Chapter 3, Problem 3.42P

(a)

To determine

The expectation value of $\langle x\rangle ,\langle x^2\rangle ,\langle p\rangle ,\langle p^2\rangle$ in the state $|\alpha \rangle$.

Answer to Problem 3.42P

The expectation value of $\langle x\rangle$ is $\underset{\_}{\sqrt{\frac{\hslash }{2m\omega }}\left(\alpha +{\alpha }^{*}\right)}$, $\langle x^2\rangle$ is $\underset{\_}{\frac{\hslash }{m\omega }\left[1+{\left(\alpha +{\alpha }^{*}\right)}^2\right]}$, $\langle p\rangle$ is $\underset{\_}{-i\sqrt{\frac{\hslash m\omega }{2}}\left(\alpha -{\alpha }^{*}\right)}$  and $\langle p^2\rangle$ is $\underset{\_}{\frac{\hslash m\omega }{2}\left[1-{\left(\alpha -{\alpha }^{*}\right)}^2\right]}$ in the state $|\alpha \rangle$.

Explanation of Solution

Write the expression for the expectation value of the position.

    $\begin{array}{c}\langle x\rangle =\sqrt{\frac{\hslash }{2m\omega }}\langle \alpha |\left(a_{+}+a_{-}\right)\alpha \rangle \\ =\sqrt{\frac{\hslash }{2m\omega }}\langle a_{-}\alpha |\alpha \rangle +\langle \alpha |a_{-\alpha }\rangle \\ =\sqrt{\frac{\hslash }{2m\omega }}\left(\alpha +{\alpha }^{*}\right)\end{array}$        (I)

Here, $\hslash$ is the reduced Planck’s constant, $m$ is the mass of the oscillator, $\omega$ is the angular frequency, $\alpha$ is the state.

Write the expression for the $\langle x^2\rangle$.

    $\begin{array}{c}\langle x^2\rangle =\frac{\hslash }{2m\omega }\langle \alpha |\left(a_{+}^2+2a_{+}{a}_{-}+1+a_{-}^2\right)\alpha \rangle \\ =\frac{\hslash }{2m\omega }\left(\langle a_{-}^2\alpha |\alpha \rangle +2\langle a_{-}\alpha |a_{-}\alpha \rangle +\langle \alpha |\alpha \rangle +\langle \alpha |a_{-}^2\alpha \rangle \right)\\ =\frac{\hslash }{2m\omega }\left[{\left({\alpha }^{*}\right)}^2+2\left({\alpha }^{*}\right)\alpha +1+{\alpha }^2\right]\\ =\frac{\hslash }{2m\omega }\left[1+{\left(\alpha +{\alpha }^{*}\right)}^2\right]\end{array}$        (II)

Write the expression for the expectation value of momentum.

    $\begin{array}{c}\langle p\rangle =i\sqrt{\frac{\hslash m\omega }{2}}\langle \alpha |\left(a_{+}-a_{-}\right)\alpha \rangle \\ =i\sqrt{\frac{\hslash m\omega }{2}}\left(\langle a_{-}\alpha |\alpha \rangle -\langle \alpha |a_{-}\alpha \rangle \right)\\ =-i\sqrt{\frac{\hslash m\omega }{2}}\left(\alpha -{\alpha }^{*}\right)\end{array}$        (III)

Write the expectation value of $p^2$.

    $\begin{array}{c}\langle p^2\rangle =-\frac{\hslash m\omega }{2}\langle \alpha |\left(a_{+}^2-2a_{+}{a}_{-}-1+a_{-}^2\right)\alpha \rangle \\ =-\frac{\hslash m\omega }{2}\left(\langle a_{-}^2\alpha |\alpha \rangle -2\langle a_{-}\alpha |a_{-}\alpha \rangle -\langle \alpha |\alpha \rangle +\langle \alpha |a_{-}^2\alpha \rangle \right)\\ =-\frac{\hslash m\omega }{2}\left[{\left(a^{*}\right)}^2-2\left(a^{*}\right)\alpha -1+{\alpha }^2\right]\\ =\frac{\hslash m\omega }{2}\left[1-{\left(\alpha -{\alpha }^{*}\right)}^2\right]\end{array}$        (IV)

Conclusion:

Therefore, the expectation value of $\langle x\rangle$ is $\underset{\_}{\sqrt{\frac{\hslash }{2m\omega }}\left(\alpha +{\alpha }^{*}\right)}$, $\langle x^2\rangle$ is $\underset{\_}{\frac{\hslash }{m\omega }\left[1+{\left(\alpha +{\alpha }^{*}\right)}^2\right]}$, $\langle p\rangle$ is $\underset{\_}{-i\sqrt{\frac{\hslash m\omega }{2}}\left(\alpha -{\alpha }^{*}\right)}$  and $\langle p^2\rangle$ is $\underset{\_}{\frac{\hslash m\omega }{2}\left[1-{\left(\alpha -{\alpha }^{*}\right)}^2\right]}$ in the state $|\alpha \rangle$.

