# What is a “picture” in quantum mechanics?

One of the basic ingredients of quantum mechanics is the possibility of working in different "pictures". Thus, while we normally work in the Schrödinger picture, in which states evolve according to the Schrödinger equation $$i\partial_t |\psi(t)\rangle = H|\psi(t)\rangle,$$ it is sometimes convenient to work in the Heisenberg picture, in which you set the state of the system as fixed, and evolve the operators themselves through the Heisenberg equation, $$i\frac{\mathrm d}{\mathrm dt}A(t) = [H,A(t)] +i\frac{\partial A}{\partial t},$$ or even in a weird hybrid of those two called the interaction picture.

Generally speaking, textbooks do a good job of explaining the commonalities and differences between those three, showing that they are equivalent, and demonstrating how one can change from one picture to another. However, there's one question that's often left unanswered, and it then hangs in the air over the whole proceedings, giving them an unwarranted air of mystery in the eyes of a first-timer:

• what is, in the abstract, a "picture" in this sense?

This is part of what makes the formalism slightly unsettling to a newcomer, because the use of the phrase "Schrödinger picture" implies that the "Schrödinger" is is an adjective or modifier on the general term "picture", but that general term is never explained. Moreover, the Heisenberg and Schrödinger pictures are usually presented as very different formalisms and it's hard for a beginner to see how they can be understood as two versions of the same thing; if you could have that, then a "picture" would be a way to specialize the general formalism - but, again, that is rarely explained in introductory texts.

• how is the interaction picture "a weird hybrid" the two? As you outline in your answer correctly the Schrödinger and Heisenberg picture are limiting/special cases of the interaction picture. – Wolpertinger Feb 9 '17 at 13:41
• @Numrok I would suggest presenting this topic to undergraduates a few times and asking them how they feel about this. The whole point is that from the top down it's all very organized, but from the bottom up it can look like quite a jumble. – Emilio Pisanty Feb 9 '17 at 13:46
• I am an undergraduate... – Wolpertinger Feb 10 '17 at 7:25

To answer this, you need a bit of language. The fundamental observables of quantum mechanics are neither of states or operators; instead, the fundamental observables are matrix elements, of the form $$\langle\varphi |A|\psi\rangle.$$ In the most general case, you might have temporal dependence in the states, of the form $$i\partial_t |\psi(t)\rangle = H_\mathrm S|\psi(t)\rangle,$$ as well as time dependence in the operators, of the form $$i\frac{\mathrm d}{\mathrm dt}A(t) = [H_\mathrm H,A(t)] +i\frac{\partial A}{\partial t}.$$ If you combine these two, the temporal evolution of the fundamental observables is of the form $$i\frac{\mathrm d}{\mathrm dt}\langle\varphi(t)|A(t)|\psi(t)\rangle = \left\langle\varphi(t)\middle|\bigg[H_\mathrm S+H_\mathrm H, A(t)\bigg] +i\frac{\partial A}{\partial t}\middle|\psi(t)\right\rangle .\tag{*}$$ This is the fundamental time-evolution equation of quantum mechanics, where $H=H_\mathrm S+H_\mathrm H$ needs to be the full hamiltonian of the system, and every construction which is consistent with this form is equivalent in terms of its experimental predictions.

In this language, then,

a picture is a choice of how to partition $\boldsymbol H$ into $\boldsymbol H_{\mathrm{\mathbf S}}$ and $\boldsymbol H_{\mathrm{\mathbf H}}$.

Thus, the Schrödinger picture corresponds to the choice of $H_\mathrm H=0$ and $H_\mathrm S=H$, the Heisenberg picture is the opposite, and you tend to call 'interaction picture' a set of intermediate choices - though it should be clear that there's no unique such picture.

