Question

A harmonic oscillator is in a state such that a measurement of the energy would yield either(1/2)$\sout{\mathbf{h}}\mathit{\omega }$or (3/2) $\sout{\mathbf{h}}\mathit{\omega }$, with equal probability. What is the largest possible value of in such a state? If it assumes this maximal value at time $\mathit{t}\mathbf{=}\mathbf{0}$ , what is $\mathit{\psi }\left(x,t\right)$ ?

Short answer

The largest possible value is $\sqrt{\sout{h}m\omega /2}$which occurs when $t=0$ and wave function will be $\psi \left(x,t\right)=\frac{1}{\sqrt{2}}{e}^{-i\omega t/2}\left({\psi }_0+i{\psi }_1{e}^{-i\omega t}\right)$.

Step-by-step solution

Diagonal elements of the matrices X and P

In a harmonic oscillator, the rising and dropping operators are used to compute <x>, and <p> , they are both zero for all stationary conditions. These measures are the diagonal elements of the matrices X and P . That is:

${<x>}_{\mathbf{n}\mathbf{n}}\mathbf{=}<n\left|x\right|n>$

From equation and equation 2.66,

$$\begin{gathered} p=i\sqrt{\frac{\sout{h}m\omega }{2}}\left(a_{+}-a_{-}\right)a_{+}\overline{)n>} \\ =\sqrt{n+1}\overline{)n+1>a\_\overline{)n>}} \\ =\sqrt{n}\overline{)n-1} \end{gathered}$$

General matrix elements for the operator

The general matrix elements for the operator p can then be calculated as:

Thus,

$P=i\sqrt{\frac{m\sout{h}\omega }{2}}\left(\begin{array}{cccccc}0& -\sqrt{1}& 0& 0& 0& 0\\ \sqrt{1}& 0& -\sqrt{2}& 0& 0& 0\\ 0& \sqrt{2}& 0& -\sqrt{3}& 0& 0\\ 0& 0& \sqrt{3}& 0& -\sqrt{4}& 0\\ 0& 0& 0& \sqrt{4}& 0& -\sqrt{5}\end{array}\right)...\left(4\right)$

Calculate <p> for wave function

Now calculate <p> for this wave function by the use of the matrix elements of (2), so the result is:

Now make this value occur at t=0 , so shift the origin of time by introducing a new time variable $\tau$such that $\tau =t+\pi /2\omega$. Make this substitution into the wave function, to get:

Consider wave function that is a combination of two states, that is:

$$\begin{gathered} \Psi \left(x,t\right)=\left(c_0{\Psi }_0\left(x\right)e^{-i{E}_{0}t/\sout{h}}+c_1{\Psi }_1\left(x\right)e^{-i{E}_{0}t/\sout{h}}\right) \\ \Psi \left(x,t\right)=\frac{1}{\sqrt{2}}\left(e^{1{\theta }_0}{\Psi }_0\left(x\right)e^{i\sout{h}\omega t/\sout{h}}+e^{1{\theta }_1}{\Psi }_1\left(x\right)e^{i\frac{3}{2}\sout{h}\omega t/\sout{h}}\right) \end{gathered}$$

Substitute ${\theta }_0=0,{\theta }_1=\pi /2$

$$\begin{gathered} \Psi \left(x,t\right)=\frac{1}{\sqrt{2}}{e}^{-i\omega t/2}\left[{\Psi }_0+{\Psi }_1{e}^{-i\pi /2}{e}^{-i\omega t}\right] \\ \Psi \left(x,t\right)=\frac{1}{\sqrt{2}}{e}^{-i\omega t/2}\left({\Psi }_0+i{\Psi }_1{e}^{-i\omega t}\right) \end{gathered}$$

The probability of getting either state is still equal to 0.5 at t=0 . Now make this substitution into the expectation value of the momentum, so:

The maximum is $\sqrt{\sout{h}m\omega /2}$which occurs when $\sin \left(\omega \tau -\pi /2\right)=-1$, that is:

$$\begin{gathered} \sin \left(\omega \tau -\frac{\pi }{2}\right)=-1 \\ \sin \left(\omega \tau -\frac{\pi }{2}\right)=\sin \left(-\frac{\pi }{2}\right) \end{gathered}$$

so,

$$\begin{gathered} \omega \tau -\frac{\pi }{2}=-\frac{\pi }{2} \\ \tau =0 \end{gathered}$$

Therefore, The largest possible value is $\sqrt{\sout{h}m\omega /2}$ which occurs when $\tau =0$and wave function will be$\frac{1}{\sqrt{2}}{e}^{-1\omega t/2}\left(\Psi +i{\Psi }_1{e}^{-i\omega t}\right)$ .