Ensembles in quantum field theory

[A. Neumaier]

Yes

To what did you here agree?

and for this a single-particle-observable expectation value is not sufficient. For this you need a many particle quantity as I said before.

I don't see how a many-particle description is needed for a process in which there is at most one particle at any time.

That's described in quantum-optics textbooks.

I also don't see how quantum optics is relevant for the nonrelativistic QFT modeling of a silver beam.

 

[Demystifier]

I don't see how a many-particle description is needed for a process in which there is at most one particle at any time.

First, you want to describe the lab. Second, if a late particle does not exist at earlier times, then it's created from something and this something is also made of particles.

 

[A. Neumaier]

First, you want to describe the lab.

No, I only wanted to describe the beam in the lab, represented by a free field; see posts #3, #19, #21. (The Heisenberg state of the lab would not be the vacuum state and its fields would have to be interacting.)

Second, if a late particle does not exist at earlier times, then it's created from something and this something is also made of particles.

That something is the silver oven, which is outside the region modeled. Its influence can therefore be solely through the boundary conditions for the free field operators at the beginning of the beam. Specifying this boundary condition at all times (together with the free field equations) fully captures the dynamics of the beam alone.

 

[vanhees71]

To what did you here agree?

I don't see how a many-particle description is needed for a process in which there is at most one particle at any time.

I also don't see how quantum optics is relevant for the nonrelativistic QFT modeling of a silver beam.

I agreed with the statement that usually one considers single-particle states and the corresponding probabilities in the standard description of the SGE. I don't know, what you mean by the statement that it's not described within QFT since of course it is described by QFT when it's necessary. For photons you find this in all details in standard textbooks on quantum optics, and not only for single-photon observations but also for two- and multi-photon observations. The latter usually is needed when dealing with entangled states of all kinds or for Hanbury-Brown twiss ("intensity interferometry") etc.

So it depends on which description you want. First you said you want to describe a situation, where every 1 second a silver atom is emitted from some source. This is described by a many-body state, i.e., for $N$ particles (to be "delivered" within $N$ seconds), i.e., an $N$-point correlation function. Which one depends on the detector. I'd say a good detector is measuring the charge/current density.

 

[A. Neumaier]

I don't know, what you mean by the statement that it's not described within QFT

I only meant that in the literature I haven't seen a discussion that treats the process as a long-time nonstationary process involving a sequence of several generates particles. In the present discussion I want to inquire about this nonstationary description in terms of QFT, which indeed exists, I believe.

First you said you want to describe a situation, where every 1 second a silver atom is emitted from some source.

Yes, that's the standing assumption.

This is described by a many-body state, i.e., for $N$ particles (to be "delivered" within $N$ seconds), i.e., an $N$-point correlation function.

Since we don't generate entangled particles and don't perform coincidence tests, the use of multiparticle correlation functions seems not warranted. For photons, the spatially integrated response rate to a single beam is proportional to the 1-point function of the energy density, which can be expressed in terms of two-point functions of the electromagnetic field. For silver, you had suggested instead the integrated 3-component of the silver current (which can be converted to an integral over silver density using the conservation of mass). Neither involves correlation functions.

Moreover, I thought we had agreed that the Heisenberg state is the vacuum state, and the dynamics is solely in the operators. Thus the information about how often particles are emitted must be in the incoming boundary conditions of the operators. How would you accommodate your many-body-state view in the operator boundary conditions?

 

[vanhees71]

Let's discuss it first within non-relativistic QFT (although there's no principle problem to do the same in relativistic QFT as far as I can see).

You just define operator-valued wave packets $\hat{\phi}^{(-)}(t,\vec{x})$ (pure creation operators) as solutions of the Schrödinger equation for particles moving in a static inhomogeneous magnetic field which have a peak every 1 second. Then you define the states
$$|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle = \hat{\phi}^{(-)}(t_1,\vec{x}_1) \hat{\phi}^{(-)}(t_2,\vec{x}_2) \cdots \hat{\phi}^{(-)}(t_N,\vec{x}_N) |\Omega \rangle.$$
Then the function
$$P(t_1,\vec{x}_1;\ldots;t_2 \vec{x}_2)=|\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle|\Omega \rangle|^2$$
describes the probability for registering $N$ particles at times $t_j$ and positions $\vec{x}_j$. These particles are as unentangled as they can be. The only entanglement is due to the bosonic or fermionic nature implemented in the field operators (usually one defines such a state as a state of unentangled indistinguishable particles).

