I get that in the Schrödinger picture the wave function evolve in time and the quantum operators are independent of time. However, in the Heisenberg picture the operators evolve in time and the wave functions remain independent. I understand that the operator must obey the commutator equation but, what does it mean to have an operator that evolve in time from a physicist's point of view?
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So in my view the Heisenberg picture is just one of the many “interaction pictures” that exists and the fundamental idea is just that we are doing a sort of quantum coordinate transform where we describe our Hamiltonian as $\hat H(t) = \hbar (\eta(t) + \xi(t))$ where in the usual case $\eta$ is the “simple” parts of the Hamiltonian and the $\xi$ is the “complicated” part having to do with interactions and what-not.
Mathematical background
So the idea is that just as $\hat H$ induces a unitary operator $$U(t) = \mathcal T \exp\left(-\frac{i}{\hbar} \int_{t_0}^t \mathrm d\tau~\hat H(\tau)\right)$$we can say that $\eta$ induces a unitary operator $$u(t) = \mathcal T \exp\left(-i \int_{t_0}^t \mathrm d\tau~ \eta(\tau)\right)$$and all predicted expectation values $$\langle A \rangle_\Psi = \langle \Psi_0 |U^\dagger \hat A U|\Psi_0\rangle$$can have $uu^\dagger = 1$ inserted into them arbitrarily to yield $$\langle A \rangle_\Psi = \langle \tilde \Psi| \tilde A | \tilde \Psi\rangle, \text{ where }\\ |\tilde \Psi\rangle = u^\dagger U |\Psi_0\rangle \text{ and }\tilde A = u^\dagger\hat A u.$$ One can then confirm that $$\frac{\mathrm d\phantom{t}}{\mathrm dt} |\tilde \Psi\rangle = u^\dagger \left(i\eta -\frac{iH}{\hbar}\right) U ~|\Psi_0\rangle = - i \tilde \xi ~|\tilde \Psi\rangle,\\ \frac{\mathrm d\tilde A}{\mathrm dt} = u^\dagger \left(i \eta \hat A + \dot A - i \hat A \eta\right) u = i[\tilde\eta, \tilde A] + \widetilde{\dot A}. $$ So these are all expressed in the tilde-basis and my interpretation of this is just that you have changed coordinates into a reference frame which is varying over time.
Actually I go maybe a step further than this generic math suggests. So if you ever deal with a bosonic annihilator $b$ you will be introduced to its quadratures $x = (b^\dagger + b)/2$ and $p = i(b^\dagger - b)/2$, and the coherent states $b|\alpha\rangle = \alpha |\alpha\rangle.$ Under these the harmonic oscillator Hamiltonian $H = \hbar \Omega~ (b^\dagger b + 1/2)$ actually just induces a time evolution $|\alpha\rangle \mapsto |e^{-i\Omega t} \alpha\rangle$ in the Schrödinger picture, and one can work out that taking $\eta = \Omega b^\dagger b$ makes this really obvious because you have $b \mapsto e^{-i\Omega t} b$ so of course that's what it does. Meanwhile $\langle x\rangle$ and $\langle p \rangle$ describe $|\alpha\rangle$ as being centered on a specific point in phase space given by the real and imaginary parts of $\alpha$.
What I am inclined to do here is to imagine that I can always create this sort of coherent-state phase space so that the Hamiltonian here can be visualized as a flow field depicting where the probability flows given where it was, and in this case it is a big flow field going around the origin in circles. What this interaction picture did was, it used rotating coordinates for phase space $\tilde x(t) = x \cos(\Omega t) + p \sin(\Omega t), \tilde p(t) = p \cos(\Omega t) - x \sin(\Omega t),$ and thus that rotating part of the flow map disappeared; we offloaded all of that onto the operators (in this case $b, b^\dagger$).
The Heisenberg picture in particular
Once you have this general recipe for the interaction picture, it is very tempting to take the special case where $\eta = \hat H/\hbar$ and thus $\xi = 0$, in which case $u = U$ and $|\tilde\Psi\rangle = |\Psi_0\rangle.$ From a physical perspective this then means that this coordinate transformation zeroes out the flow field so that the state never evolves across phase space, rather what evolves is how we measure these various quadratures of our system.
So like I imagine just watching a classical bouncy ball describing several perfect parabolas as it loses no energy bouncing on a perfectly flat frictionless floor, this coordinate transform physically means using a camera which is totally fixed on the ball, and in this system the force of gravity on the ball disappears altogether, the floor moves off to the side at a pretty constant speed and flies up to meet the ball, and we introduce a fictitious force which exactly balances out the floor-force on the ball whenever the floor is in contact with the ball. We are co-moving with the ball and we describe the entire world around the ball as changing instead of the ball, and so things like "what is the dynamical height of the ball?" become questions like "what is the dynamical distance of the floor from the ball?" which must ultimately be solved by similar equations, transferred to the floor.
An interaction picture is some sort of hybrid visualization, for example just using free-fall coordinates to zero out the effect of gravity, but not zeroing out the forces of the floor. Meanwhile, the floor experiences a constant acceleration upwards, due to that transformation. So the ball starts off going upwards at speed $0$ and then after one impact it starts traveling upwards at speed $v$ and then after another impact it starts traveling upwards at speed $2v$ and then $3v$ and so on, the ball has these long constant-speed trajectories interrupted by the floor which each time I think gives it a constant dose of momentum upwards.
