Question

a) Compute $\mathbf{⟨}\mathbf{x}\mathbf{⟩}$, $\mathbf{⟨}\mathbf{p}\mathbf{⟩}$, $\mathbf{⟨}{\mathbf{x}}^{\mathbf{2}}\mathbf{⟩}$, $\mathbf{⟨}{\mathbf{p}}^{\mathbf{2}}\mathbf{⟩}$, for the states ${\mathbf{\psi }}_{\mathbf{0}}$and${\mathbf{\psi }}_{\mathbf{1}}$ , by explicit integration. Comment; In this and other problems involving the harmonic oscillator it simplifies matters if you introduce the variable $\mathbf{\xi }\mathbf{\equiv }\sqrt{\frac{\mathbf{m\omega }}{\mathbf{\hslash }}}\mathbf{x}$and the constant $\mathbf{\alpha }\mathbf{\equiv }{\left(\frac{\mathrm{m\omega }}{\pi \hslash }\right)}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{4}$}\right.}$.

b) Check the uncertainty principle for these states.

c) Compute $\mathbf{⟨}\mathbf{T}\mathbf{⟩}$(the average kinetic energy) and $\mathbf{⟨}\mathbf{V}\mathbf{⟩}$ (the average potential energy) for these states. (No new integration allowed). Is their sum what you would expect?

Short answer

(a) The value of $⟨x⟩$ is $\int \mathrm{x}{\left|\mathrm{\psi }\right|}^2\mathrm{dx}=0$ and the value of $⟨p⟩$ is role="math" localid="1658998623381" $\mathrm{m}\raisebox{1ex}{$\mathrm{d}⟨\mathrm{x}⟩$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.=0$ and The value of ${⟨x⟩}^2$ is $\frac{3\hslash }{2m\omega }$ and${⟨P⟩}^2$ is $\frac{3m\hslash \omega }{2}$.

(b)The uncertainty principal for these states is greater than $\frac{\hslash }{2}$.

(c) The value of the average kinetic energy is role="math" localid="1658998806318" $\left\{{}_{\frac{3}{4}\omega \hslash (n=1)}^{\frac{1}{4}\omega \hslash (n=0)}\right\}$ and The average potential energy is $\left\{{}_{\frac{3}{4}\omega \hslash (n=1)}^{\frac{1}{4}\omega \hslash (n=0)}\right\}$.The sum of these two is same as expected value.

Step-by-step solution

Define explicit integration

In this method, the accelerations and velocities at a specific time point are taken to be constant during a time interval and are utilised to find the location of the following time point.

:The value of ⟨x⟩ and  ⟨p⟩

(a)

The function ${\mathrm{\psi }}_0$ is even, and ${\psi }_1$is odd. In either case ${\left|\mathrm{\psi }\right|}^2$is even, so the value of $⟨x⟩=\int x{\left|\psi \right|}^{2}dx=0$. Therefore $⟨\mathrm{p}⟩=\frac{\mathrm{md}⟨\mathrm{x}⟩}{\mathrm{dt}}=0$

Thus, the value of $⟨x⟩$ is $\int \mathrm{x}{\left|\mathrm{\psi }\right|}^2\mathrm{dx}=0$ and the value of $⟨p⟩$is$\frac{\mathrm{md}⟨\mathrm{x}⟩}{\mathrm{dt}}=0$.

Now, evaluate the value of $⟨{\mathrm{x}}^2⟩$and $⟨{\mathrm{p}}^2⟩$.

The below values are known that

$\begin{array}{l}{\mathbf{\psi }}_{\mathbf{0}}\mathbf{=}{\mathbf{\alpha e}}^{\raisebox{1ex}{$\mathbf{-}{\mathbf{\xi }}^{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\\ {\mathbf{\psi }}_{\mathbf{1}}\mathbf{=}\sqrt{\mathbf{2}}{\mathbf{\alpha \xi e}}^{\raisebox{1ex}{$\mathbf{-}{\mathbf{\xi }}^{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\end{array}$

For $\mathrm{n}=0$, the value of $⟨{\mathrm{x}}^2⟩$ will be

$⟨{\mathrm{x}}^2⟩={\mathrm{\alpha }}^2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{x}}^2{\mathrm{e}}^{-{\mathrm{\xi }}^2/2}\mathrm{dx}$ ...(1)

For given equation-

$\begin{array}{c}\mathrm{\xi }\equiv \sqrt{\frac{\mathrm{m\omega }}{\mathrm{\hslash }}}\mathrm{x}\\ \mathrm{x}=\mathrm{\xi }\sqrt{\frac{\mathrm{\hslash }}{\mathrm{m\omega }}}\\ {\mathrm{x}}^2={\mathrm{\xi }}^2\left(\frac{\mathrm{\hslash }}{\mathrm{m\omega }}\right)\end{array}$

And

$\begin{array}{c}\mathrm{x}=\mathrm{\xi }\sqrt{\frac{\mathrm{\hslash }}{\mathrm{m\omega }}}\\ \mathrm{dx}=\mathrm{d\xi }\times \sqrt{\frac{\mathrm{\hslash }}{\mathrm{m\omega }}}\end{array}$

Substitute the value of $x^2$ and $dx$ in equation (1).