 

[A. Neumaier]

Let's discuss it first within non-relativistic QFT (although there's no principle problem to do the same in relativistic QFT as far as I can see).

Since silver is heavy, there is indeed no need for relativistic QFT.

You just define operator-valued wave packets $\hat{\phi}^{(-)}(t,\vec{x})$ (pure creation operators) as solutions of the Schrödinger equation for particles moving in a static inhomogeneous magnetic field which have a peak every 1 second.

This seems conceptual progress on our question, since now the dynamics is in the operators. But in the setting under discussion we do not have a magnetic field, just occasional silver particles. So what does have a peak every second?

Then you define the states
$$|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle = \hat{\phi}^{(-)}(t_1,\vec{x}_1) \hat{\phi}^{(-)}(t_2,\vec{x}_2) \cdots \hat{\phi}^{(-)}(t_N,\vec{x}_N) |\Omega \rangle.$$
Then the function
$$P(t_1,\vec{x}_1;\ldots;t_2 \vec{x}_2)=|\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N |\Omega \rangle|^2$$
describes the probability for registering $N$ particles at times $t_j$ and positions $\vec{x}_j$.

The typical lab experiment does not resolve the times and positions of each impacting silver particle but only checks the areas where a silver spot forms after some time; thus only an integrated version of this is needed and everything should reduce to single particle matrix elements. But let us consider the probabilities later. At the moment I am primarily interested in getting a correct description of the beam before it hits the screen.

These particles are as unentangled as they can be. The only entanglement is due to the bosonic or fermionic nature implemented in the field operators (usually one defines such a state as a state of unentangled indistinguishable particles).

Agreed.

 

[vanhees71]

Let's discuss it first within non-relativistic QFT (although there's no principle problem to do the same in relativistic QFT as far as I can see).

You just define operator-valued wave packets $\hat{\phi}^{(-)}(t,\vec{x})$ (pure creation operators) as solutions of the Schrödinger equation for particles moving in a static inhomogeneous magnetic field which have a peak every 1 second. Then you define the states
$$|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle = \hat{\phi}^{(-)}(t_1,\vec{x}_1) \hat{\phi}^{(-)}(t_2,\vec{x}_2) \cdots \hat{\phi}^{(-)}(t_N,\vec{x}_N) |\Omega \rangle.$$
Then the function
$$P(t_1,\vec{x}_1;\ldots;t_2 \vec{x}_2)=|\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle|\Omega \rangle|^2$$
describes the probability for registering $N$ particles at times $t_j$ and positions $\vec{x}_j$. These particles are as unentangled as they can be. The only entanglement is due to the bosonic or fermionic nature implemented in the field operators (usually one defines such a state as a state of unentangled indistinguishable particles).

Correction: This is of course nonsense, because it's zero ;-). Right is
$$P(t_1,\vec{x}_1;\ldots;t_2,\vec{x}_2)=\langle t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N|t_1,\vec{x}_1;\ldots;t_N,\vec{x}_N \rangle.$$
This can be written in terms of the $N$-point correlation function of the density $\hat{\phi}^{-}(t,\vec{x}) \hat{\phi}^{(+)}(t,\vec{x})$.

 

[vanhees71]

Since silver is heavy, there is indeed no need for relativistic QFT.This seems conceptual progress on our question, since now the dynamics is in the operators. But in the setting under discussion we do not have a magnetic field, just occasional silver particles. So what does have a peak every second?

The field operators. Without magnetic field you can just use the solution of the free-particle Schrödinger field. It's easy to construct Gaussian wave packets peaking around arbitrary positions and times.

