In the last post, we deal with the rotations in quantum mechanics. In the post, we focus on yet another translation, not in space, but in time. From time-evolution of quantum mechanical states, we will meet the Schrödinger equation and the Hamiltonian, the operator corresponding to the total energy of the system.

The Time-Evolution Operator and the Schrödinger Equation

Consider a time-dependent state $|\psi(t)\rangle$, we’d like a time-dependent operator $\hat U(t)$, such that:

What properties should this time-evolution operator have? It turns out it should be very similar to a rotational operator $\hat R(d\psi)$:

1. It should act like the identity operator when $t=0$: $\hat U(t = 0) = \mathbf 1$

2. It should be dimensionless.

3. It should be unitary, a.k.a norm-preserving, i.e.

   $$\langle\psi(t)|\psi(t)\rangle = \langle\psi(0)|\hat U^\dagger (t) \hat U (t)|\psi(0)\rangle = \langle\psi(0)|\psi(0)\rangle \Longleftrightarrow U^\dagger (t) \hat U = \pmb 1$$

It should be reasonable to motivate that the time-evolution operator have a similar form as the rotational operator. And just like the rotational operator has a generator of rotation, spin operator, the time-evolution operator also has its generator. We assert that—and it will be apparent that this is equivalent to the 5th postulate, the time-dependent Schrödinger equation—that:

where $\hat H$ is called the Hamiltonian, the generator of time translation.

Now just like what we did for the rotational operator in the second post, we can obtain a closed form solution for the time-evolution operator by taking a series of infinitesimal time translations. Assuming the Hamiltonian is time-independent:

The time-evolution operator also satisfies a first-order differential equation w.r.t time:

Then,

And take the infinitesimal limit on time,

Finally, apply the operator equation on the initial state $|\psi(0)\rangle$:

There we have the time-dependent Schrödinger equation.

The Hamiltonian

Let’s come back to the generator of time translation, the Hamiltonian. Like the generator of rotation, the Hamiltonian is a Hermitian operator. It’s reasonable to postulate that it corresponds to some measurable quantity. Let’s first consider its unit, $[\hat H] = [\hbar/t]$, which is the unit of energy. Moreover, assuming Hamiltonian is time-independent, its expectation value is time independent.

To see this, we first Taylor-expand the time-evolution operator:

The expansion shows that $\hat H$ commutes with $\hat U$, since $\hat U$ is the sum of powers of $\hat H$. Therefore,

These results suggest, with our classical intuition, the Hamiltonian is the energy operator. The eigenstates of the Hamiltonian is the energy eigenstates $|E\rangle$, and the eigenvalues are measurable energy $E$ of the system:

The time-evolution of an energy eigenstate is especially convenient. Again, we use the Taylor-expansion of the time-evolution operator,

Time Dependence of Expectation

With the Schrödinger equation, we can also find the expectation of a measurement.

where in the second to last step we apply the Schrödinger equation and its adjoint complement $-i\hbar \frac{d}{dt} \langle\psi(t)| = \langle\psi(t)| \hat H$.

This relationship tells us that an expectation value can be time-dependent if the operator does not commute with the Hamiltonian or if the operator time-dependent.

Precession of Spin-1/2 Particles in a B-field

Indeed, we will have an example of a time-dependent operator. In fact a time-dependent Hamiltonian, which will result in a time-dependent expectation value for the energy of the system.

Suppose we have a homogeneous magnetic field $\pm B = B_0 \hat k$. Assume we have an electron with no kinetic energy in the field. In the classical mechanics, the total energy will be its potential energy,

But this is erroneous, since we cannot even comprehend spin $\pmb S$ in the classical mechanics. In the quantum mechanics, we need to express the Hamiltonian with the spin operator:

The hamiltonian has the same eigenstates as $\hat{S_z}$, $|\pm z\rangle$. The energy eigenvalues can be calculated:

So the energy eigenvalues are $E_\pm = \pm \frac{\hbar \omega_0}{2}$ for $|\pm z\rangle$.

What will happen as time evolves? At time $t$, the time-evolution operator is:

The time-evolution operator is the rotation operator about the z axis! This means that the electron’s spin will precess about the z-axis with the period $T=2\pi/\omega_0$.

For an electron originally in the states $|+ x\rangle = \frac{1}{\sqrt{2}} |+ z\rangle + \frac{1}{\sqrt{2}} |- z\rangle$, we can calculate the expectation value of $S_z$ at time $t$,

The result is as expected. Intuitively, we can think of the spin precessing about the z-axis in the x-y plane, and the measurement of $S_z$ should be time-independent and 0 in expectation.

Similarly, we can calculate $\langle S_x\rangle$,

which has a clear time-dependence.

Energy-Time Uncertainty

From the last post, we know from the the generalized uncertainty principle that

For a time-independent operator $\hat A$, we have

and from the time dependence of expectation,

These two equations combined tell us that

where $\Delta t = \frac{\Delta A}{|\frac{d\langle A\rangle}{dt}|}$ is the time for $\langle A\rangle$ to change by $\Delta A$.

References

[1] Zajc, William. PHYS GU4021-4022 Quantum Mechanics I-II, Department of Physics, Columbia University. 2019-2020.

[2] Townsend, John. A Modern Approach to Quantum Mechanics. 2nd Edition.