Imagine we have a system consisting of an electron in an external magnetic field. The electrons magnetic dipole moment will interact with this magnetic field, and the energy of the interaction can be modelled as $$ H = \omega_0 S_z, $$ where $\omega_0 = \frac{eB_0}{2mc}$, just a constant. I have also assumed that the magnetic field is in the $z$ direction, i.e. $\vec{B} = B_0 \hat{k}$, so it only picks out the component of the spin in the $z$ direction.

Thus, it is very easy to find the energy values of this Hamiltonian, since the energy eigenstates are just the same states of the operator $S_z$. We find $$ H |\pm z\rangle = \pm \omega_0 \frac{\hbar}{2}|\pm z\rangle.$$ These are the energy levels for the system. You can note that there is a negative energy level $E_{-} = -\frac{\hbar \omega}{2}$. My question is what is the significance of a negative energy level, because I am not so sure what that means. I understand that negative energies in the presence of a potential (in the S.E.) signify bound states, but here we are not concerned with that. So, what does an energy level mean here, and in general what do negative energies represent physically?


1 Answer 1


Usually$^1$, the sign (or more generally the value) of the energy of a single state is on its own meaningless. The only measurable quantities are energy differences between states.

Classically, this can be seen from the fact that adding a constant to the Lagrangian, or equivalently the Hamiltonian$^2$, does not affect the equations of motion, and so the dynamics is exactly the same.

Quantum mechanically, we can see this as follows. Suppose we have states $|\psi\rangle$ satisfying a time-dependent Schroedinger equation with Hamiltonian $H$. Now define a new Hamiltonian $\tilde{H}\equiv H+c$ where $c$ is a constant. The time-dependent Schoedinger equation would read (in units where $\hbar=1$), $$ i\frac{\partial}{\partial t}|\tilde{\psi}\rangle=\tilde{H}|\tilde{\psi}\rangle=(H+c)|\tilde{\psi}\rangle\,. $$ Consider taking out a phase factor, $|\tilde{\psi}\rangle \equiv e^{-ict}|\psi\rangle$. Then substituting this into our equation we have, $$ e^{-ict}\left(i\frac{\partial}{\partial t}|\psi\rangle+c|\psi\rangle\right)=(H+c)e^{-ict}|\psi\rangle\,. $$ We see the two terms proportional to $c$ cancel, and we can also drop the phase. Hence we see that the states in the theory with the new Hamiltonian $\tilde{H}$ are in one-to-one correspondence with the states in the old theory with Hamiltonian $H$, with just a trivial common phase factor multiplication. (Since the phase factor is the same for all states, it is not observable.)

In the case of bound states that you describe, the minus sign of the energy is relative to (at least classically speaking) the state of a particle infinitely far away and motionless, which is defined to have zero energy, and so the precise energy values of other states have been made meaningful.

1. This is no longer the case in general relativity, where energy values themselves (not just energy differences) are important, as they determine the curvature of spacetime through Einstein's field equations.

2. When the Lagrangian does not explicitly depend on time, the Hamiltonian gives the energy of the system. When the Lagrangian does explicitly depend on time, I'm not sure how to make the argument for the unimportance of the zero of the energy in this way.

  • $\begingroup$ when $L$ does it depend explicitly on $t$, the Hamiltonian is conserved but not necessarily equal to the energy. $H$ is the energy in natural systems and - roughly speaking - when $\sum_k \dot q_k p_k=2T$, i.e. twice the kinetic energy. $\endgroup$ Commented Feb 16, 2017 at 22:38
  • $\begingroup$ So basically the dynamics are invariant when you add a constant to $H$? $\endgroup$ Commented Feb 16, 2017 at 22:41
  • $\begingroup$ @JoshPilipovsky yes. $\endgroup$ Commented Feb 16, 2017 at 22:45

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