Suppose I have starting qubit like the one described on the Bloch sphere below. enter image description here

Also, denote the state $|0\rangle \equiv |+\rangle$. So my starting state would be:

$$ |\psi(0)\rangle=\frac{1}{\sqrt{2}}(|+\rangle-|-\rangle) $$

Now suppose I apply some magnetic field $\vec{B}(t)=B_0\cos(\omega t)\hat{k}$. Thus the hamiltonian of my system would be:

$$ \hat{H}=-\hat{\vec{\mu}}\cdot\hat{\vec{B}}=-\gamma B_0\cos(\omega t)\hat{S_z} $$

where $\gamma$ denotes the gyromagnetic ratio and $\hat S_z$ is the $z$ Pauli matrix multiplied by $\hbar/2$. My goal is to find the probability as a function of time of finding the particle on the other eigenvector of $\hat S_x$ corresponding to the eigenvalue of $-\hbar/2$ which is$|\alpha\rangle=\frac{1}{\sqrt{2}}(|-\rangle-|-\rangle)$. Here's my attempt:

The evolution operator is:

$$ \hat U(t;0)=\exp \bigg(\frac{i}{\hbar}\gamma B_0 \cos (\omega t)\hat S_z t\bigg) \tag{*} $$

Hence the state that is time dependent is given by:

$$ |\psi(t)\rangle=\hat U(t;0)|\psi(0)\rangle=\exp \bigg(\frac{i}{\hbar}\gamma B_0 \cos (\omega t)\hat S_z t\bigg)\frac{1}{\sqrt{2}}(|+\rangle-|-\rangle) $$

which in the eigenbasis of $\hat S_z$ translates to: $$ |\psi(t)\rangle=\frac{\exp\bigg(\frac{i\gamma B_0cos(\omega t)t}{2}\bigg)}{\sqrt{2}} \bigg[|+\rangle+\exp\bigg(-i\gamma B_0 \cos(\omega t)t\bigg)|-\rangle \bigg] $$

Finnaly the probability of finding the particle at the state $|\alpha\rangle$ is:

$$ P_\alpha (t)=|\langle \alpha |\psi(t)\rangle|^2=(...)=\frac{1}{2}-\frac{1}{2}\cos\bigg[\gamma B_0 \cos(\omega t)t] $$

Now here's my question:

From a quick "playing around with a graphing software we see that the function is something like the image below.

enter image description here

So we see that the behaviour on the bloch sphere would be quite erratic, starting by wobbling around the eigenstate that it started in but quickly changing to the opposite pole of the sphere and it keeps changing between poles at a forever increasing rate. Why is this behaviour like so? My intuition would be: If it started on an eigenstate it should remain in it forever, are my calculations wrong? Does the basis that I chose reflect on the result? If so, how should I have proceeded?


3 Answers 3


A key point which the other answers miss: Your equation (*) giving the time evolution is incorrect! For a time dependent Hamiltonian, you have to integrate the Schroedinger equation, which is more complicated than just giving an exponential of $iHt$. Thus, your final result is incorrect.

  • $\begingroup$ You are damn right, thanks $\endgroup$ Dec 30, 2019 at 9:42
  • $\begingroup$ Didn't even notice that... Could you help me solve it? This is from my studies on QM I where we deal only with time independent hamiltonians, so I have no idea on how to solve the equation, that is material for QM II. $\endgroup$
    – Bidon
    Dec 30, 2019 at 16:29
  • $\begingroup$ @Bidon You'll have to study how to deal with time dependent Hamiltonians. This is a topic which is well covered in most textbooks. $\endgroup$ Dec 30, 2019 at 21:28

"If it started on an eigenstate it should remain in it forever, are my calculations wrong?" - The state you started with is not an eigenstate of the Hamiltonian. This is also clear from the result you got after evaluation of the time evolved state. If your magnetic field had been in the $ \hat{i} $ direction, then the initial state would be an eigenstate and would have remained in that state.

The entire problem is related to the case of Rabi Oscillations where a two-level system oscillates between eigenstates (of the unperturbed Hamiltonian) after the introduction of non-diagonal terms in the Hamiltonian.

  • $\begingroup$ "The state you started with is not an eigenstate of the Hamiltonian" I should have seen that, I feel stupid I didn't spot it... But it is the first time I hear of Rabi Oscillations. Could you link me to some material a bit lighter than wikipedia? If not ignore this. It's ok $\endgroup$
    – Bidon
    Dec 29, 2019 at 23:25
  • 1
    $\begingroup$ This is a useful link but involves a lot of math - ocw.mit.edu/courses/electrical-engineering-and-computer-science/… $\endgroup$
    – Hari
    Dec 29, 2019 at 23:33

The initial state is an eigenstate of $S_x$, so it is not an eigenstate of the Hamiltonian (which is proportional to $S_z$); therefore, the spin will precess around the direction of the field (i.e. the $z$ axis) half of the time in a clockwise fashion (when $\cos \omega t>0$) and the other half in a counterclockwise fashion.

If the field was constant, the probability would just have been a periodic sinusoidal function of time, but because of the nested cosines the profile is a little more involved and less intuitive; you can try to first figure out what happens with a constant B field in the z direction, and then try to upgrade your intuition by imagining the B field decay to zero at some pace and then go to -B st the same pace;

You're right saying that if it started as an eigenstate of the Hamiltonian it would remain in that state, but this is not the case;


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