The typical lab experiment does not resolve the times and positions of each impacting silver particle but only checks the areas where a silver spot forms after some time; thus only an integrated version of this is needed and everything should reduce to single particle matrix elements. But let us consider the probabilities later. At the moment I am primarily interested in getting a correct description of the beam before it hits the screen.

Sure, that's much simpler. There you only need the current operator [EDIT: technical typo corrected]
$$
\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m}
\hat{\psi}^{\dagger}(t,\vec{x}) \overleftrightarrow{\nabla}
\hat{\psi}(t,\vec{x})
$$
changing the notation for the usual Schrödinger field operators, which are pure annihilation operators $\hat{\psi}$ and creation operators $\hat{\psi}^{\dagger}$ and as initial state an appropriate one-particle state. Then you integrate the expectation value over the time of observation to get the accumulated probability distribution on the screen.

 

[A. Neumaier]

The field operators. Without magnetic field you can just use the solution of the free-particle Schrödinger field. It's easy to construct Gaussian wave packets peaking around arbitrary positions and times.

Ok, this looks more satisfying.

Sure, that's much simpler. There you only need the current operator
$$\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m} \hat{\psi}^{\dagger}(t,\vec{x}) \leftrightarrow{\nabla} \hat{\psi}(t,\vec{x})$$
changing the notation for the usual Schrödinger field operators, which are pure annihilation operators $\hat{\psi}$ and creation operators $\hat{\psi}^{\dagger}$ and as initial state an appropriate one-particle state. Then you integrate the expectation value over the time of observation to get the accumulated probability distribution on the screen.

Your latex formula doesn't display as it should since $\leftrightoverarrow$ is not recognized and a curly bracket is missing. Is my correction what you had intended? It doesn't quite make sense...

I'll think about your formulas to see what they mean in terms of my intuitive picture of the situation.

 

[Fra]

Of course, QFT is not restricted to relativistic QFT but can be as well formulated for non-relativistic QM as well. In the cases, where you only deal with systems, where the conserved current has the meaning of a particle-number current, this QFT formulation of non-relativistic QM is equivalent to the first-qantization formulation since then you always stay in the subspace of the Fock space with the particle number that was fixed in the beginning.

The reason, why there is no viable 1st-quantization formulation for relativistic QT and interacting particles is that the conserved currents of relativistic wave equations, which follow from invariance under proper orthochronous Poincare transformations as the symmetry group of Minkowski space, provide no positive definite densities and thus are charge rather than particle-number densities. Further to ensure causality in terms of the microcausality constraint in connection with local realizations of the Poincare group you need both annihilation and creation operators in the mode expansion of the free fields, and in the interacting case not particle number but the charges are conserved, and thus you need the full Fock space to get a consistent description.

My perspective was more the foundational one - ie the issues of interpretation and mapping elements of theory to elements of "reality" (which by I mean elements of the the observer/agent) is existing already in normal QM. But the problems become more accentuated when you consider a quantum field theory.

Suppose you have something that a copenhagen interpretation of QM, what is the copenhagen interpretation of QFT?

I think a bit like this: If we think the quantum state, represents the "state of the knowledge" the observer has about a "particle", then the fock space corresponds to the "state of the knowledge" the observer has about the "field", but in "second quantization" the "field" is not a classical field, but reprents just one step in an induction chain - it's a higher level knowledge about the lower level knowlwedge; ie the observer starts (as part of internal processes, which are physical of course) to "reflect" over it's own information, noting that say probabilities are not conserved, and he lower level of logical processing seems to not be viable, so a higher level construction is to consider the "probability" for the "probability following from a simplere inference system" combined with a changed of dependent variables, which can be conceptually thought of as a change of coding/representation, into something more efficient (that is say easier to truncate). Change of variables from KG -> Dirac equation is I think an "example".

Once you get the point of this, there is notthing to stop you also from a third quantisation and thus n'th quantization; each level represents more complex inference system withint the agent, affecting also it's interaction properties. At with n does this stop? - possible when the additional computational complexity of higher orders stalls the agent, more than it helps it?

