Now that we have a Hamiltonian, let's return to the problem of how the state of our electron bar magnet changes—or evolves—with time. If we apply a magnetic field, there must be some **time evolution** unitary operator $U(t)$ that tells us how the state of the electron evolves after a time $t$, $$ \vert \psi(t)\rangle = U(t)\vert \psi(0)\rangle, $$ where $\vert \psi(0)\rangle$ is the initial state of the electron. What is the unitary $U(t)$? It turns out that we can calculate it from the Hamiltonian! This was discovered by Erwin Schrödinger in 1926, so it is called **Schrödinger's equation**: $$ U(t) = e^{-it\hat{H}}. \tag{3} $$ (Hold on!... Did we just exponentiate a matrix? Yes, this is a legitimate operation. If you are not familiar with it, please read the appendix of this section). In the case with the electron in the vertical magentic field, the time evolution operator turns out to be $$ U(t) = e^{i\alpha t Z} = R_z(-2\alpha t). $$ This is just a $Z$ rotation (see Appendix)! Once again, we've already seen this in the context of quantum computing. We can implement the time evolution of an electron state $\vert \psi\rangle$ as a circuit:

Let's see what this does to our electron in the simple case where the qubit starts in state $\vert 0\rangle$, i.e. it is aligned with the field in the $z$ direction. It turns out that the effect of $U(t)$ on this state is kind of boring... Indeed, using the definition of the matrix exponential: $$ e^{i\alpha t Z} = I + i\alpha t Z + \frac{(i\alpha t)^2}{2!} Z^2 + \cdots = \sum_{k=0}^\infty \frac{(i\alpha t)^k}{k!}Z^k. $$ Acting on $\vert 0\rangle$ with any number of powers of $Z$ just gives us $\vert 0\rangle$ back, so $$ e^{i\alpha tZ}\vert 0\rangle = \sum_{k=0}^\infty \frac{(i\alpha t)^k}{k!}Z^k\vert 0\rangle = \sum_{k=0}^\infty \frac{(i\alpha t)^k}{k!}\vert 0\rangle = e^{i\alpha t}\vert 0\rangle. $$ All we get is a phase! Making the time-dependence a bit more explicit, if $\vert \psi (0)\rangle = \vert 0\rangle$, then $\vert \psi(t)\rangle = e^{iBet/2m_e}\vert 0\rangle$. --- ***Exercise H.3.1.*** Show that $e^{i\alpha t Z}\vert 1\rangle = e^{-i\alpha t}\vert 1\rangle$.

Solution.

Now we use $Z^k\vert 1\rangle = (-1)^k\vert 1\rangle$: $$ e^{i\alpha tZ}\vert 1\rangle = \sum_{k=0}^\infty \frac{(i\alpha t)^k}{k!}Z^k\vert 1\rangle = \sum_{k=0}^\infty \frac{(-i\alpha t)^k}{k!}\vert 1\rangle = e^{-i\alpha t}\vert 1\rangle. $$ This is also a phase, but rotating in the opposite direction. ▢

--- Let's try a more interesting case, where instead of starting in an eigenstate of the $Z$ operator, we start in state $\vert \psi(0)\rangle = \vert +\rangle$. What does this mean for the spin vector? We can check this by evaluating $(\langle X\rangle, \langle Y\rangle, \langle Z\rangle).$ Since $\langle X\rangle = 1$ and $\langle Y\rangle=\langle Z\rangle = 0$, the state $\vert + \rangle$ corresponds to the spin vector pointing in the $x$ direction. To see how this state changes with time upon exposing it to a magnetic field, let's write out the $Z$ rotation more explicitly: $$ U = e^{i\alpha tZ} = \cos(\alpha t) \cdot I + i \sin (\alpha t) \cdot Z. \tag{4} $$ Hence, the state of the qubit is $$ \begin{align*} \vert \psi\rangle = U\vert +\rangle &= \left[\cos(\alpha t) \cdot I + i \sin(\alpha t) \cdot Z\right]\vert +\rangle\\ &= \cos(\alpha t) \vert +\rangle + i \sin(\alpha t) \vert -\rangle \tag{5} \end{align*} $$ since $Z\vert +\rangle = \vert -\rangle$. This corresponds to a spin pointing in the direction $$ (\langle X\rangle, \langle Y\rangle, \langle Z\rangle) = (\cos(2\alpha t), \sin(2\alpha t), 0). $$ The tiny bar magnet is just spinning around in the $xy$-plane! Let's check the $x$ component, for instance. Using the fact that $X\vert \pm\rangle = \pm \vert \pm\rangle$, we find that $$ \begin{align*} \langle \psi \vert X \vert \psi\rangle & = \left(\cos(\alpha t) \langle +\vert - i \sin(\alpha t) \langle-\vert \right) X \left(\cos(\alpha t) \vert +\rangle + i \sin(\alpha t) \vert -\rangle\right) \\ & = \left(\cos(\alpha t) \langle +\vert - i \sin(\alpha t) \langle -\vert \right) \left(\cos(\alpha t) \vert +\rangle - i \sin(\alpha t) \vert -\rangle\right) \\ & = \cos^2(\alpha t) \langle +\vert +\rangle - \sin^2(\alpha t) \langle -\vert -\rangle + i \cos(\alpha t)\sin(\alpha t) (\langle-\vert +\rangle - \langle +\vert -\rangle) \\ & = \cos^2(\alpha t) - \sin^2(\alpha t) \\ & = \cos(2\alpha t). \end{align*} $$ Similar manipulations show $\langle Y\rangle = \sin(2\alpha t)$ and $\langle Z\rangle =0$, with the angle changing at a rate $2\alpha$. A stronger field won't align the spin any better, but it will rotate the spin vector faster! This is called *Larmor precession*:

