# Time-dependence in probability amplitudes

• I
Hi all, I am rather confused about the following concept. Assistance is greatly appreciated!

A time-dependent probability amplitude can be written as
$$\langle a_k| e^{-\frac{i}{\hbar}\hat{H}t} |\psi\rangle$$
where ##a_k## is an eigenvalue. Suppose I want the x-representation of the ket, I can then choose to write this as
$$\langle x,t |\psi\rangle$$
or as
$$\langle x |\psi(t)\rangle$$

While I understand that either way is equally valid, and makes no difference to the physics, how does it change the way I interpret things?

For example, if I was observing how some atom prepared in state ##|\psi\rangle## propagates in space, could I interpret the ##\langle x |\psi(t)\rangle## case as me having my apparatus fixed at one point in space, while the particle evolves (propagates relative to my apparatus) as time evolves?

What then, would be the interpretation for the ##\langle x,t |\psi\rangle## case? Is this related to the idea that the operator/observable evolves, rather than the state (though I don't really understand what this means either).

Last edited:

vanhees71
Gold Member
2021 Award
It doesn't change anything. What you discovered here is the possibility to quite arbitrarily put the time dependence on the state (statistical) and observable operators. What you discuss are the two "extreme cases", i.e., the Schrödinger picture (only the statistical operators are time dependent, and the observable operators are time-independent) and the Heisenberg picture (only the observable operators are time independent, while the statistical operators are time independent). The probabilities to find a certain value of an observable as a function of time, which are the physical observational outcomes of the formalism, are of course independent on this choice of the "picture of time evolution". For the Schrödinger and Heisenberg pictures it's easy to see. Take a time-independent Hamiltonian. Then the Heisenberg picture is defined by the equations of motion of the statistical operator and the observable operators (for simplicity we take an observable ##A## with only non-degenerate eigenstates):

Schrödinger picture

$$\frac{\mathrm{d}}{\mathrm{d} t} \hat{\rho}_S(t)=\mathrm{i} [\hat{\rho}_S,\hat{H}_S], \quad \hat{A}_S(t)=\hat{A}_S(0).$$
The solution of the equation of motion for the statistical operator obviously is
$$\hat{\rho}(t)=\exp(-\mathrm{i} \hat{H}_S t) \hat{\rho}_S(0) \exp(\mathrm{i} \hat{H}_S t).$$
For a pure state, you can also use the state ket, and then up to a relevant phase the time dependence can be chosen as
$$|\psi_S(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi_S(0) \rangle.$$
The probability to find the eigenvalue ##a## of the self-adjoint observable operator ##\hat{A}_S## is given by (the eigenvectors of \hat{A}_S can be chosen as time independent up to an irrelvant phase):
$$P(t,a)=\langle a_S|\hat{\rho}_S(t)|a_S \rangle=\langle a_S|\exp(-\mathrm{i} \hat{H} t) \hat{\rho}_S(0) \exp(\mathrm{i} \hat{H} t)|a_S \rangle.$$

Heisenberg picture

To get the Heisenberg picture we take a shortcut, i.e., we use the Schrödinger picture to derive it via a time-dependent unitary transformation. The aim is to make the statistical operator time independent by defining
$$\hat{\rho}_H(t)=\exp(\mathrm{i} \hat{H}_S t) \hat{\rho}_{S}(t) \exp(-\mathrm{i} \hat{H}_S t)=\hat{\rho}_S(0)=\text{const}.$$
Then of course we must also transform the observable operators with the same unitary transformation,
$$\hat{A}_H(t)=\exp(\mathrm{i} \hat{H}_S t) \hat{A}_S \exp(-\mathrm{i} \hat{H}_S t).$$
Applied to ##\hat{H}_S## itself, you immediately see that
$$\hat{H}_H(t)=\hat{H}_S=\text{const}$$
in this case.

Now it's clear that in the Heisenberg picture the eigenvectors of ##\hat{A}(t)## for any eigenstate ##a## is given by
$$|a_H(t) \rangle=\exp(\mathrm{i} \hat{H}_S t) |a_S \rangle,$$
and thus the probability to find the value ##a## when measuring ##A## can again be evaluated with the Heisenberg picture objects:
$$P(t,a)=\langle a_H(t) |\hat{\rho}_H(t)|a_H(t) \rangle=\langle a_S|\exp(-\mathrm{i} \hat{H}_S t) \hat{\rho}_S(0) \exp(\mathrm{i} \hat{H}_S t|a_S \rangle,$$
which is the same as given at the end of our calculation within the Schrödinger picture.

The most general case, where you can more or less freely choose the time dependence of the statistical and the observation operators is called the Dirac picture of time evolution. The most important application is to use the socalled interaction picture to derive time-dependent perturbation theory.

kent davidge
Hi, thank you for your response.

I think I'm beginning to see the equivalence in both these pictures, however I do still have an issue with the interpretation.

My lecturer stated that the Schrodinger picture, one can imagine that an experimenter prepares atom in some state ##|\psi(0)\rangle## and allows it to evolve under some Hamiltonian ##\hat{H}## to ##|\psi(t)\rangle##. The obervable/apparatus does not change with time.