The conventional method that you get taught isn't like this, it's more

Often, the first step in introducing QFT is "trying" to just plug operators into the klein gordon equation in order to illustrate that, we get the strange negative energy solutions and conservation of probability just goes down the drain - in short, it does not seem to make any sense. So we make say make the ansatz of the linear dirac equation, "relabel" some of the "particle states" as one beeing the anti-particle etc.

But when doing this change of "dependent variable", even mixing it up with the prior spacetime dynamics - what is happening to the "interpretation" of the information of the agent?

I think the conventional method is to just conclude that the 1st quantization doesn't make sense, and neither does the above, so instead one just arbitrarily (by ansatz) introduces more dependent variables, whose "interpretation" to the prior non-viable level is ignored.

One can also just say that the "ansatz" of the original observer was "wrong". ie the observer was convinved it was a one-particle system, but as the predictions based one that ended up non-correctable by changing the intial conditions, he was "wrong" as it was a multi-particle system.

This disturbs me alot as it avoids the problems. I think one can not questions the observers "best knowledge" unless it's of the "ignorance/Bell type" the incompleteness that is due to that is seems physically impossible for an agent to infer with certainty, say initial conditions and the effective laws, will unavoidable I think give predictable consequences.

We just have a collection of $N$-point functions for each Heisenberg state.

If by "we" here mean the "observer/agent", the question I ask is: How did the agent infer the baggage implied here, without adding in "external information" (which at least is what I strive to avoid, but which is of course very difficult it not impossible).

Or if we by "we" instead mean the collection of all "classical" measurementdevices, distributed throughout the universer that can build the records post-aquistion, then that line of thinking deviates from the path I try to stay on

/Fredirk

 

[A. Neumaier]

If by "we" here mean the "observer/agent", the question I ask is: How did the agent infer the baggage implied here, without adding in "external information"

"we" is the user of QFT.

We use a lot of external information from books, journals, discussions, and experiments to create the model from which to calculate the N-point functions.

For example, in this thread we try to figure it out for a simple silver beam on the way from the source to the screen.

 

[vanhees71]

Ok, this looks more satisfying.

Your latex formula doesn't display as it should since $\leftrightoverarrow$ is not recognized and a curly bracket is missing. Is my correction what you had intended? It doesn't quite make sense...

I'll think about your formulas to see what they mean in terms of my intuitive picture of the situation.

I corrected it:
$$
\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m}
\hat{\psi}^{\dagger}(t,\vec{x}) \overleftrightarrow{\nabla}
\hat{\psi}(t,\vec{x})$$
It's a usual symbol in QFT it means
$$\hat{\psi}^{\dagger}\overleftrightarrow{\nabla} \hat{\psi} = \hat{\psi}^{\dagger} \nabla \hat{\psi} -(\nabla \hat{\psi}^{\dagger}) \psi.$$
It's the standard probability current of the Schrödinger equation. From the Noether symmetry under multiplication with space-time independent phase factors it follows that
$$\partial_t \rho +\nabla \cdot \vec{j}=0, \quad \rho =\psi^{\dagger} \psi.$$
I find it much more intersting to include the magnet in the analysis. It's one of the few examples, where you can solve the Schrödinger equation quasi analytically, treating the spin-flipping part of the magnetic field (which must exist because of $\vec{\nabla} \cdot \vec{B}=0$) as a perturbation. Then you can put the screen at a place before the magnet, where you have a single beam of unpolarized Ag atoms (provided you initialize the state in this way) or behind it, where you have two beams with the position entangled with the spin component in direction of the magnetic field.

 

[A. Neumaier]

I corrected it:
$$
\hat{\vec{j}}(t,\vec{x})=\frac{-\mathrm{i}}{2m}
\hat{\psi}^{\dagger}(t,\vec{x}) \overleftrightarrow{\nabla}
\hat{\psi}(t,\vec{x})$$
It's a usual symbol in QFT it means
$$\hat{\psi}^{\dagger}\overleftrightarrow{\nabla} \hat{\psi} = \hat{\psi}^{\dagger} \nabla \hat{\psi} -(\nabla \hat{\psi}^{\dagger}) \psi.$$
It's the standard probability current of the Schrödinger equation. From the Noether symmetry under multiplication with space-time independent phase factors it follows that
$$\partial_t \rho +\nabla \cdot \vec{j}=0, \quad \rho =\psi^{\dagger} \psi.$$

OK. That's what I thought; I just didn't know how to write it in latex.

