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## Main Question or Discussion Point

We have a 1 dimensional infinite well (from x=0 to x=L) and the time dependent solution to the wavefunction is the product of the energy eigenstate multiplied by the complex exponential:

[tex] \Psi_n(x, t) = \sqrt{\frac{2}{L}} \sin(\frac{n\pi x}{L}) e^{-\frac{iE_n}{\hbar}} [/tex]

Now, I want to create a superposition of two states, n = 1 and n = 2. I can write this as:

[tex] \Psi(x, t) = \frac{1}{\sqrt{2}}\sqrt{\frac{2}{L}} \sin(\frac{\pi x}{L}) e^{-\frac{iE_1}{\hbar}} + \frac{1}{\sqrt{2}}\sqrt{\frac{2}{L}} \sin(\frac{2\pi x}{L}) e^{-\frac{iE_2}{\hbar}} [/tex]

Before measurement, the energy is ##E_1 + E_2##.

I make a measurement of the energy and we find that the energy is ##E_1##.

Here's my question: What happened to the rest of the energy that was in the system (ie, the ##E_2## energy)? Where does it go?

[tex] \Psi_n(x, t) = \sqrt{\frac{2}{L}} \sin(\frac{n\pi x}{L}) e^{-\frac{iE_n}{\hbar}} [/tex]

Now, I want to create a superposition of two states, n = 1 and n = 2. I can write this as:

[tex] \Psi(x, t) = \frac{1}{\sqrt{2}}\sqrt{\frac{2}{L}} \sin(\frac{\pi x}{L}) e^{-\frac{iE_1}{\hbar}} + \frac{1}{\sqrt{2}}\sqrt{\frac{2}{L}} \sin(\frac{2\pi x}{L}) e^{-\frac{iE_2}{\hbar}} [/tex]

Before measurement, the energy is ##E_1 + E_2##.

I make a measurement of the energy and we find that the energy is ##E_1##.

Here's my question: What happened to the rest of the energy that was in the system (ie, the ##E_2## energy)? Where does it go?