# Heisenberg picture

1. Mar 26, 2012

### zonde

I would like to find out how popular is Heisenberg picture. Is there someone who finds Heisenberg picture useful?

And as well - are there any ideas how photon double-slit should be treated in Heisenberg picture?

2. Mar 26, 2012

### Matterwave

Heisenberg picture is especially useful in QFT where you allow the field operators to change in time, but the states (e.g. vacuum state) to be fixed.

3. Mar 27, 2012

### f95toli

It is not really a question of "popularity". You can change between different "pictures" just by a change of basis. Hence, you should work in whatever picture is the most convenient (mathematically).

It is not really any different from choosing say Cartesian or cylindrical coordinates when solving a PDE.

4. Mar 27, 2012

### naffin

From Wikipedia - "Heisenberg picture" :
"By the Stone-von Neumann theorem, the Heisenberg picture and the Schrödinger picture are unitarily equivalent."

As far as I know that theorem proves the equivalence between Heisenberg matrix representation and Schrodinger wave representation, time pictures are not involved: the two pictures are equivalent because the knowledge of a vector is the same thing as the knowledge of all its mean values on the observables.
Moreover, Stone-von Neumann theorem is about representations of x,p, while the equivalence of the two time pictures is more general.

5. Mar 27, 2012

### zonde

I understand that if you look at simple example consisting of one state vector and one operator it does not really matter which one is rotated and which one stays in place.
But what happens when you take more complicated examples? Therefore my question about double-slit.

In double-slit experiment state is split in two parts and when we measure them together there appears phase difference between two parts of the state. This phase difference makes measurable difference.
Now if we do not attribute time evolution to state we can not have phase difference between two parts of the state. But then time evolution is attributed to operator and so we should have this phase difference between two operators (that describe the same physical measurement device).
It does not seem like two pictures are physically equivalent.

And if we talk about photons Heisenberg picture seems more "natural" because photons moving at speed of light can not undergo any evolution along their trajectories. Only when you project photon ensemble on timelike surface you can talk about some evolution of photon ensemble.

6. Mar 28, 2012

### f95toli

But they are. They have to be, since you are always free to move between pictures while doing calculations. Note also that one of the most common pictures is the interaction picture, which is somewhere in-between the S and H pictures. It tends "pops out" quite naturally if you start with say the driven J-C Hamiltonian and move to a rotating frame.

Again, I don't' think this is any different from choosing your coordinate system when solving problems in classical mechanics: although different choices might appear to yield different physics (some choices leading to the appearance of fictitious forces such as the Coriolis force) we can presumably agree that this is just an illusion.

7. Mar 29, 2012

### zonde

I am trying to test this statement.
I took Mach-Zehnder interferometer with this as a hint how to do calculations.

For beam-splitter I have matrix:
$$\frac{1}{\sqrt{2}}\begin{pmatrix}1 & i\\i & 1\end{pmatrix}$$
right?

And evolution for state:
$$\begin{pmatrix}e^{i\alpha} & 0\\0 & e^{i\beta}\end{pmatrix}$$

So how do you move to Heisenberg picture?
If I leave out evolution of state but rotate beam-splitter matrix like that:
$$\frac{1}{\sqrt{2}}\begin{pmatrix}e^{i\theta} & e^{i(\theta+\pi/2)}\\ e^{i(\theta+\pi/2)} & e^{i\theta}\end{pmatrix}$$
nothing happens as it seems. So ... ?

8. Mar 30, 2012

### tom.stoer

It's rather simple: in experiments you deal with expectation values of observables A for states ψ. Of course these expectation values can change in time, so we have something like

$$\langle A \rangle_\psi (t)$$

The time-evolution is generated by the Hamiltonian H and implemented via a unitary time-evolution operator U which reads

$$U(t,t_0) = e^{-iH(t-t_0)}$$

This operator will propagate states or operators in time.

The time-dependence for <A> with the simple choice t0=0 reads

$$\langle A \rangle_\psi (t) = \langle\psi|\,U^\dagger(t)\,A\,U(t)\,|\psi\rangle$$

Now there are two options:

1) Schrödinger picture

In the Schrödinger picture one choses time-dependent states and time-independent operators, i.e.

$$\langle A \rangle_\psi (t) = \langle\psi|\,U^\dagger(t)\,A\,U(t)\,|\psi\rangle = \left(\langle\psi|\,U^\dagger(t)\right)\,A\,\left(U(t)\,|\psi\rangle\right) = \langle\psi(t)|_S\,A_S\,|\psi(t)\rangle_S$$

where 'S' means that we are now in the Schrödinger picture; of course AS is simply A and ψS(t) is nothing else but U(t)ψ.

2) Heisenberg picture

In the Heisenberg picture nothing changes but the brackets and the index 'H'. That means we are now talking about fixed states (fixed means no time evolution) and the dynamics is encoded in the operators.

$$\langle A \rangle_\psi (t) = \langle\psi|\,U^\dagger(t)\,A\,U(t)\,|\psi\rangle = \langle\psi|\,\left(U^\dagger(t)\,A\,U(t)\right)\,|\psi\rangle = \langle\psi|_H\,A_H(t)\,|\psi\rangle_H$$

Of course both expressions are identical, the only difference is which object - A or ψ - will contain the time dependence.

Think about a rotating object like the earth and the trajectory of a particle in the gravitational field of the earth. Already in classical mechanics you can change between two reference frames, one for a fixed observer and a rotating earth, one for a co-rotating observer. The difference is that in classical mechanics there is no difference between states and observables; the observable is just r(t) and this is identical to the state of the particle, so the analogy is not perfect. But in QM you have two different objects, namely states and operators, and therefore it's your choice which object is propagating in time and which one stays fixed. There is another big difference to classical mechanics: the operator U can be rather formal b/c H may be awfully complex and therefore U is even more complicated. So the expresseions above do not solve anything! They just shift the complexity from states to operators or vice versa w/o solving anything. It's like defining a reference frame using a rather complex or even unknown trajectory of a particle (not simply a rotating reference frame).

9. Mar 31, 2012

### zonde

So if take Mach-Zehnder interferometer as an example we can write time dependent operator of second beam-splitter like this, right?
$$\begin{pmatrix}e^{-i\alpha} & 0\\0 & e^{-i\beta}\end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix}1 & i\\i & 1\end{pmatrix} \begin{pmatrix}e^{i\alpha} & 0\\0 & e^{i\beta}\end{pmatrix}= \frac{1}{\sqrt{2}} \begin{pmatrix}1 & e^{i(\beta-\alpha+\pi/2)}\\e^{i(\alpha-\beta+\pi/2)} & 1\end{pmatrix}$$

Last edited: Mar 31, 2012
10. Mar 31, 2012

### Staff: Mentor

This only holds for systems with a finite number of degrees of freedom. In the case of infinite degrees of freedom, in QFT, there are known examples of unitarily inequivalent representations--for example, the representation natural to inertial observers is inequivalent to the representation natural to accelerated observers. This is the reason for the Unruh effect:

http://en.wikipedia.org/wiki/Unruh_effect

This is not specifically an inequivalence of the Heisenberg and Schrodinger pictures, but it certainly is an inequivalence of different representations of the same field state: the state that looks like a ground state (vacuum state) to inertial observers does *not* look like a vacuum state to accelerated observers.