How observation leads to wavefunction collapse?

[masudr]

The path integral forumlation of quantum mechanics states that the wavefunction is every possible path (trajectory) from one state to the next.

It does no such thing: it states that the transition amplitude (which is not the wavefunction) is the integral of a specific function of a specific functional of the path over every possible path.

It reduces to the classical trajectory when the phases cancel almost everywhere except the classical path.

The wavefunction does not reduce to the classical trajectory for large-scale phenomena. The transition amplitude tends to one for the classical trajectory for large-scale phenomena.

 

[masudr]

The path integral speaks about the paths that a particle takes, and therefore takes into account trajectories. The path integral is also another definition of the wavefunction.

No. Specifically, the path integral approach states that the integral of

$$e^{i S[x] / \hbar}$$

over all paths gives the transition amplitude, not the wavefunction, for the particle to actually take the path that is fed into S[x].

Therefore, the wavefunction is related to trajectories, right?

It is true to say that they have some level of dependence. To say they are related is pushing it a bit. To say that they are qualitatively equivalent (which is the source of the original dispute between Shahin and I) is, I feel, completely incorrect.

 

[meopemuk]

In fact, I do not know what would be your main objections against THIS paper. Could you briefly indicate them?

I believe we are talking about http://www.arxiv.org/quant-ph/0602024

In this paper you are trying to find a probabilistic interpretation for quantum fields in QED. As I argued in this thread (see, for example, my last post) and elsewhere, I don't see any need to do that, because n-particle wavefunctions in the Fock space can be easily defined by standard quantum mechanical means without any involvement of quantum fields. These wavefunctions have standard probabilistic interpretation and they satisfy all requirements that can be demanded from them by Rules of Quantum Mechanics.

Let me just show that your wavefunction definition (3) leads to a controversy. For simplicity, let us assume that we are interested only in one time instant, and set t=0 in this formula. In quantum mechanics the position-space wavefunction $\psi(\mathbf{r})$ corresponding to the 1-particle state $| 1 \rangle$ is defined as an inner product of $| 1 \rangle$ with eigenvectors of the position operator, which I denote $| \mathbf{r} \rangle$

$$\psi(\mathbf{r}) = \langle \mathbf{r} | 1 \rangle$$

Comparing this to your eq. (3) I may conclude that

$$\langle \mathbf{r} | = \langle 0 | \phi(\mathbf{r})$$

where

$$\phi(\mathbf{r}) = \int \frac{d^3p}{\sqrt{E_p}} \Bigl( e^{\frac{i}{\hbar} \mathbf{pr}} a_{\mathbf{p}}^{\dag} + e^{-\frac{i}{\hbar} \mathbf{pr}} a_{\mathbf{p}} \Bigr)$$

is the quantum field at $t=0$, and $E_p$ is the one-particle energy (I omitted some unimportant factors to simplify formulas). Then I can write

$$\langle \mathbf{r} |= \int \frac{d^3p}{\sqrt{E_p}} e^{-\frac{i}{\hbar} \mathbf{p} \mathbf{r}} \langle 0 | a_{\mathbf{p}}$$ (1)

and

$$|\mathbf{r}' \rangle= \int \frac{d^3p'}{\sqrt{E_{p'}}} e^{\frac{i}{\hbar} \mathbf{p' r}} a_{\mathbf{p}'} ^{\dag}|0 \rangle$$ (2)

Eigenvectors of position must be orthogonal, therefore their inner product must be proportional to the delta function

$$\langle \mathbf{r} |\mathbf{r}' \rangle \propto \delta (\mathbf{r-r'})$$

Let us see if this important property is satisfied in your approach. Using formulas (1) and (2) and commutation relation

$$a_{\mathbf{p}'} ^{\dag}a_{\mathbf{p}} -a_{\mathbf{p}} a_{\mathbf{p}'} ^{\dag} = \delta (\mathbf{p-p}')$$

I obtain

$$\langle \mathbf{r} |\mathbf{r}' \rangle = \int \frac{d^3p}{E_p} e^{\frac{i}{\hbar} \mathbf {p} (\mathbf{r-r'})}$$


which is not proportional to the delta function because of the factor $E_p$ in the denominator. So, in your approach a particle localized at point $\mathbf{r}$ has a probability of being found in any other point $\mathbf{r}'$. I think this is unacceptable.

