How to add time variation to a Schrodinger operator?

[snoopies622]

TL;DR Summary

I'm not clear on exactly how one switches from the Schrodinger picture to the Heisenberg picture.

I'm looking at Dirac's "Lectures on Quantum Field Theory" and I have a question about the basic mathematics of something that's part of ordinary quantum mechanics. On page 3, he says,

The two pictures are connected in this way: any Schrodinger dynamical variable is connected with the corresponding Heisenberg dynamical variable by the transformation

$$u_{S} = e^{-iHt/\hbar} u_{H} e^{iHt/ \hbar}$$

and state vectors are correspondingly connected by
$$| A_{S}> = e^{-iHt/\hbar} | A_{H}>$$

In the first equation, are the $u_{S}$ and $u_{H}$ matrices while the exponential terms are simply (time dependent) complex numbers? If so, how does one multiply a matrix by a scalar like a complex number? It can't be just multiplying every term in the matrix by both of the exponential terms, because in this case they are inverses of each other and would cancel out.

If the exponential terms aren't complex numbers, what are they?

 

[PeterDonis]

If the exponential terms aren't complex numbers, what are they?

The $H$ in the exponentials is the Hamiltonian, which is an operator. So the exponentials are exponentiated operators, which can be understood as the corresponding power series in the operators. So $\exp(i H t) = 1 + i H t + ( i H t )^2 / 2 + ...$.

 

[snoopies622]

Oh wow, thanks Peter. More complicated than i expected. I thought in this case H was simply the energy eigenvalue. (This is why i wish textbooks would consistently use a "hat" over operators or something else to distinguish them from actual numbers.)

 

[vanhees71]

Although in almost all QM 1 lectures you take a lot of the time to evaluate energy eigenvectors and eigenvalues, because these are the stationary states of the quantum system under investigation, the more general and conceptually important case is the dynamics of the states (statistical operators) and observables (or rather their representing self-adjoing operators).

Further QM is invariant under arbitrary time-dependent unitary transformations, i.e., you can choose an arbitrary "picture of time evolution". You can describe the time dependence of the observable physics in quite arbitrary ways without changing them. Obviously Dirac discusses the two cases of the Heisenberg and Schrödinger pictures.

In the Heisenberg picture the entire time dependence is in the self-adjoint operators that describe observables, and the statistical operator (or for pure states the state vector) is constant. This is, in my opinion, the most intuitive "picture": The state represents the preparation of the system before a measurement is made, i.e., the state is given by an "initial condition". For a pure state it's given by a normalized state vector $|\psi_{\text{H}} \rangle=\text{const}.$ The time evolution is for the observables and the corresponding eigenvectors of their representing operators. For a complete set of compatible observables these eigen kets are uniquely determined (up to phase factors), $|a_1,\ldots,a_n,t \rangle=\exp(\mathrm{i} \hat{H} t/\hbar) |a_1,\ldots,a_n,t=0 \rangle$ (assuming that $\hat{H}$ is not explicitly time dependent). Then the probability (density) to measure the values $(a_1,\ldots,a_n)$ when measuring the complete compatible set of observables at time time is given by
$$P_{\psi}(a_1,\ldots,a_n,t)=|\langle a_1,\ldots,a_n,t|\psi_{\text{H}} \rangle|^2= |\langle a_1,\ldots,a_n,t=0|\exp(-\mathrm{i} \hat{H} t/\hbar)|\psi_{\text{H}} \rangle|^2.$$
Now in the Schrödinger picture you define
$$|\psi_{\text{S}}(t) \rangle=\exp(-\mathrm{i} \hat{H} t/\hbar)|\psi \rangle_{\text{H}}, \quad |a_1,\ldots a_n \rangle_{\text{S}}=|a_{1},\ldots,a_n,t=0 \rangle=\text{const},$$
and then obviously
$$P_{\psi}(a_1,\ldots,a_n,t)=|{_{\text{S}}} \langle a_1,\ldots,a_n|\psi_{\text{S}}(t) \rangle|^2.$$
So the physics is independent whether you choose to evaluate the time evolution in the Heisenberg or Schrödinger picture.

