91影视

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,

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

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

Thus,

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

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 $蟿$such that $蟿=t+蟺/2蝇$. Make this substitution into the wave function, to get:

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

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

Substitute ${胃}_0=0,{胃}_1=蟺/2$

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

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 $\sqrt{\sout{h}m蝇/2}$which occurs when $\sin \left(蝇蟿-蟺/2\right)=-1$, that is:

$$\begin{gathered} \sin \left(蝇蟿-\frac{蟺}{2}\right)=-1 \\ \sin \left(蝇蟿-\frac{蟺}{2}\right)=\sin \left(-\frac{蟺}{2}\right) \end{gathered}$$

so,

$$\begin{gathered} 蝇蟿-\frac{蟺}{2}=-\frac{蟺}{2} \\ 蟿=0 \end{gathered}$$

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