I find it much more interesting to include the magnet in the analysis.

We'll come to that, but first I want to be sure that we agree on the simpler setting.

you can just use the solution of the free-particle Schrödinger field. It's easy to construct Gaussian wave packets peaking around arbitrary positions and times.

Please give details. I couldn't find a Gaussian solution of the free Schrödinger equation.

 

[vanhees71]

Here are some details, but not in QFT language but as a simple one-particle treatment with wave functions (in German):

https://itp.uni-frankfurt.de/~hees/faq-pdf/quant.pdf

The Gaussian wave packet for free particles is in Sect. 1.2. A simplified treatment of the idealized SGE is in Sect. 6.12. Here the "spin-flipping" is neglected. I guess, one could treat it in time-dependent perturbation theory or refer to the numerical solutions by Potel et al:

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.71.052106

 

[A. Neumaier]

Here are some details, but not in QFT language but as a simple one-particle treatment with wave functions (in German):

https://itp.uni-frankfurt.de/~hees/faq-pdf/quant.pdf

The Gaussian wave packet for free particles is in Sect. 1.2.

Ah, a Gaussian in momentum space, giving a complex Gaussian in position space that flattens with time. Thanks.

 

[A. Neumaier]

Here are some details, but not in QFT language but as a simple one-particle treatment with wave functions (in German):

https://itp.uni-frankfurt.de/~hees/faq-pdf/quant.pdf

The Gaussian wave packet for free particles is in Sect. 1.2.

You just define operator-valued wave packets $\hat{\phi}^{(-)}(t,\vec{x})$ (pure creation operators) as solutions of the Schrödinger equation for particles moving in a static inhomogeneous magnetic field which have a peak every 1 second.

I want to cast this in terms of the general setting sketched in your your post #16. In the Heisenberg picture, the Hamiltonian would then seem to be be $H(t)=\int dx H(t,x)$ with a normally ordered Hamiltonian density
$$(1)~~~~H(t,x):=a(x)^*\frac{\widehat p^2}{2m}a(x)+\delta(x_3)\Big(a(x)^*V(t,x)+V(t,x)^*a(x)\Big)$$
involving a purely ingoing potential. Here I simplified your notation, using stars in place of daggers and dropping hats and vec's except for the momentum operator, and $V(t,x)$ is the coefficient function of the time-dependent operator solution $\hat{\phi}^{(-)}(t,\vec{x})=a(x)^*V(t,x)$, composed of a superposition of narrow complex Gaussians with a peak every second.

But this cannot be quite true since silver is fermionic and a physical Hamiltonian must be even in fermionic creation and annihilation operators. I propose to use
$$(2)~~~H(t,x):=a(x)^*\frac{\widehat p^2}{2m}a(x)+\delta(x_3)\Big[\Big(a(x)^*V(t,x)\Big)^2+\Big(V(t,x)^*a(x)\Big)^2\Big].$$
Do you agree, or what is your proposal for $H(t,x)$?

 

[vanhees71]

Something along these lines it should be.

 

[A. Neumaier]

Something along these lines it should be.

OK. Now we are ready to go to the interaction picture. Do you agree that the Hamltonian density (2) in the Heisenberg picture becomes the Hamiltonian density
$$(3)~~~~H(t,x):=\Big(a(x)^*V(t,x)\Big)^2+\Big(V(t,x)^*a(x)\Big)^2$$
in the interaction picture, where $V(t,x)$ is a solution of the free 1-particle Schrödinger equation composed of a superposition of narrow complex Gaussians with a peak every second? If not, please suggest what should replace (3) it in your view.

Of course, the initial state is still the vacuum state, and since it transforms under the free Hamiltonian, it is the vacuum state at all times. Thus vacuum expectation values still apply.

 

[vanhees71]

In the interaction picture, the state evolves with the interaction part of the Hamiltonian, the field operators and observables built from them by the free part of the Hamiltonian.

