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When a particle is in superposition of energy eigenstates and has a probability of being found in either state, what does that say about the energy of the particle and conservation of energy.

What I mean is, since the energy eigenstates have different energy values, where's the rest of the energy of the particle, if it is found to be in the lower energy eigenstate?

Let's say [itex]|\psi> = c_1 |E_1> + c_2 |E_2>[/itex]

Take the system to be a free particle in a box where [itex]\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})[/itex] and

[itex]V(x)=\{

\begin{array}{1 1}

0 \quad |x|<a/2\\

V_0 \quad |x|>a/2

\end{array}[/itex]

I guess my questions could be rephrased as, how do you calculate the total energy of a quantum particle?

What I mean is, since the energy eigenstates have different energy values, where's the rest of the energy of the particle, if it is found to be in the lower energy eigenstate?

Let's say [itex]|\psi> = c_1 |E_1> + c_2 |E_2>[/itex]

Take the system to be a free particle in a box where [itex]\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})[/itex] and

[itex]V(x)=\{

\begin{array}{1 1}

0 \quad |x|<a/2\\

V_0 \quad |x|>a/2

\end{array}[/itex]

I guess my questions could be rephrased as, how do you calculate the total energy of a quantum particle?

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