Consider the wave function that is a combination of two different infinite well stationary states the $n^{th}$ and the $m^{th}$

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

1. Show that the $\psi \left(x,t\right)$is properly normalized.
2. Show that the expectation value of the energy is the average of the two energies:$\overline{E}=\frac{1}{2}\left(E_n+E_m\right)$
3. Show that the expectation value of the square of the energy is given by .
4. Determine the uncertainty in the energy.

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

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:

${\int }_0^L\psi \left(x,t\right){\psi }^{*}\left(x,t\right)dx=1$

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

Replacethe values in the above equation’s left hand side.

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

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

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

$\overline{E}={\int }_0^L\psi *\left(x,t\right)\left(i\hslash \frac{\partial }{\partial t}\right)\psi *\left(x,t\right)dx$

Replaceall the values in the above equation.

$\begin{array}{rcl}\overline{E}& =& {\int }_0^L\left(\frac{1}{\sqrt{2}}{\psi }_n\left(x\right)e^{i\left(E_n/\hslash \right)t}+\frac{1}{\sqrt{2}}{\psi }_m{e}^{i\left(E_m/\hslash \right)t}\right)\left(i\hslash \frac{\partial }{\partial t}\right)\left(\frac{1}{\sqrt{2}}{\psi }_n\left(x\right)e^{-i\left(E_n/\hslash \right)t}+\frac{1}{\sqrt{2}}{\psi }_m{e}^{-i\left(E_m/\hslash \right)t}\right)dx\\ & =& \frac{1}{2}{\int }_0^L\left({\psi }_n\left(x\right)e^{i\left(E_n/\hslash \right)t}+{\psi }_m{e}^{i\left(E_m/\hslash \right)t}\right)\left(E_n{\psi }_n*\left(x\right)e^{-i\left(E_n/\hslash \right)t}+E_m{\psi }_m*\left(x\right)e^{-i\left(E_m/\hslash \right)t}\right)dx\\ & =& \frac{1}{2}{\int }_0^L{E}_n\left({\psi }_n\left(x\right)e^{i\left(E_n/\hslash \right)t}{\psi }_n*\left(x\right)e^{-i\left(E_n/\hslash \right)t}\right)dx+\frac{1}{2}{\int }_0^L{E}_m\left({\psi }_m\left(x\right)e^{i\left(E_m/\hslash \right)t}{\psi }_m*\left(x\right)e^{-i\left(E_m/\hslash \right)t}\right)dx\\ & =& \frac{1}{2}\left(E_n+E_m\right)\\ & & \end{array}$

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:

${\overline{E}}^2={\int }_0^L\psi *\left(x,t\right)\left(-{\hslash }^2\frac{{\partial }^2}{\partial t^2}\right)\psi *\left(x,t\right)dx$

Replaceall the values in the above equation.

$\begin{array}{rcl}{\overline{E}}^2& =& \frac{1}{2}{\int }_0^L\left({\psi }_n\left(x\right)e^{i\left(E_n/\hslash \right)t}+{\psi }_m\left(x\right)e^{i\left(E_m/\hslash \right)t}\right)\left(-{\hslash }^2\frac{{\partial }^2}{\partial t^2}\right)\left({\psi }_n*\left(x\right)e^{-i\left(E_n/\hslash \right)t}+{\psi }_m*\left(x\right)e^{-i\left(E_m/\hslash \right)t}\right)dx\\ & =& \frac{1}{2}{\int }_0^L\left({\psi }_n\left(x\right)e^{i\left(E_n/\hslash \right)t}+{\psi }_m\left(x\right)e^{i\left(E_m/\hslash \right)t}\right)\left(E_n^2{\psi }_n*\left(x\right)e^{-i\left(E_n/\hslash \right)t}+E_m^2{\psi }_m*\left(x\right)e^{-i\left(E_m/\hslash \right)t}\right)dx\\ & =& \frac{1}{2}\left(E_n^2+E_m^2\right).\\ & & \end{array}$

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:

$\Delta E=\sqrt{\overline{E^2}-{\overline{E}}^2}$

Replace all the values in the above equation.

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

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.