Question

A particle of mass m is in the state:

$\mathbf{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{a}\mathbf{[}\mathbf{(}{\mathbf{mx}}^{\mathbf{2}}\mathbf{/}\sout{\mathbf{h}}\mathbf{)}\mathbf{+}\mathbf{it}\mathbf{]}}$

where A and a are positive real constants.

(a) Find A.

(b) For what potential energy function, V(x), is this a solution to the Schrödinger equation?

(c) Calculate the expectation values of x,${\mathit{x}}^{\mathbf{2}}$ , p, and${\mathit{p}}^{\mathbf{2}}$ .

(d) Find σ_x and σ_p. Is their product consistent with the uncertainty principle?

Short answer

a. $\mathrm{A}={\left(\frac{2\mathrm{am}}{\mathrm{\pi }\sout{\mathrm{h}}}\right)}^{1/4}$

b.${\mathrm{V}}_{(\mathrm{x},\mathrm{t})}=2{\mathrm{ma}}^2{\mathrm{x}}^2$

c. $⟨\mathrm{x}⟩=0$, $⟨{\mathrm{x}}^2⟩=\frac{\sout{\mathrm{h}}}{4\mathrm{am}}$, $⟨\mathrm{p}⟩=0$,$⟨{\mathrm{p}}^2⟩=\sout{\mathrm{h}}\mathrm{am}$

d. Yes, it’s consistent with the uncertainty principle

Step-by-step solution

Normalizing the wave equation (a)

Calculating for A by normalizing the wave function.

$\begin{array}{c}1=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{|{\mathrm{\psi }}_{(\mathrm{x},\mathrm{t})}|}^2\mathrm{dx}\\ 1=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{\psi }{*}_{(\mathrm{x},\mathrm{t})}{\mathrm{\psi }}_{(\mathrm{x},\mathrm{t})}\mathrm{dx}\\ 1=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left({\mathrm{Ae}}^{-\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})+\mathrm{it}]}\right)\left({\mathrm{Ae}}^{-\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})-\mathrm{it}]}\right)\mathrm{dx}\\ 1={\mathrm{A}}^2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-2{\mathrm{amx}}^2/\sout{\mathrm{h}}}\mathrm{dx}\end{array}$

Further solving above equation,

Thus, the value of A is ${\left(\frac{2\mathrm{am}}{\mathrm{\pi }\sout{\mathrm{h}}}\right)}^{1/4}$.

Finding the potential energy for which it is a solution of the Schrödinger equation (b)

Schrodinger equation is given by,

$\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}=\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}^2}-\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{V}}_{(\mathrm{x},\mathrm{t})}{\mathrm{\psi }}_{(\mathrm{x},\mathrm{t})}$

Solving for ${\mathrm{V}}_{(\mathrm{x},\mathrm{t})}$,

Further solving above equation,

Thus, the potential energy function is${\mathrm{V}}_{(\mathrm{x},\mathrm{t})}=2{\mathrm{ma}}^2{\mathrm{x}}^2$ .

Calculating the expectation values of x (c)

Solving for the expectation values.

Since, integral of an odd function over a symmetric interval is zero.

Therefore, the expectation value is zero $⟨\mathrm{x}⟩=0$.

Now, for $⟨{\mathrm{x}}^2⟩$

$\begin{array}{c}⟨{\mathrm{x}}^2⟩=\frac{{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x}}^2|{{\mathrm{\psi }}_{(\mathrm{x},\mathrm{t})}|}^2\mathrm{dx}}}{{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|{{\mathrm{\psi }}_{(\mathrm{x},\mathrm{t})}|}^2\mathrm{dx}}}\\ ⟨{\mathrm{x}}^2⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x}}^2{\mathrm{\psi }}_{(\mathrm{x},\mathrm{t})}\mathrm{\psi }{*}_{(\mathrm{x},\mathrm{t})}\mathrm{dx}\\ ⟨{\mathrm{x}}^2⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x}}^2\left[{\left(\frac{2\mathrm{am}}{\mathrm{\pi }\sout{\mathrm{h}}}\right)}^{1/4}{\mathrm{e}}^{-\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})+\mathrm{it}]}\right]\left[{\left(\frac{2\mathrm{am}}{\mathrm{\pi }\sout{\mathrm{h}}}\right)}^{1/4}{\mathrm{e}}^{-\mathrm{a}\left[\right({\mathrm{mx}}^2/\sout{\mathrm{h}})-\mathrm{it}]}\right]\mathrm{dx}\\ ⟨{\mathrm{x}}^2⟩=\sqrt{\frac{2\mathrm{am}}{\mathrm{\pi }\sout{\mathrm{h}}}}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x}}^2{\mathrm{e}}^{-2{\max }^2/\sout{\mathrm{h}}}\mathrm{dx}\end{array}$