I don't understand, why $V(t,x)$ should be a solution of a field equation. A very simple model along these lines (but not in QFT lingo) is here:

https://arxiv.org/abs/2207.04898

 

[A. Neumaier]

In the interaction picture, the state evolves with the interaction part of the Hamiltonian,

Sorry, of course. Thus we have a time-dependent state, with the vacuum state as initial condition.

the field operators and observables built from them by the free part of the Hamiltonian.

I don't understand, why $V(t,x)$ should be a solution of a field equation.

As you say, the field operators and observables are built by the free part of the Hamiltonian. This makes $V(t,x)$ a solution of the free 1-particle Schrödinger equation, as I had asserted, not of a field equation as you misread it!

 

[A. Neumaier]

A very simple model along these lines (but not in QFT lingo) is here:

https://arxiv.org/abs/2207.04898

Interesting. We will come back to this later. once we have agreed on a final QFT model for the beam.

 

[vanhees71]

Sorry, of course. Thus we have a time-dependent state, with the vacuum state as initial condition.

As you say, the field operators and observables are built by the free part of the Hamiltonian. This makes $V(t,x)$ a solution of the free 1-particle Schrödinger equation, as I had asserted, not of a field equation as you misread it!

No. $V(t,x)$ is an arbitrary external potential. It has nothing to do with a field equation. You can choose it as you like to model what you want to describe.

 

[A. Neumaier]

No. $V(t,x)$ is an arbitrary external potential. It has nothing to do with a field equation. You can choose it as you like to model what you want to describe.

If one could prepare it, yes.

But if we model a beam generated by an outside source then $V$ must look in the Heisenberg picture like (2), hence in the interaction picture like (3), where $V$ satisfies the free 1-particle Schrödinger equation. (Note that we assumed that there is no external field. If an external field with potential $U(x)$ is present, the free part of the Hamiltonian density (2) must get an extra $a^*(x)U(x)a(x)$ term, and $V$ in (3) satisfies instead the 1-particle Schrödinger equation with potential $U(x)$.)

The reason is that a time-independent creation operator in the Heisenberg picture turns in the interaction picture into a creation operator whoise coefficient satisfy the free 1-particle Schrödinger equation, and in (2), the creation operators are time-independent except at the incoming boundary $x_3=0$, where they model the effect of the source.

The same can also be seen from a physical point of view. If you have the field specified everywhere at some time you cannot specify it everywhere at a slightly later time since the signal moves and you can induce change only at the source.

 

[A. Neumaier]

No. $V(t,x)$ is an arbitrary external potential. It has nothing to do with a field equation. You can choose it as you like to model what you want to describe.

Ok. We agreed that in quantum field theory, (3) from post #54 is a suitable general form for the interaction Hamiltonian of a silver beam, where $V(t,x)$ is an external potential that depends on how the beam is prepared in the lab. Since silver has spin 1/2, both $a(x)$ and $V(t,x)$ are vectors with two spin indices.

Let us move closer to the traditional Stern-Gerlach experiment by considering the following preparation of the silver beam:

A silver source outside the beam, switched on at time $t=0$, produces silver particles, approximately one particle per second that quickly (in much less than a second) proceed along a beam initially along the $z$ axis. The beam is subject to a weak external magnetic field $B(t,x)$ that may be switched on gradually. No other forces are assumed to be present; in particular no air or obstacle. Since silver is heavy, we consider a nonrelativistic quantum field description.

For this particular preparation procedure, the potential $V(t,x)$ is given by a solution of the single-particle Pauli equation
$$i\hbar \partial_t V(t,x)=H_1(t,x) V(t,x)$$
with the single-particle interaction
$$H_1(t,x):=-\mu B(t,x)\cdot I$$
used by Potel et al., where $\mu$ is the magnetic moment and $I$ is the spin operator?

Before the magnetic field is switched on, the preparation also implies that $V(t,x)$ is composed of a superposition of narrow complex Gaussians that flatten with time (as described in Section 1.2 of your lecture notes), with a peak every second.