He then seemed to state that in the Heisenberg picture - where as you've quite helpfully pointed - the time dependence lies in the eigenkets and thus the Observable/Apparatus. The experimenter then prepares some state ##|\psi\rangle##, but this doesn't change with time, but instead it is our Observable/Apparatus that evolves under the specified Hamiltonian. He also states that this is the better way to think about things, rather than the state of the atoms evolving in time.

Is this a valid way of thinking about the pictures?

DrClaude
Mentor
The two pictures can be thought (in a simplified way) as a change in frame of reference. In the Schrödinger picture, you are putting yourself in the reference frame of the lab, where the apparatus is fixed, and it is the quantum system that changes with time. In the Heisenberg picture, you put yourself in the reference frame of the quantum system, so the state is fixed, but you see the lab around the atom as changing with time.

He also states that this is the better way to think about things, rather than the state of the atoms evolving in time.
That's hogwash. One could argue that, depending on context, one approach is more useful than the other, but that doesn't necessarily make it "a better way to think about things." Also, I would argue that in most cases, the Schrödinger picture is more "natural," as it corresponds to the usual situation of the experimenter being in the reference frame of the aparatus.

PeroK
vanhees71
Gold Member
2021 Award
Hi, thank you for your response.

I think I'm beginning to see the equivalence in both these pictures, however I do still have an issue with the interpretation.

My lecturer stated that the Schrodinger picture, one can imagine that an experimenter prepares atom in some state ##|\psi(0)\rangle## and allows it to evolve under some Hamiltonian ##\hat{H}## to ##|\psi(t)\rangle##. The obervable/apparatus does not change with time.

He then seemed to state that in the Heisenberg picture - where as you've quite helpfully pointed - the time dependence lies in the eigenkets and thus the Observable/Apparatus. The experimenter then prepares some state ##|\psi\rangle##, but this doesn't change with time, but instead it is our Observable/Apparatus that evolves under the specified Hamiltonian. He also states that this is the better way to think about things, rather than the state of the atoms evolving in time.

Is this a valid way of thinking about the pictures?

This is a very strange statement, I cannot make any sense of. It's simply that the choice of the "picture of time evolution" is pretty arbitray, since you can always transform from one picture of time evolution to any other by using an arbitrary time-dependent unitary transformation of both the statistical operator (representing the state) (or the Hilbert-space vectors for the case of pure states) and the observable operators.

You cannot distinguish this transformation in any physics way. In other words you cannot define it as an "active transformation". It's more like a gauge transformation in classical elecrodynamics, where different electromagnetic potentials only different by a gauge transformation, represent perfectly the same physical situation.

Note that the self-adjoint operators representing observables do not represent measurement devices. A measurement device is rather a macroscopic system coupled to the measured (quantum) system, i.e., it is describe quantum mechanically by an effective coarse-grained description since you cannot resolve all its microscopic details.

Hi @vanhees71 and @DrClaude , thank you both for your replies.

I am unable to piece together what the both of you have said. If I understood it correctly, @DrClaude 's post says that the pictures can be seen as swapping frames of reference between that of the experimenter and the quantum system. However @vanhees71 's says that there isn't such a distinction...?

vanhees71
Gold Member
2021 Award
I don't agree with @DrClaude here. The operations with operators on Hilbert spaces cannot be mapped one-to-one with what an experimenter does in the lab. In the lab an experimenter works with macroscopic objects like particle accelerators and detectors (sometimes both huge as at the LHC).

The mathematical operations on Hilbert space are calculational tools to predict probabilities for the outcome of measurements and preparation procedures. E.g., making a beam of neutrons split in an inhomogeneous magnetic field in a Stern-Gerlach experiment and then blocking out one of the partial beams is what the experimenter does when he wants do prepare the neutron to have a spin component "up" in the direction given by his/her magnetic field. He uses real-world material, no vectors in Hilbert space and self-adjoint or unitary operators.

As a theorist he/she describes this in the Hilbert-space formalism as "I have prepared the spin-3 component of a neutron in the state represented by the normalized eigenvector ##|\sigma_z=+\hbar/2 \rangle## of the spin-3 component represented by the operator ##\hat{s}_3##".

Concerning the choice of the picture of time evolution, it's just a matter of convenience for a given task to calculate probabilities, expectation values, etc. All these quantities are independent on this choice. There is no difference in the processes and observations in the real world by just changing the description of these processes and observations in the formalism.

An analogy in everyday life is how to tell where you live. You can give the city, street and house number or, e.g., GPS coordinates. This doesn't change where your home is located; it's just two different and equivalent ways to describe your location.

Mentz114
Concerning the choice of the picture of time evolution, it's just a matter of convenience for a given task to calculate probabilities, expectation values, etc. All these quantities are independent on this choice. There is no difference in the processes and observations in the real world by just changing the description of these processes and observations in the formalism.

An analogy in everyday life is how to tell where you live. You can give the city, street and house number or, e.g., GPS coordinates. This doesn't change where your home is located; it's just two different and equivalent ways to describe your location.

Aah, I see. This cleared things up a little, many thanks!