$\begin{array}{c}⟨{\mathrm{x}}^2⟩={\mathrm{\alpha }}^2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left({\mathrm{\xi }}^2\left(\frac{\mathrm{\hslash }}{\mathrm{m\omega }}\right)\right)\times {\mathrm{e}}^{-{\mathrm{\xi }}^2/2}\times \mathrm{d\xi }\times \sqrt{\frac{\mathrm{\hslash }}{\mathrm{m\omega }}}\\ ⟨{\mathrm{x}}^2⟩={\mathrm{\alpha }}^2{\left(\frac{\mathrm{\hslash }}{\mathrm{m\omega }}\right)}^{3/2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\xi }}^2{\mathrm{e}}^{-{\mathrm{\xi }}^2}\mathrm{d\xi }\\ =\frac{1}{\sqrt{\mathrm{\pi }}}\left(\frac{\mathrm{\hslash }}{\mathrm{m\omega }}\right)\frac{\sqrt{\mathrm{\pi }}}{2}\\ =\frac{\mathrm{\hslash }}{2\mathrm{m\omega }}\end{array}$

And the value of $⟨P^2⟩$ is

For $n=1$, the value of ${⟨x⟩}^2$will be

Substitute the value of $x^2$ and $dx$ in the above equation. So,

And the value of $⟨{\mathrm{P}}^2⟩$ is

Therefore, the value of $⟨x^2⟩$is $\frac{3\hslash }{2m\omega }$ and is$\frac{3m\hslash \omega }{2}$ .

The uncertainty principal of above states

(b)

The value of ${\sigma }_x$ will be for $n=0:$

$\begin{array}{c}{\mathrm{\sigma }}_{\mathrm{x}}=\sqrt{⟨{\mathrm{x}}^2)-{⟨\mathrm{x}⟩}^2}\\ =\sqrt{\frac{\mathrm{\hslash }}{2\mathrm{m\omega }}}\\ {\mathrm{\sigma }}_{\mathrm{p}}=\sqrt{⟨{\mathrm{p}}^2⟩-{⟨\mathrm{p}⟩}^2}\\ =\sqrt{\frac{\mathrm{m\omega }\mathrm{\hslash }}{2}}\\ {\mathrm{\sigma }}_{\mathrm{x}}{\mathrm{\sigma }}_{\mathrm{p}}=\sqrt{\frac{\mathrm{\hslash }}{2\mathrm{m\omega }}}\sqrt{\frac{\mathrm{m\omega }\mathrm{\hslash }}{2}}\\ =\frac{\mathrm{\hslash }}{2}\text{(Rightattheuncertaintymoment)}\end{array}$

For $n=1$, the value will be :

$\begin{array}{c}{\mathrm{\sigma }}_{\mathrm{x}}=\sqrt{\frac{3\mathrm{\hslash }}{2\mathrm{m\omega }}}\\ {\mathrm{\sigma }}_{\mathrm{p}}=\sqrt{\frac{3\mathrm{m\omega }\mathrm{\hslash }}{2}}\\ {\mathrm{\sigma }}_{\mathrm{x}}{\mathrm{\sigma }}_{\mathrm{p}}=3\frac{\mathrm{\hslash }}{2}>\frac{\mathrm{\hslash }}{2}\end{array}$

The uncertainty principal for these states is greater than $\frac{\hslash }{2}$.

The value of the average kinetic energy and the average potential energy 

c)

The average kinetic energy is:

$\begin{array}{c}⟨T⟩=\frac{1}{2m}⟨P^2⟩\\ =\left\{{}_{\frac{3}{4}\omega \hslash (n=1)}^{\frac{1}{4}\omega \hslash (n=0)}\right\}\end{array}$

The average potential energy is

$\begin{array}{c}⟨V⟩=\frac{1}{2}m{\omega }^2⟨x^2⟩\\ =\left\{{}_{\frac{3}{4}\omega \hslash (n=1)}^{\frac{1}{4}\omega \hslash (n=0)}\right\}\end{array}$

$⟨T⟩+⟨V⟩=⟨H⟩=\left\{{}_{\frac{3}{2}\omega \hslash (n=1)=E_1}^{\frac{1}{2}\omega \hslash (n=0)=E_0}\right\}$ , as expected.

So, the value of the average kinetic energy is $\left\{{}_{\frac{3}{4}\omega \hslash (n=1)}^{\frac{1}{4}\omega \hslash (n=0)}\right\}$ and the average potential energy is $\left\{{}_{\frac{3}{4}\omega \hslash (n=1)}^{\frac{1}{4}\omega \hslash (n=0)}\right\}$.The sum of these two is same as expected value.