Do you agree?

 

[vanhees71]

I still don't get, what is so difficult.

The standard description in the Heisenberg picture of the Potel paper is that the field operators move according to the Heisenberg operator equations of motion. Since it's a linear system in this simplified treatment, the equations are solved as if the operators where simple real-valued functions, i.e., it looks like the solution of the classical equations.

The state in this description is time-independent and a single-particle state describing a silver atom emitted from the oven through a hole.

Of course, the result is the same as in the first-quantization formalism described in the paper by Potel et al.

 

[A. Neumaier]

I still don't get, what is so difficult.

I don't think that anything here is difficult. I just want to get your agreement on what seems to be the findings of the discussion so far, or an explanation of where you think my description in post #60 needs to be changed.

The standard description in the Heisenberg picture

In post #60 I had already switched to the interaction picture since in this picture no separate discussion of boundary conditions is needed. Please confirm agreement or let me know where you don't agree.

of the Potel paper [...]

The state in this description is time-independent and a single-particle state describing a silver atom emitted from the oven through a hole.

But this thread is about a quantum field description, not a 1-particle description, and I just want to know whether you agree that #60 is an adequate QFT formulation.

 

[vanhees71]

I don't understand, why you think there are equations of motion for some potential. I can't say, whether I agree or agree not with something, I don't understand.

In the interaction picture the state evolves according to
$$\hat{\rho}(t)=\hat{C}(t) \hat{\rho}(0) \hat{C}^{\dagger}(t),$$
where
$$\dot{\hat{C}}(t)=-\mathrm{i} \hat{H}_I(t) \hat{C}(t) \; \Rightarrow \; \hat{C}(t) =\mathcal{T}_c \exp[-\mathrm{i} \int_{0}^{t} \mathrm{d} t' \hat{H}_I(t')]$$
and the operators representing observables
$$\hat{O}(t) = \hat{A}(t) \hat{O}(0) \hat{A}^{\dagger}(t),$$
where
$$\dot{\hat{A}}(T) = \mathrm{i} \hat{H}_0(t) \hat{A}(t) \; \Rightarrow \; \hat{A}(t) = \mathcal{T}_c \exp[+\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_0(t')].$$
You have
$$\hat{H}=\hat{H}_0+\hat{H}_I=\frac{\vec{p}^2}{2m} + \mu_{\text{B}} g_s \hat{\vec{s}} \cdot \vec{B}(\hat{\vec{x}}).$$
A nice model field is
$$\vec{B}=(B_0+\beta z) \vec{e}_3 -\beta y \vec{e}_2.$$
It's realized by the usual magnet depicted in the Wikipedia article

https://commons.wikimedia.org/wiki/...#/media/File:Stern-Gerlach_experiment_svg.svg

for particles close to the axis marked by the beam in that picture.

A good decomposition here is not the naive "interaction-picture" one but
$$\hat{H}_0=\frac{\hat{\vec{p}}^2}{2m} + \mu_{\text{B}} g_s \hat{s}_z (B_0+\beta z)$$
and
$$\hat{H}_I=-mu_{\text{B}} g_s \hat{s}_y \beta y.$$
You can solve the equation of motion for $\hat{H}_0$ analytically and treat the "interaction" $\hat{H}_I$ as perturbation, because for a large $B_0$ the $y$-component of the spin is rapidly precessing and thus the influence on the much less rapid momentum change is negligible, since it averages out to 0 approximately, and that's the interesting case to describe the SGE.

 

[A. Neumaier]

I don't understand, why you think there are equations of motion for some potential. I can't say, whether I agree or agree not with something, I don't understand.