(b)

To determine

The value of ${\sigma }_x$ and ${\sigma }_p$ and show that that ${\sigma }_x{\sigma }_p=\hslash /2$.

Answer to Problem 3.42P

The value of ${\sigma }_x$ is $\underset{\_}{\sqrt{\frac{\hslash }{2m\omega }}}$ and ${\sigma }_p$ is $\underset{\_}{\sqrt{\frac{\hslash m\omega }{2}}}$ and it is showed that that ${\sigma }_x{\sigma }_p=\hslash /2$.

Explanation of Solution

Write the expression for the ${\sigma }_x^2$.

    ${\sigma }_x^2=\langle x^2\rangle -{\langle x\rangle }^2$        (V)

Use equation (I) and (II) to solve for ${\sigma }_x$.

    $\begin{array}{c}{\sigma }_x^2=\frac{\hslash }{2m\omega }\left[1+{\left(\alpha +{\alpha }^{*}\right)}^2-{\left(\alpha +{\alpha }^{*}\right)}^2\right]\\ =\frac{\hslash }{2m\omega }\\ {\sigma }_x=\sqrt{\frac{\hslash }{2m\omega }}\end{array}$        (VI)

Write the expression for ${\sigma }_p^2$.

    ${\sigma }_p^2=\langle p^2\rangle -{\langle p\rangle }^2$        (VII)

Use equation (III) and (IV) to solve for ${\sigma }_p$.

    $\begin{array}{c}{\sigma }_p^2=\frac{\hslash m\omega }{2}\left[1-{\left(\alpha -{\alpha }^{*}\right)}^2+{\left(\alpha -{\alpha }^{*}\right)}^2\right]\\ =\frac{\hslash m\omega }{2}\\ {\sigma }_p=\sqrt{\frac{\hslash m\omega }{2}}\end{array}$        (VIII)

Use equation (VII) and (VIII) to find ${\sigma }_x{\sigma }_p$.

    $\begin{array}{c}{\sigma }_x{\sigma }_p=\sqrt{\frac{\hslash }{2m\omega }}\sqrt{\frac{\hslash m\omega }{2}}\\ =\frac{\hslash }{2}\end{array}$        (IX)

Conclusion:

Therefore, the value of ${\sigma }_x$ is $\underset{\_}{\sqrt{\frac{\hslash }{2m\omega }}}$ and ${\sigma }_p$ is $\underset{\_}{\sqrt{\frac{\hslash m\omega }{2}}}$ and it is showed that that ${\sigma }_x{\sigma }_p=\hslash /2$.

(c)

To determine

Show that the expansion coefficients are $c_n=\frac{{\alpha }^n}{\sqrt{n!}}{c}_0$.

Answer to Problem 3.42P

It is showed that the expansion coefficients are $c_n=\frac{{\alpha }^n}{\sqrt{n!}}{c}_0$.

Explanation of Solution

Write the expression for the $c_n$.

    $\begin{array}{c}{c}_n=\langle {\psi }_n|\alpha \rangle \\ =\frac{1}{\sqrt{n!}}\langle {\left(a_{+}\right)}^n{\psi }_0|\alpha \rangle \\ =\frac{1}{\sqrt{n!}}{\alpha }^n\langle {\psi }_0|\alpha \rangle \\ =\frac{{\alpha }^n}{\sqrt{n!}}{c}_0\end{array}$        (X)

Conclusion:

Therefore, it is showed that the expansion coefficients are $c_n=\frac{{\alpha }^n}{\sqrt{n!}}{c}_0$.

(d)

To determine

The value of $c_0$ by normalizing $|\alpha \rangle$.

Answer to Problem 3.42P

The value of $c_0$ by normalizing $|\alpha \rangle$ is $\underset{\_}{e^{-{\left|\alpha \right|}^2/2}}$.

Explanation of Solution

Write the expression for the normalization of $|\alpha \rangle$.

