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I am aware of the two pictures, namely the Schrödinger and Heisenberg pictures, where the time dependence is carried by the state in the former and by the operator in the latter. However, why does it have to be either one xor the other? In other words, I'm trying to gain an intuition from first principles as to how such a dichotomy in time dependence does not exist in classical mechanics but arises in quantum mechanics.

I came to ask myself the question while watching a lecture in the context of quantum optics. In minute 9:46, the ladder operators pop up out of the blue from a time-dependent variable $\alpha(t)$ related to the amplitude of electric field while the ladder operators themselves are time-independent.

EDIT: In view of the answers, I understand how there may not necessarily be a dichotomy between Schrödinger and Heisenberg and that many other pictures are conceivable. However, I should have shifted the emphasis in my question as to why these pictures suddenly come into play in the quantization procedure. For example, considering the quantization of momentum $\vec{p}(t) \rightarrow \hbar \vec{\nabla}$, I just don't understand why time-dependence suddenly ceases to be part of the observable and has to live in the state (or vice-versa). Why does time-dependence have to be shuffled around (in whatever way) when going from classical to quantum? Or is this simply one of those dreaded "postulates"?

(PS: I haven't studied QFT, so intuitive answers from first principles would be appreciated.)

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The same two pictures also exist in classical mechanics.$^\dagger$ Classical physics can be regarded as a special case of quantum physics in which all observables commute with each other. Just like quantum physics, classical physics can be expressed either in the "Heisenberg picture" or in the "Schrödinger picture", and the two pictures are equivalent: they're just two different ways of thinking about the same thing.

$^\dagger$ As explained in other answers posted here, it's not really a dichotomy, because there are other pictures, too. The point of my answer is that classical physics has the same "pictures" that quantum physics does, not that there are only two pictures.

Perspective

In classical physics, all observables commute with each other, so we can (and do) always take the state to be an eigenstate of all of the observables.

For this reason, we don't really need to bother distinguishing between the state and the observables in classical physics, but they are logically distinct: the state is what tells us the values of the observables. Observables represent the kinds of things we can measure, and the state tells us what the results of those measurements will be.

(In quantum physics, this distinction is essential, because most observables do not commute with each other, so they cannot all have predictable measurement outcomes. The state tells us what we will get, but only statistically, when we measure observables.)

Two equivalent pictures

To illustrate the two pictures in classical mechanics, consider the classical mechanics of a system of objects interacting with each other as in Newton's model of gravity:

  • In the Heisenberg picture, the observables are the fact that we can measure the objects' positions at any desired time, and those positions are related to each other by the equations of motion. The state specifies a particular solution of the equations of motion, which endows all of those observables (at each time) with specific values.

  • In the Schrödinger picture, the observables are the fact that we can measure the object's positions and momenta. The state specifies a particular set of values for the positions and momenta at any given time, and the time-evolution of the state tells us how the positions and momenta evolve in time.

If the distinction seems insignificant, that's because the two pictures are indeed equivalent. Either one by itself accounts for all of the time-dependence. In classical physics, we subconsciously switch back and forth between these two equivalent pictures. After enough experience, we do this subconsciously in quantum physics, too.

The example in the question

Regarding the ladder-operators example that is mentioned in the question: I didn't watch the lecture, but the notation $\alpha(t)$ indicates that the lecturer is working in the Heisenberg picture. The operator $\alpha(t)$ is a time-dependent observable. The Heisenberg equations of motion are probably Maxwell's equations (just guessing because I didn't watch the lecture), but with operator-valued components $\alpha(t)$ of the fields.

Even though the observable is time-dependent (a different operator at different times), the fact that the time-dependence is governed by the Heisenberg equation of motion implies that we can write all of these observables in terms of a common set of operators, the ladder operators. In this context, the ladder operators are not associated with any particular time. They're just operators on the Hilbert space that we can use to express all of the observables $\alpha(t)$.

The classical-physics analogue is that the general time-dependent electromagnetic field that satisfies Maxwell's equations can be written in terms of a fixed set of unspecified coefficients. The time-dependent components of the electromagnetic field are the observables. The state selects a specific solution by specifying the values of the coefficients in that general solution. The ladder operators are analogous to the coefficients, except that we can't completely "specify their values" in quantum physics, because they don't commute with each other.

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  • $\begingroup$ Just to add a little clarification, that $\alpha(t)$ in the lecture I linked to isn't an operator, but merely a quantity proportional to the time evolution of the classical electric field (min 9:46 to min 10:01). It is then decomposed into its real and imaginary parts, namely position and momentum. The leap of faith I can't wrap my head around is the fact that the time dependence disappears outright when $\alpha(t)$ and $\alpha^{*}(t)$ are transformed into the creation and annihilation operators, respectively. $\endgroup$ – Tfovid May 6 at 22:16
  • $\begingroup$ @Tfovid I watched that segment of the video. Apparently the lecturer combined two things into one step, namely going from classical to quantum and changing where the time-dependence is handled, but he didn't show where he put the time-dependence and didn't even acknowledge that he moved it. It was just carelessness on the lecturer's part. $\endgroup$ – Chiral Anomaly May 6 at 23:39
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However, why does it have to be either one xor the other?

It doesn't. It's also possible to work in mixed schemes, known generally as the Interaction Picture, where different parts of the hamiltonian are responsible for the time evolution of the operators and the states.

This is mostly used when the hamiltonian has an 'easy' part (which you solve first, and use to drive the time evolution of operators), and a harder part that you may need to solve numerically or perturbatively, with both of those approaches becoming easier once the simpler components have been solved.

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You can define other pictures in which the operators get some of the time dependence and the states get some as well. The interaction picture is one of those and is probably the only useful implementation of this. You split the Hamiltonian in a 'simple' and 'hard' to solve part. The states get the simple part of the time evolutation while the operators get the hard-to-solve part.

But why does this not occur in classical mechanics? This is kind of subtle since quantum/classical mechanics are both different mathematical frameworks. Classical mechanics describes trajectories in state space while quantum mechanics describes evolution of the full state vector $|\psi\rangle$. From $|\psi\rangle$ you then still have to extract its observables.

There are several formulas in terms of Poisson brackets that look similar to time evolution in QM. For example $$\frac d{dt}A(p,q,t)=\{A,H\}+\frac{\partial A}{\partial t}$$ Where $A$ is just a function (the classical equivalent of an operator). Another example taken straight from Wikipedia: \begin{align} q(t)&=\exp(-t\{H,.\})q(0)\\ p(t)&=\exp(-t\{H,.\})p(0) \end{align} These 'observables' are all regular functions, so when you ask why doesn't classical mechanics have this you have to be careful what you mean by observables, states etc. But I think personally the biggest reason that this doesn't work is that QM is linear in time evolution while CM is not. For example in CM $$U(t)( q_1(0) + q_2(0)) \neq U(t)q_1(0) + U(t)q_2(0)$$ Because the classical equations of motion are not linear you can't just take the time evolution from the states and give it to the observables.

You can (to some extent) put CM and QM on equal footing. For example this paper described classical mechanics in terms of wavefunctions so it is possible to define classical operators. But the classical equations are still not linear so this problem remains. Note that the classical wave equation has some issues so it is not exactly equivalent to the classical equations of motions.

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