\psi =\frac{\hslash }{i}(\cos \varphi )[\sqrt{\frac{3}{8\pi }}(-\cos \theta \cos \varphi -i\sin \theta )]-\cot \theta \sin \varphi \sqrt{\frac{3}{8\pi }}(\sin \theta \sin \varphi )$$ $$⟹\sqrt{\frac{3}{8\pi }}\frac{\hslash }{i}(-\cos \theta -i\sin \theta \cos \varphi )$$ $$⟹\hslash \sqrt{\frac{3}{8\pi }}(i\cos \theta -\sin \theta \cos \varphi )$$ $$L_{yop}\psi =\hslash \sqrt{\frac{3}{8\pi }}\psi$$ $$\therefore \lambda =\hslash \sqrt{\frac{3}{8\pi }}$$

$\psi$ is an eigenfunction with eigenfunctions given by $\lambda$ above.

6.17

The energy for a rigid rotator constrained in the x-y plane is given by

$$E=\frac{L_z^2}{2I}$$

Where the moment of inertia is constant.

a. What is the Hamiltonian? What are the allowed energies and normalized energy eigenfunctions?

$$\hat{H}=\frac{L_z^2}{2I}=\frac{-{\hslash }^2}{2I}\frac{d^2}{d^2\varphi }$$ $$[\hat{H},{\hat{L}}_z]=\frac{1}{2I}[L_z^2,L_z]=\frac{1}{2I}0=0$$

Therefore these two operators commute

$$\hat{H}\psi =E\psi ⟹\frac{d^2\psi }{d{\varphi }^2}+\left(\frac{2IE}{t^2}\right)\psi =0$$ $$⟹\psi (\varphi )=A{e}^{i\omega \varphi }$$

The allowed energies are given by:

$$e^{i{\omega }_2\pi }=1⟹\omega =n⟹{\omega }^2=n^2$$ $$E_n=\frac{n^2{\hslash }^2}{2I}$$

Normalized eigenfunctions are given by

$${\psi }_n(\varphi )=\frac{1}{\sqrt{2\pi }}{e}^{in\varphi }$$

b. What values of $L_z$ would be obtained if a measurement of $L_z$ were carried out on

$$\psi (\varphi )=\sqrt{\frac{4}{3\pi }}{\sin }^2\frac{\varphi }{2}=\sqrt{\frac{4}{3\pi }}\left(\frac{1-\cos \varphi }{2}\right)=\frac{1}{\sqrt{3\pi }}\left[1-\frac{1}{2}{e}^{i\varphi }-\frac{1}{2}{e}^{-i\varphi }\right]$$

What are their probabilities?

Lz	n=0	n=1	n=2
P	1/3	1/6	1/6

c. Time evolution of $\psi$

$$\psi (\varphi ,t)=\sqrt{\frac{4}{3\pi }}\left(\frac{1-\cos \varphi }{2}\right)e^{-it/2I\hslash }$$

d.

$$P(n=1)=\frac{1}{2}$$

e.

$$<E>=<H>={\int }_0^{2\pi }{\psi }^{\ast }\hat{H}\psi d\varphi =-\frac{1}{3\pi }{\int }_0^{2\pi }(1-\cos \varphi e^{it/2I\hslash })\frac{{\hslash }^2}{2I}(1-\cos \varphi e^{-it/2I\hslash })d\varphi$$ $$=-\frac{{\hslash }^2}{6\pi I}(-\pi )=\frac{{\hslash }^2}{6I}$$ $$<E^2>=<H^2>={\left(\frac{\hslash }{2I}\right)}^2\frac{1}{3\pi }\pi =\frac{4\hslash }{12I^2}$$ $$\Delta E\sqrt{<E^2>-<E{>}^2}=\frac{{\hslash }^2\sqrt{2}}{6I}$$