# Can quantum mechanics be formulated without any reference to pictures?

NOTE: in the following with the word "picture" I refer to Schroedinger, Heisenberg, Interaction pictures, i.e. to the way the time-evolution is "distributed" between states and operators.

We often switch, according to what is more suitable to the problem under consideration, between various pictures (Schroedinger, Heisenberg, Interaction, various combinations of the above). This is purely, to my understanding, to ease formal manipolations, and the physical content of the theory is always independent of the picture used to describe it.

Is it possible to formulate quantum mechanics in a way that completely avoids the use of various pictures? The time evolution of the matrix elements $$\langle \alpha | \hat{O} | \beta \rangle (t)$$ is enough (is it?) to characterize the system at any time. Is it possible, at least in principle, to carry out calculations using just this and without any reference to pictures?

• Not even pure mathematics avoids pictures, since many important results are of the form: "If you are looking at the structures A over X, that is exactly the same as looking at the structures B over Y". (Examples are many abstract algebra results: To find the morphisms between two varieties is exactly the same as finding the morphisms between their rings of regular functions in the inverse direction). I guess what I'm saying is: What would it mean not to have a picture, if is it perfectly possible to have different mathematical theories be equivalent? – ACuriousMind Oct 16 '14 at 19:10
• A "picture" is merely there to help humans develop an intuition. Ideally you don't have one picture of a physical process ready but half a dozen or more, and you need to know where the usefulness of these pictures ends. Mathematics alone is meaningless, there is no more an algorithm which would allow us to reason our way trough physics than there is an algorithm which allows mathematicians to do mathematics. The mathematicians even have a proof for that: look up formal logic and Goedel. – CuriousOne Oct 16 '14 at 19:58
• But the pictures are exactly the same. You can derive one from the other. Every result that holds in one holds in the others. There is no result (known to me, at least) about physically observable things that holds in one but not the others. And why would "studying time averages" (I'm not sure what precisely you mean) be qualitatively different from the other pictures, i.e. what makes it not just another picture? – ACuriousMind Oct 16 '14 at 20:08
• These "pictures" (terrible word) are just coordinate systems in a way. There's no lack of formality or rigour. Using different pictures is analogous to changing variables to do an integral. – DanielSank Oct 16 '14 at 20:13
• @glance: the fact that you ask if a more rigorous way to do QM without these 'pictures' is what gives me the idea that you think the Heisenberg and Schrodinger are flawed. – Kyle Kanos Oct 16 '14 at 20:16

As you say, pictures in QM are merely different ways to compute the exact same thing - the expectation value of observables $\mathcal{O}$. Now, in all pictures, we can derive Ehrenfest's theorem
$$\frac{\mathrm{d}}{\mathrm{d}t}\langle \mathcal{O} \rangle = \frac{1}{\mathrm{i}\hbar}\langle [\mathcal{O},H]\rangle + \langle\frac{\partial}{\partial t}\mathcal{O}\rangle$$
• 2 related questions: 1. Is it enough to talk about average values $\langle \mathcal{O} \rangle$? Shouldn't also non-diagonal matrix elements $\langle\alpha|\mathcal{O}|\beta\rangle$ be characterized? 2. Writing Ehrenfest's theorem this way, aren't we again making reference to a particular picture in which the operators are expressed? But if I had to express QM only using average values, I shouldn't talk about operators anymore, as these are necessarily expressed in some picture. This raises some problems: how should then something like $\langle [\mathcal{O},H]\rangle$ be interpreted? – glS Oct 17 '14 at 4:46
• @glance: There's no "picture" in which operators or states are expressed. All that the "picture" you choose does is to tell you "where" time evolution goes. Essentially, you know that $\langle \alpha \rvert U^\dagger(t,t_0) \mathcal{O} U(t,t_0) \lvert \beta \rangle$ is the time evolution of a matrix element, and it is equivalent to consider the operator evolving or the states evolving or some mixture ala interaction picture. (Since Ehrenfest theorem is equivalent to the Schrödinger equation, it already contains the evolution of such matrix elements.) – ACuriousMind Oct 17 '14 at 15:03