In general, the spin vector $\mathbf{S}$ of the electron will simply rotate around the $z$-axis with angular velocity $Be/m_e$, as you can check in the next exercise. It may seem odd that the magnetic field doesn't push the spin into alignment, but it turns out that in quantum as in classical physics, a *uniform* magnetic field cannot cause tiny magnets to align. Iron filings only arrange themselves along field lines because the strength of the field at the top and bottom of the filing is different! --- ***Exercise H.3.2.*** Suppose that $\vert \psi(0)\rangle$ has expectations $$ (\langle X\rangle_0, \langle Y\rangle_0,\langle Z\rangle_0) = (x, y, z). $$ Show that $\vert \psi(t)\rangle = U(t)\vert \psi(0)\rangle$ has expectations $$ (\langle X\rangle_t, \langle Y\rangle_t,\langle Z\rangle_t) = (\cos(2\alpha t) x + \sin(2\alpha t) y, -\sin(2\alpha t) x + \cos(2\alpha t) y, z), $$ where for an operator $\mathcal{O}$, $\langle\mathcal{O}\rangle_t = \langle\psi(t)\vert \mathcal{O}\vert \psi(t)\rangle$. This means a qubit starting in any state will simply precess clockwise around the $z$-axis with angular rate of change $2\alpha$.

Solution.

We'll do $\langle X\rangle_t$ and leave the rest to the diligent reader. We will need to take the adjoint of $U(t)$ in order to apply it to the bra version of the state we use in the expectations. Making use of ($4$) and the identity $$ U(t)^\dagger = \left(\cos(\alpha t) \cdot I + i \sin (\alpha t) \cdot Z\right)^\dagger = \cos(\alpha t) \cdot I - i \sin (\alpha t) \cdot Z, $$ we obtain after some algebra $$ \begin{align*} \langle X\rangle_t & = \langle \psi(t)\vert X \vert \psi(t)\rangle \\ & = \langle \psi(0)\vert U(t)^\dagger X U(t) \vert \psi(0)\rangle \\ & = \langle \psi(0)\vert (\cos(\alpha t) \cdot I - i \sin (\alpha t) \cdot Z)X (\cos(\alpha t) \cdot I + i \sin(\alpha t) \cdot Z) \vert \psi(0) \\ & = \cos^2(\alpha t) \langle X\rangle_0 + i\cos(\alpha t)\sin(\alpha t) (\langle XZ\rangle_0 - \langle ZX\rangle_0) + \sin^2(\alpha t) \langle ZXZ\rangle_0. \end{align*} $$ From the algebra of Pauli matrices, $ZX = -XZ = iY$ and $ZXZ = iYZ = -X$, and hence $$ \langle X\rangle_t = \big(\cos^2(\alpha t) - \sin^2(\alpha t)\big)\langle X_0\rangle + 2\cos(\alpha t)\sin(\alpha t) \langle Y \rangle_0 = \cos(2\alpha t) x + \sin(2\alpha t) y. $$ This is what we wanted! The other components are similar. ▢

--- We now have a nice simple example where the unitary evolution is connected to the physics. Time to consider something more interesting!

Codercise H.3.1a - Rotation as exponential

Codercise H.3.1b - Evolving the magnet Hamiltonian

Codercise H.3.1d - Plotting the evolution

Codercise H.3.1e - Evolving a superposition

Magnets and spin

The quantum description

Time evolution

Appendix—The Matrix Exponential