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I've solved an exercise of a given quantum system with 3 given states. We had to find the energy expectation value, when we put the system in the "second starting quantum state".

So I did the necessary calculations, and found out that $\langle\hat{H}\rangle=0$, which was the right answer.

The expectation value is what we'll get if we measure the energy an infinite amount of times, and then take the average.

Doesn't that answer mean that since energy can't be negative (well, can it?), the system's energy must be equal to zero?

Does that mean that since it can't be that way, that you can't put the system in the second state as a starting state?

Edit: I was given $$ \hat{H}= \begin{pmatrix} -1 & a & 0 \\ a & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$ when the 3 states are $$ |1\rangle=\begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} |2\rangle=\begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} |3\rangle=\begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix} $$

Definitely Not a Mathematician
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2 Answers2

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energy can't be negative (well, can it?)

Energies are never absolute ─ the only thing that is ever accessible is relative energy differences between two different states of the world. Since the sign of an energy difference will switch if you flip the roles of the two states you're comparing, it's no surprise at all that energy differences can be negative.

Whenever we talk about some 'absolute' energy scale, as in the problem you're solving, we're basically fixing some arbitrary reference state and measuring 'absolute' energy as the energy difference to that reference state. So long as you don't change what reference you're using, it will look like an 'absolute' scale, though of course it isn't.

If the 'absolute' energy comes out negative, it just means that it is possible to harvest work from the system while you take it from the reference state to the state you're interested, just like positive-energy states require work to be put in to take them there from the reference. Zero-energy states are in the middle - they don't require any net work transfers for that transition.

Emilio Pisanty
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Yes, it could be negative as well. One example is an electron in an atom. The energy turns out to be negative. Also, Griffiths has a section on bound states and scattering states, that might give you a better insight.

Also, take a look at this: https://physics.stackexchange.com/a/16752/183683 and this.