91影视

$蠄\left(x,t\right)=\frac{1}{\sqrt{2}}{蠄}_n\left(x\right)e^{-i\left(E_n/\mathrm{鈩�}\right)t}+\frac{1}{\sqrt{2}}{蠄}_m{e}^{-i\left(E_m/\mathrm{鈩�}\right)t}$

The expected value of an experiment based on a probability is known in quantum mechanics as anexpected value. This value is not always the most likely measurement result.

Anexpectation value may have a probability of zero. It can be thought of as the average of all conceivable measurement results weighted by their likelihood.

The normalization of the wave function is mathematically presented as:

${鈭�}_0^L蠄\left(x,t\right){蠄}^{*}\left(x,t\right)dx=1$

Here, $蠄\left(x,t\right)$is the wave function of the particle and$蠄*\left(x,t\right)$ .

Replacethe values in the above equation鈥檚 left hand side.

$\begin{array}{rcl}L.H.S& =& {鈭�}_0^L\frac{1}{\sqrt{2}}{蠄}_n\left(x\right)e^{-i\left(E_n/\mathrm{鈩�}\right)t}+\frac{1}{\sqrt{2}}{蠄}_m{e}^{-i\left(E_m/\mathrm{鈩�}\right)t}\left(\frac{1}{\sqrt{2}}{蠄}_n\left(x\right)e^{i\left(E_n/\mathrm{鈩�}\right)t}+\frac{1}{\sqrt{2}}{蠄}_m{e}^{i\left(E_m/\mathrm{鈩�}\right)t}\right)dx\\ & =& \frac{1}{2}{鈭�}_0^L\left({蠄}_n{蠄}_n*\left(x\right)e^{i\left(E_n-E_n/\mathrm{鈩�}\right)t}+\left({蠄}_m{蠄}_m*\left(x\right)e^{i\left(E_m-E_m/\mathrm{鈩�}\right)t}\right)\right)dx\\ & =& \frac{1}{2}\left(1+1\right)\\ & =& 1=R.H.S.\\ & & \end{array}$

Hence, it is proved that wave function is properly normalized.

The expectation value of the energy of the wave function is:

$\stackrel{炉}{E}={鈭�}_0^L蠄*\left(x,t\right)\left(i\mathrm{鈩�}\frac{鈭�}{鈭�t}\right)蠄*\left(x,t\right)dx$

Replaceall the values in the above equation.

$\begin{array}{rcl}\stackrel{炉}{E}& =& {鈭�}_0^L\left(\frac{1}{\sqrt{2}}{蠄}_n\left(x\right)e^{i\left(E_n/\mathrm{鈩�}\right)t}+\frac{1}{\sqrt{2}}{蠄}_m{e}^{i\left(E_m/\mathrm{鈩�}\right)t}\right)\left(i\mathrm{鈩�}\frac{鈭�}{鈭�t}\right)\left(\frac{1}{\sqrt{2}}{蠄}_n\left(x\right)e^{-i\left(E_n/\mathrm{鈩�}\right)t}+\frac{1}{\sqrt{2}}{蠄}_m{e}^{-i\left(E_m/\mathrm{鈩�}\right)t}\right)dx\\ & =& \frac{1}{2}{鈭�}_0^L\left({蠄}_n\left(x\right)e^{i\left(E_n/\mathrm{鈩�}\right)t}+{蠄}_m{e}^{i\left(E_m/\mathrm{鈩�}\right)t}\right)\left(E_n{蠄}_n*\left(x\right)e^{-i\left(E_n/\mathrm{鈩�}\right)t}+E_m{蠄}_m*\left(x\right)e^{-i\left(E_m/\mathrm{鈩�}\right)t}\right)dx\\ & =& \frac{1}{2}{鈭�}_0^L{E}_n\left({蠄}_n\left(x\right)e^{i\left(E_n/\mathrm{鈩�}\right)t}{蠄}_n*\left(x\right)e^{-i\left(E_n/\mathrm{鈩�}\right)t}\right)dx+\frac{1}{2}{鈭�}_0^L{E}_m\left({蠄}_m\left(x\right)e^{i\left(E_m/\mathrm{鈩�}\right)t}{蠄}_m*\left(x\right)e^{-i\left(E_m/\mathrm{鈩�}\right)t}\right)dx\\ & =& \frac{1}{2}\left(E_n+E_m\right)\\ & & \end{array}$

Hence, the expectation value of the energy is same as the average of the two energies whose value is$\frac{1}{2}\left(E_n+E_m\right)$ .

The expectation value of the square of the energy is:

${\stackrel{炉}{E}}^2={鈭�}_0^L蠄*\left(x,t\right)\left(-{\mathrm{鈩�}}^2\frac{{鈭�}^2}{鈭�t^2}\right)蠄*\left(x,t\right)dx$

Replaceall the values in the above equation.

$\begin{array}{rcl}{\stackrel{炉}{E}}^2& =& \frac{1}{2}{鈭�}_0^L\left({蠄}_n\left(x\right)e^{i\left(E_n/\mathrm{鈩�}\right)t}+{蠄}_m\left(x\right)e^{i\left(E_m/\mathrm{鈩�}\right)t}\right)\left(-{\mathrm{鈩�}}^2\frac{{鈭�}^2}{鈭�t^2}\right)\left({蠄}_n*\left(x\right)e^{-i\left(E_n/\mathrm{鈩�}\right)t}+{蠄}_m*\left(x\right)e^{-i\left(E_m/\mathrm{鈩�}\right)t}\right)dx\\ & =& \frac{1}{2}{鈭�}_0^L\left({蠄}_n\left(x\right)e^{i\left(E_n/\mathrm{鈩�}\right)t}+{蠄}_m\left(x\right)e^{i\left(E_m/\mathrm{鈩�}\right)t}\right)\left(E_n^2{蠄}_n*\left(x\right)e^{-i\left(E_n/\mathrm{鈩�}\right)t}+E_m^2{蠄}_m*\left(x\right)e^{-i\left(E_m/\mathrm{鈩�}\right)t}\right)dx\\ & =& \frac{1}{2}\left(E_n^2+E_m^2\right).\\ & & \end{array}$

Hence, the expectation value of the square of the energy is$\frac{1}{2}\left(E_n^2+E_m^2\right)$

The uncertainty in the energy of the wave function is:

$螖E=\sqrt{\stackrel{炉}{E^2}-{\stackrel{炉}{E}}^2}$

Replace all the values in the above equation.

$\begin{array}{rcl}螖E& =& \sqrt{\frac{1}{2}\left(E_n^2+E_m^2\right)-{\left(\frac{1}{2}\left(E_n+E_m\right)\right)}^2}\\ & =& \sqrt{\frac{1}{4}{E}_n^2+\frac{1}{4}{E}_m^2-\frac{1}{4}{E}_n{E}_m}\\ & =& \frac{1}{2}\sqrt{{\left(E_n-E_m\right)}^2}\\ & =& \frac{\left|E_n-E_m\right|}{2}.\\ & & \end{array}$

Hence, the uncertainty of the energy is$\frac{\left|E_n-E_m\right|}{2}$. The uncertainty of the combined state energy will increase with an increase in the energy difference between the states.