When doing time-dependent perturbation theory, you choose some picture in between, where the state kets time evolve with the "perturbation part" of the Hamiltonian and the observable operators (and thus their eigenvectors) with the "unperturbed part" of the Hamiltonian. A special case is the "interaction picture": There the observable-operators evolve with the Hamiltonian of the free particles and the state kets with the interaction (potential) part of the Hamiltonian. These "mixed pictures" of time evolution are also called the "Dirac pictures", because it was in fact Dirac who developed the representation-independent formulation of QT in terms of abstract operators and (rigged-)Hilbert-space vectors.

 

[snoopies622]

The state represents the preparation of the system before a measurement is made, i.e., the state is given by an "initial condition".

Thanks vanhees71, this is helpful. Up to now, for me the Schrodinger picture has always been the only one that had any intuitive meaning, but reading the next page or so of this Dirac book, he basically says that it must be abandoned for quantum field theory - it just doesn't work. One of the problems for me has simply been referring to the "state vector" as just that, as if it represents the state of a system at any moment, like a point in phase space in a classical system. But to think of it as just "the starting point" changes everything.

 

[Haborix]

Maybe this isn't a useful observation, but the notion of pictures isn't unique to quantum mechanics. In fluid mechanics, there is the Eulerian and Lagrangian "pictures" of the flow. Although, I can't imagine the equivalent of the interaction picture in fluid mechanics. My hunch is that a mixed description in fluid mechanics fails because of the underlying non-linearity of evolution equation. Anyway, I share this observation because I always found it helpful to recognize when quantum mechanics wasn't any more strange than other areas of physics. Quantum mechanics is weird enough.

 

[snoopies622]

I always want a picture, but with QM it seems one can only go so far. I remember another quote from Dirac that went something like, "there can't be a 'picture' in quantum mechanics because when people say 'picture' what they have in mind is something classical."

 

[vanhees71]

Thanks vanhees71, this is helpful. Up to now, for me the Schrodinger picture has always been the only one that had any intuitive meaning, but reading the next page or so of this Dirac book, he basically says that it must be abandoned for quantum field theory - it just doesn't work. One of the problems for me has simply been referring to the "state vector" as just that, as if it represents the state of a system at any moment, like a point in phase space in a classical system. But to think of it as just "the starting point" changes everything.

That's not true either. You can formulate also QFT in the Schrödinger picture. It's only not very natural, and the fundamental principles are easier to express in the Heisenberg picture, where time and space are treated on the same footing, i.e., as parameters, labelling the causal structure and the field-degrees of freedom, respectively. A good QFT book, which treats also the Schrödinger picture, is

B. Hatfield, Quantum Field Theory of Point Particles and
Strings, Addison-Wesley, Reading, Massachusetts, 10 edn.
(1992).

 

[snoopies622]

Thanks for the book recommendation. Check out Dirac's comments 25:35 - 27:04 here, which might explain partly why he's so hard on Schrodinger in his QFT book.

 

[Demystifier]

Up to now, for me the Schrodinger picture has always been the only one that had any intuitive meaning, but reading the next page or so of this Dirac book, he basically says that it must be abandoned for quantum field theory - it just doesn't work.

The Schrodinger picture in QFT is problematic due to the infinite number of degrees of freedom. However, the Heisenberg picture in QFT (or any other picture, for that matter) is also problematic, for the same reason. The infinite number of degrees of freedom manifests as UV and IR divergences. In practical calculations those divergences should be regulated somehow, and when this is done, the Schrodinger (or any other) picture works fine again. At the end one wants a result which does not explicitly depend on the regularization, and the technical procedure that achieves that is called renormalization.

 

[andresB]

I recommend the OP to take a look at the solution of the Harmonic oscillator in the Heisenberg picture given in Sakurai. It is as clear as it gets.

 

[snoopies622]

Thanks andresB, checking it out now. Looks similar to what Dirac presents in his The Principles of Quantum Mechanics (section 34), though a little more accessible.