OK; then I'll explain again. I take the interaction picture with respect to the QFT Hamiltonian associated with the 1-particle Hamiltonian
$$\frac{\vec{p}^2}{2m}-\mu B(t,x)\cdot I$$
used by Potel et al., where $\mu$ is the magnetic moment and $I$ is the spin operator. Then

$$\hat{O}(t) = \hat{A}(t) \hat{O}(0) \hat{A}^{\dagger}(t),$$
where
$$\dot{\hat{A}}(T) = \mathrm{i} \hat{H}_0(t) \hat{A}(t) \; \Rightarrow \; \hat{A}(t) = \mathcal{T}_c \exp[+\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_0(t')].$$

can be evaluated with (2) from #52 and gives (3) from #54, where $V$ the potential $V$ satisfies an equation of motion because it is transformed in this way by your general recipe. This follows from the general properties of how linear combinations of creation and annihilation operators transform under conjugation with $\hat{A}(t)$. Doing the calculations shows that $V$ is as claimed in #60.

 

[A. Neumaier]

I don't understand, why you think there are equations of motion for some potential.

A more intuitive way to understand this is to first consider in (3) a general potential. It is surely completely determined by the way the beam is prepared.

Since a local source can prepare only some boundary condition at the source itself (at some fixed $z<0$), so how the potential changes with $z$ must be determined by the forces acting on the beam. These forces therefore determine the potential at all other $z$, and the only sensible way to do so is by means of some differential equation. Working out the details when the source is an electromagnetic field leads necessarily to what I wrote in #60.

 

[vanhees71]

OK; then I'll explain again. I take the interaction picture with respect to the QFT Hamiltonian associated with the 1-particle Hamiltonian
$$\frac{\vec{p}^2}{2m}-\mu B(t,x)\cdot I$$
used by Potel et al., where $\mu$ is the magnetic moment and $I$ is the spin operator. Then

can be evaluated with (2) from #52 and gives (3) from #54, where $V$ the potential $V$ satisfies an equation of motion because it is transformed in this way by your general recipe. This follows from the general properties of how linear combinations of creation and annihilation operators transform under conjugation with $\hat{A}(t)$. Doing the calculations shows that $V$ is as claimed in #60.

But the potential is given by plugging in the operators with the time-dependence solved exactly with $\hat{H}_0$ as the generator for the time-evolution of the "observable operators".

I've unfortunately not the time right now to do this additional calculation for the time-evolution of the states. There it should be sufficient to use the first-order perturbation theory because of the rapid precession of the $\sigma_y$ component (using the simplified magnetic field given above), but this should be straight forward.

 

[A. Neumaier]

But the potential is given by plugging in the operators with the time-dependence solved exactly with $\hat{H}_0$ as the generator for the time-evolution of the "observable operators".

Yes. That's what I did, with the results I claimed. No approximation is involved.

I've unfortunately not the time right now to do this additional calculation for the time-evolution of the states.

I'll give explicit details for deriving #60, later today or tomorrow.

 

[vanhees71]

Then it should be sufficient to use the 1st-order approximation for $\hat{C}$:
$$\hat{C}(t)=1-\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_I(t').$$
The 1st-order correction term should stay hopefully "small".

 

[A. Neumaier]

I'll give explicit details for deriving #60, later today or tomorrow.

See the attached pdf. I gave a complete derivation of all formulas used, but I guess that one can find them also in books such as

• A.L. Fetter and J.D. Walecka, Quantum theory of many-particle systems. Courier Corporation (2012).

While writing up the details I noticed that I had mixed up bosons and fermions in part of the earlier arguments. As a result, the operators (2) and (3) vanish since squares of odd operators are zero. To model a nontrivial source, one therefore needs to use in the interaction picture the more general form
$$V_I(t)=\int dxdx' w(t,x,x')a(x)a(x')+ h.c.$$
for the interaction, where $w(t,x,x')$ satisfies the 2-particle Schrödinger equation for two independent particles in an external magnetic field.

Then it should be sufficient to use the 1st-order approximation for $\hat{C}$:
$$\hat{C}(t)=1-\mathrm{i} \int_0^t \mathrm{d} t' \hat{H}_I(t').$$
The 1st-order correction term should stay hopefully "small".

Since the results are nice and exact, I derived everything without using perturbation theory. Of course you may check the validity by comparing with the results of first order perturbation theory.

 

[vanhees71]

What should be given at the end of the calculation is the distribution function of the silver atoms at the plate. I don't see this in your pdf.

I understand sect. 1, and it looks correct. I have no idea what's done in sect. 2.