About spin precession in a magnetic field

[Francisco Dahab]

So I was trying to see what the result for the spin precession would be if the magnetic field pointed in the y-direction instead of z. I feel like either something with what I'm doing is wrong or, I'm just overlooking something because I keep getting complex energy eigenvalues. So what I'm doing is:
Initially, we need to find the Hamiltonian which is just
$$H=-\gamma (B \cdot S)=-\gamma B_0S_y=\frac{-\gamma B_0 \hbar}{2}
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}
$$

Then I proceed to find the eigenvalues, taking the determinant of this matrix and setting it equal to zero

$$
\begin{pmatrix}
-E & \frac{i \gamma B_0 \hbar}{2} \\
\frac{-i \gamma B_0 \hbar}{2} & -E
\end{pmatrix} \Rightarrow E^2+\left(\frac{\gamma B_0 \hbar}{2}\right)^2=0 \Leftrightarrow E=\pm\frac{i\gamma B_0 \hbar}{2}
$$
which is complex, but what's the meaning of complex energies? And if this isn't wrong why won't the spin precess (because once you apply time evolution you won't have an exponential with an imaginary exponent you'll have an $e^{- t}$

 

[king vitamin]

It looks like you simply made an algebra mistake in your second equation. Perhaps you are forgetting the minus signs which occur in the definition of the determinant?

As an aside, sometimes people consider non-hermitian Hamiltonians in order to effectively describe unstable particles (there is time dependence like $e^{-t/T}$). But if your Hamiltonian is hermitian like in this case, your eigenvalues are always real (there is an exact proof of this which I recommend you review, it should be in most quantum mechanics textbooks).

 

[Francisco Dahab]

It looks like you simply made an algebra mistake in your second equation. Perhaps you are forgetting the minus signs which occur in the definition of the determinant?
As an aside, sometimes people consider non-hermitian Hamiltonians in order to effectively describe unstable particles (there is time dependence like $e^{-t/T}$). But if your Hamiltonian is hermitian like in this case, your eigenvalues are always real (there is an exact proof of this which I recommend you review, it should be in most quantum mechanics textbooks).

Oh god, you're right thanks, this is embarassing... I got caught up in all of the minus' that are there and I went with it because of the i's in the $\sigma_y$ matrix. As for the proof I can think of one off the top of my head, which uses the commutativity of a hermitian operator (when doing an inner product) and therefore the eigenvalue must be equal to it's complex conjugate, which implies it's real. Although I'm unsure as to whether or not it's the proof you are referring to.
As for the thing about unstable particles, how can a particle have complex energy? Is it just the case that because it is non-hermitian the actual energies we measure are real?

 

[king vitamin]

When people use non-Hermitian Hamiltonians with complex eigenvalues, the usual interpretation is that the real part of the eigenvalue is the energy while the complex part is half of the decay rate of that eigenstate. For example, if you have two complex eigenvalues
$$E_1 - i \Gamma_1/2, \qquad E_2 - i \Gamma_2/2,$$
and you prepare an initial state as a superposition of the two eigenstates
$$|\psi(t=0)\rangle = a |E_1,\Gamma_1 \rangle + b |E_2,\Gamma_2 \rangle$$
then at a later time, we have
$$|\psi(t)\rangle = a e^{-i E_1 t} e^{- \Gamma_1 t/2} |E_1,\Gamma_1 \rangle + b e^{-i E_2 t} e^{- \Gamma_2 t/2} |E_2,\Gamma_2 \rangle$$
The factor of 1/2 is because the probability of the state is given by the magnitude squared, so if you start in an eigenstate it will decay exponentially with a mean time given by $\Gamma_{1,2}$.

This is largely just a mathematical trick used to describe unstable states, but it can be useful sometimes. Whenever you are describing a complete system your Hamiltonian must always have only real eigenvalues.

 

[Francisco Dahab]

When people use non-Hermitian Hamiltonians with complex eigenvalues, the usual interpretation is that the real part of the eigenvalue is the energy while the complex part is half of the decay rate of that eigenstate. For example, if you have two complex eigenvalues
$$E_1 - i \Gamma_1/2, \qquad E_2 - i \Gamma_2/2,$$
and you prepare an initial state as a superposition of the two eigenstates
$$|\psi(t=0)\rangle = a |E_1,\Gamma_1 \rangle + b |E_2,\Gamma_2 \rangle$$
then at a later time, we have
$$|\psi(t)\rangle = a e^{-i E_1 t} e^{- \Gamma_1 t/2} |E_1,\Gamma_1 \rangle + b e^{-i E_2 t} e^{- \Gamma_2 t/2} |E_2,\Gamma_2 \rangle$$
The factor of 1/2 is because the probability of the state is given by the magnitude squared, so if you start in an eigenstate it will decay exponentially with a mean time given by $\Gamma_{1,2}$.

This is largely just a mathematical trick used to describe unstable states, but it can be useful sometimes. Whenever you are describing a complete system your Hamiltonian must always have only real eigenvalues.

Oh ok, that makes sense. It's actually really cool haha. Thanks :)