    $\begin{array}{c}1={\displaystyle \sum _{n=0}^{\infty }{\left|c_n\right|}^2}\\ ={\left|c_0\right|}^2{\displaystyle \sum _{n=0}^{\infty }\frac{{\left|\alpha \right|}^{2n}}{n!}}\\ ={\left|c_0\right|}^2{e}^{{\left|\alpha \right|}^2}\\ c_0=e^{-{\left|\alpha \right|}^2/2}\end{array}$        (XI)

Conclusion:

Therefore, the value of $c_0$ by normalizing $|\alpha \rangle$ is $\underset{\_}{e^{-{\left|\alpha \right|}^2/2}}$.

(e)

To determine

Show that $|\alpha \left(t\right)\rangle$ remains an eigenstate of $a_{-}$.

Answer to Problem 3.42P

It is showed that $|\alpha \left(t\right)\rangle$ remains an eigenstate of $a_{-}$.

Explanation of Solution

Write the expression for $|\alpha \left(t\right)\rangle$.

    $\begin{array}{c}|\alpha \left(t\right)\rangle ={\displaystyle \sum _{n=0}^{\infty }{c}_n{e}^{-i{E}_{n}t/\hslash }|n\rangle }\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{{\alpha }^n}{\sqrt{n!}}}{e}^{-{\left|\alpha \right|}^2/2}{e}^{-i\left(n+\frac{1}{2}\right)\omega t}|n\rangle \\ =e^{-i\omega t/2}{\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\alpha e^{-i\omega t}\right)}^n}{\sqrt{n!}}}{e}^{-{\left|\alpha \right|}^2/2}|n\rangle \end{array}$        (XII)

Apart from the overall phase factor $e^{-i\omega t/2}$ (which doesn’t affect its status as an eigenfunction of ($a_{-}$, or its eigenvalue), $|\alpha \left(t\right)\rangle$ is the same as $|\alpha \rangle$ , but with eigenvalue $\alpha \left(t\right)=e^{-i\omega t}\alpha$.

Conclusion:

Therefore, it is showed that $|\alpha \left(t\right)\rangle$ remains an eigenstate of $a_{-}$.

(f)

To determine

The value of $\langle x\rangle$ and ${\sigma }_x$ as a function of time.

Answer to Problem 3.42P

The value of $\langle x\rangle$ is $\underset{\_}{C\cos \left(\omega t-\varphi \right)}$ and ${\sigma }_x$ is $\underset{\_}{\sqrt{\frac{\hslash }{2m\omega }}}$ as a function of time.

Explanation of Solution

It is given that the value of $\alpha \left(t\right)=e^{-i\omega t}\alpha$.

Use equation (I) to solve for the value of $\langle x\rangle$ as a function of time.

    $\begin{array}{c}\langle x\rangle =\sqrt{\frac{\hslash }{2m\omega }}\left(\alpha \left(t\right)+{\alpha }^{*}\left(t\right)\right)\\ =\sqrt{\frac{\hslash }{2m\omega }}\left(\alpha e^{-i\omega t}+{\alpha }^{*}{e}^{i\omega t}\right)\\ =\sqrt{\frac{\hslash }{2m\omega }}\left(C\sqrt{\frac{m\omega }{2\hslash }}{e}^{i\varphi }{e}^{-i\omega t}+C\sqrt{\frac{m\omega }{2\hslash }}{e}^{-i\varphi }{e}^{i\omega t}\right)\\ =\frac{1}{2}C\left(e^{-i\left(\omega t-\varphi \right)}+e^{i\left(\omega t-\varphi \right)}\right)\\ =C\cos \left(\omega t-\varphi \right)\end{array}$        (XIII)

Conclusion:

Therefore, the value of $\langle x\rangle$ is $\underset{\_}{C\cos \left(\omega t-\varphi \right)}$ and ${\sigma }_x$ is $\underset{\_}{\sqrt{\frac{\hslash }{2m\omega }}}$ as a function of time.

(g)

To determine

Whether the ground state $\left(|n=0\rangle \right)$ itself is a coherent state.

Answer to Problem 3.42P

Yes, the ground state $\left(|n=0\rangle \right)$ itself is a coherent state.

Explanation of Solution

Write the expression for the given value of $a_{-}|{\psi }_0\rangle$.

    $a_{-}|{\psi }_0\rangle =0$        (XIV)

From equation (XIV), it is known that the ground state $\left(|n=0\rangle \right)$ is a coherent state with eigenvalue $\alpha =0$.

Conclusion:

Therefore,  the ground state $\left(|n=0\rangle \right)$ itself is a coherent state.