• I don't think that you have answered anything that wasn't included in the question by the author of the question who specifically said that the textbooks do a good job to explain the equivalence and other things you have described in your answer. Also, I am surprised why your name is exactly the same as his. ;-) – Luboš Motl Feb 7 '17 at 18:51
• @LubošMotl Perhaps you have a better answer to the specific question posed, then. Or maybe you think that it's just a word with no meaning? – Emilio Pisanty Feb 7 '17 at 18:54
• A bit less contentiously, the point is that even though the Schrödinger and Heisenberg pictures do look very different (here you do this thing, over there you do this other very different thing), they are indeed encased within a bigger framework where they are different versions of the same formalism (as opposed to completely different but formally equivalent frameworks). – Emilio Pisanty Feb 7 '17 at 18:57
• Dear Emilio, please don't take it personally. I was just amused by your posting both question and answer - it may often be a good idea to spread wisdom, don't get me wrong. The question and the answer were more similar than just by the author name. You've basically asked for something in the question that you didn't fulfill. But otherwise: the "picture" is just the English translation of "Bild" - Heisenberg Bild was arguably the first phrase of this kind - and it was used as a poetic synonym for a "representation". – Luboš Motl Feb 9 '17 at 12:20
• So Picasso draws a crazy cubist painting and says that it represents a female horserider, or whatever. So the painting is a "representation" of her and her horse. At the same time, the painting is a "Bild", or a "picture". This is an ideal informal enough word for a "representation" of the theory - there are some operations that represent the time evolution (either transformations of the states, or the operators). "Picture" sounds less rigorous than "representation" to it may allow that the pictures aren't exact representations in math sense and they're not quite equivalent. – Luboš Motl Feb 9 '17 at 12:22

The Heisenberg and Schroedinger pictures are actually not equivalent in an abstract setting.

The Schroedinger picture corresponds to studying the evolution of quantum states, the Heisenberg picture the evolution of quantum observables. Quantum observables form a topological $*$-algebra, and quantum states are elements of their dual. For simplicity, let us restrict to bounded observables, so they form a Banach space (a C$*$-algebra), as well as the states.

Now, the evolution is required to be a group of continuous linear maps in one of the two Banach spaces (states for Schr, observables for Heis). On observables, it is actually required to be a group of automorphisms of the algebra. Nonetheless, in both pictures we define a map $t\mapsto U (t)$, where the $U (t)$ are continuous linear operators in the respective Banach space. It is then natural to ask the continuity properties of the aforementioned map with respect to the different topologies available on the space of continuous linear operators on a Banach space. The group is then said to be e.g. weakly, strongly or uniformly continuous. Such continuity properties are very important to determine the structure of the group. For example, if we want states to obey the Schroedinger equation, we need the group to be strongly continuous.

The duality between the Schroedinger and Heisenberg pictures is supposedly given by the fact that the group of evolution on observables induces a group of evolution on the states by duality of topological spaces. In turn, a group of evolution on states induces a group of evolution on the double dual of observables (that is usually bigger than the space of observables). The problem is that the continuity properties of the group are not preserved by the duality. Uniform continuity is the only one that is preserved, but the generator of a uniformly continuous group is bounded (and we know that physically relevant Hamiltonians are not). Strong continuity is not preserved, and therefore even if we have a strongly continuous group in the Heisenberg picture, in general the dual group in the Schroedinger picture may fail to be so, and so the Schroedinger equation may fail to hold. On the other hand, the dual group on observables is not guaranteed to map observables into observables, for it may map to objects of the double dual that are not observables.

The word "picture" is the English translation of the term in the first language where this concept was discussed, namely the German word "Bild" (just like the German magazine). The "Heisenberg Bild" was the first picture that was known.

And this term isn't supposed to be anything else than an informal synonym of the word "representation". When an artist makes a painting, it's a representation of the object that he painted. So the picture is a representation among artists, too. The usage of this artists' jargon allows certain "inaccuracies". The picture doesn't have to be a "representation" in some particular rigorous sense defined by mathematicians. And the different pictures don't really need to be exactly and reliably and universally equivalent – even though I would choose the answer that the pictures are exactly equivalent.

But the Heisenberg, Schrödinger, and interaction representations are representations in the sense that particular operations that we know "phenomenologically", like the evolution by time $\Delta t$ (waiting for some time), are represented by different actual transformations on the mathematical objects in the theory – either by a transformation of the state vector or the operators or both.

In this sense, the word "picture" just indicates "a translation of the concepts that we know experimentally, independently of theories, into some particular mathematical operations and symbols". The translation to the language of mathematics is analogous to the artists' creation of a painting. So when you fully specify such a translation to mathematical symbol, you have painted a picture, a representation, of the world around you.

The pictures first emerged in quantum mechanics. Classical physicists have never really talked about "pictures". It's because all classical theories of physics were "pictures" by themselves. They were assumed to directly reflect what is out there, and the "translation" mentioned above was therefore unique and trivial. Only quantum mechanics has realized that the translation between mathematical objects in our theory and the observations may be a bit more subtle which is why it's desirable to talk about it, to admit all translations that are possible, and to study their equivalences if any.

• Sorry, but this doesn't actually address the question as posed. – Emilio Pisanty Feb 9 '17 at 13:13