$\begin{array}{l}⟨{\mathrm{x}}^2⟩=2\sqrt{\frac{2\mathrm{am}}{\mathrm{\pi }\sout{\mathrm{h}}}}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x}}^2{\mathrm{e}}^{-{\mathrm{x}}^2/{\left[\sqrt{\sout{\mathrm{h}}/\left(2\mathrm{am}\right)}\right]}^2}\mathrm{dx}\\ ⟨{\mathrm{x}}^2⟩=2\sqrt{\frac{2\mathrm{am}}{\mathrm{\pi }\sout{\mathrm{h}}}}.\sqrt{\mathrm{\pi }}\frac{2!}{1}{\left(\frac{\sqrt{\sout{\mathrm{h}}/2\mathrm{am}}}{2}\right)}^3\\ ⟨{\mathrm{x}}^2⟩=\frac{\sout{\mathrm{h}}}{4\mathrm{am}}\end{array}$

Therefore, the value of $⟨{\mathrm{x}}^2⟩$ is $\frac{\sout{\mathrm{h}}}{4\mathrm{am}}$.

Calculating the expectation values for p (c)

Using the Ehrenfest’s theorem,

$\begin{array}{l}⟨\mathrm{p}⟩=\mathrm{m}⟨\mathrm{v}⟩\\ ⟨\mathrm{p}⟩=\mathrm{m}\frac{\mathrm{d}⟨\mathrm{x}⟩}{\mathrm{dt}}\end{array}$

Since $⟨\mathrm{x}⟩=0$.

Therefore, the value of $⟨\mathrm{p}⟩$ is zero.

Now, solving for $⟨{\mathrm{p}}^2⟩$.

Further solving above equation,

Further solving above equation,

$\begin{array}{l}⟨{\mathrm{p}}^2⟩={\sout{\mathrm{h}}}^2\left(\frac{4\mathrm{am}}{\sout{\mathrm{h}}}\right)\left(\frac{1}{2}-\frac{1}{4}\right)\\ ⟨{\mathrm{p}}^2⟩=\sout{\mathrm{h}}\mathrm{am}\end{array}$

Therefore, the value of $⟨{\mathrm{p}}^2⟩$ is $\sout{\mathrm{h}}\mathrm{am}$.

Checking the consistency with Heisenberg Uncertainty Principle

The standard variation of an observable gives the uncertainty, hence,

$\begin{array}{l}{\mathrm{\sigma }}_{\mathrm{x}}=\mathrm{\Delta x}\\ \mathrm{\Delta x}=\sqrt{⟨{\mathrm{x}}^2⟩-{⟨\mathrm{x}⟩}^2}\\ \mathrm{\Delta x}=\sqrt{\frac{\sout{\mathrm{h}}}{4\mathrm{am}}-0}\\ \mathrm{\Delta x}=\frac{1}{2}\sqrt{\frac{\sout{\mathrm{h}}}{\mathrm{am}}}\end{array}$

The calculation of standard deviation ${\mathrm{\sigma }}_{\mathrm{p}}$ is,

$\begin{array}{c}{\mathrm{\sigma }}_{\mathrm{p}}=\mathrm{\Delta p}\\ \mathrm{\Delta p}=\sqrt{⟨{\mathrm{p}}^2⟩-{⟨\mathrm{p}⟩}^2}\\ \mathrm{\Delta p}=\sqrt{\sout{\mathrm{h}}\mathrm{am}-0}\\ \mathrm{\Delta p}=\sqrt{\sout{\mathrm{h}}\mathrm{am}}\end{array}$

The uncertainty can be written as,

$\begin{array}{l}\mathrm{\Delta x\Delta p}=\frac{1}{2}\sqrt{\frac{\mathrm{h}}{\mathrm{am}}}\sqrt{\mathrm{ham}}\\ \mathrm{\Delta x\Delta p}=\frac{\mathrm{h}}{2}\end{array}$

And Heisenberg Uncertainty Principle states that

$\mathrm{\Delta x\Delta p}\ge \frac{\mathrm{h}}{2}$

Therefore, it is consistent with the Heisenberg Uncertainty Principle.