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 $$ |1\rangle=\begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} |2\rangle=\begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} |3\rangle=\begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix} $$


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.

| cite | improve this answer | |
  • $\begingroup$ But isn't this energy, the one we compute through the Hamiltonian is kinetic energy, and so should always be positive? and also, what happens then when you cool a particle until it reaches 0 kelvin? $\endgroup$ – Definitely Not a Mathematician Jan 21 '19 at 16:23
  • $\begingroup$ @DefinitelyNotaMathematician The hamiltonian need not be equal to the kinetic energy - it is generally the kinetic energy plus a potential energy, and the potential energy can be negative. In general, the only thing that you can say about hamiltonians is that we typically require $\hat H$ to be bounded below (i.e. $\exists h_0\geq 0$ such that $\hat H+h_0$ is positive-definite). For your case, it's impossible to tell as you've not provided enough details about the hamiltonian and the states and what it is you're actually doing, but it's unlikely that $\hat H$ is definite. $\endgroup$ – Emilio Pisanty Jan 21 '19 at 18:17
  • $\begingroup$ I was given the hamiltonian as a matrix, but with a parameter a $\endgroup$ – Definitely Not a Mathematician Jan 21 '19 at 18:57
  • $\begingroup$ @DefinitelyNotaMathematician Sure. And without seeing that matrix explicitly, there is nothing more that we can say about it. $\endgroup$ – Emilio Pisanty Jan 21 '19 at 19:22
  • $\begingroup$ I've added the data given. do you mean it's connected to the fact that when I find the eigenvalues and eigenvectors (I think that's the name in english) I can get a negative eigenvalue, and that "fixes" it? $\endgroup$ – Definitely Not a Mathematician Jan 21 '19 at 19:35

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.

| cite | improve this answer | |
  • $\begingroup$ So, because the system is in it's lowest energy state possible (after |x|), does it mean that the system won't ever change? or it will need a ton of energy to "want" to change? $\endgroup$ – Definitely Not a Mathematician Jan 21 '19 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.