Regards.
Eugene.

 

[Demystifier]

I
I obtain

$$\langle \mathbf{r} |\mathbf{r}' \rangle = \int \frac{d^3p}{E_p} e^{\frac{i}{\hbar} \mathbf {p} (\mathbf{r-r'})}$$


which is not proportional to the delta function because of the factor $E_p$ in the denominator. So, in your approach a particle localized at point $\mathbf{r}$ has a probability of being found in any other point $\mathbf{r}'$. I think this is unacceptable.

This is exactly why it is said that particle localization in RELATIVISTIC QM does not make sense. But this is not really "my" approach, this is a standard definition of the wave function in relativistic QM/QFT. The problem, of course, is that you cannot interpret it in the usual way. You want to save the standard interpretation, but the price is that you must then crucify relativistic invariance.

 

[Fra]

So, in your approach a particle localized at point $\mathbf{r}$ has a probability of being found in any other point $\mathbf{r}'$. I think this is unacceptable.

Is it more acceptable that suddenly one particle becomes several particles?

I guess the answer is that we didn't know how many particles we had to start with, but by the same token maybe we had insufficient confidence in the single particle concept in the first place?

If we are considering indistinguishable particles, there is no way of knowing wether we detect the same particle in two different places, or two alike particles. It seems to be different views of the same thing, and it could be argued that some inerpretations makes far more sense and that observation of a particle in two space-like separated positions suggest to us that we are in fact witnessing many particle systems. Because if you consider a black box, that you didn't prepare yourself, how many particles are in there? or what their nature is? If you say that "we have a system composed of 3 particles" then, where did thsi prior information come from? That's cheating :)

But I agree that either way there is something that isn't satisfactory.

The problem I have with the standard interpretation is that it requires completness to be consistent. And I don't like this, because it will probably lead to an unmanagable amount of parameters and possibilities that are never realized, but yet the formal possibility clogs our memories.

In reality we don't know everything. A theory that can't handle that in a consistent manner is akward to me.

Suppose that to the best of our knowledge we have a system with one particle. This is our prior information. Then we suddenly observe particles at space-like separated detectors. Now we have to revise our opinion, how do we do so?

I figure the standard answer would be that our prior information was wrong - ie we thought we hade one particle, when we in fact had a many particle system, which then consistently explains our observations.

But of course this is nonsense, because it doesn't help one bit. The information we have is all we have got! We can't possibly know that it's "right" (that it will never need revisions). I wnat a model of reality that explains in a coherent manner, how our information is updated, even when we have largely underestimated our own ignorance.

So how is it possible to underestimate your ignorance? If something is simply out of your experience and imagination, this wouldn't be part of your predictions. When you are first exposed to this, you will most probably face a contradiction, that we must respond to. Thinking "I was wrong" doesn't make any sense at all to me. It doesn't mean we were wrong, it just means that we acted upon all possible evidence at hand, and yet fail to predict the future, but that's life.

I know Demystifier has other ideas than me, and I think I see Eugene's point too, but I still share D's opinon that something is missing in the understanding.

/Fredrik

 

[Shahin]

You can't actually say that. The particle in classical mechanics needs 6 quantities for a complete description. If you can describe it in less than 6, then you have done something quite remarkable.

3 quantities do not describe a particle in quantum mechanics. The value of the function at all possible co-ordinates (or all possible energy, or all possible momenta etc.) are required to fully describe just one single state.



Not only that, but you also need to have a complex-valued function of those 4 variables (i.e. an infinite set of complex numbers). However, in classical mechanics, all the information is contained in 6 numbers, and there is no need for a function. This is why a classical mechanical state space is 6-dimensional, whereas quantum mechanical state space is infinite dimensional.



I'm glad you agree that they are different, but you initially had a problem with QM because 2 particles require 6 variables, and I picked up on that. All I am saying is that, at any moment in time, in the single particle case, CM requires 6 numbers for a complete description, and QM requires an infinite set of numbers, but they can be indexed by 3 variables. Can you see why the two aren't even qualitatively equivalent?



Apart from the odd spelling mistake (e.g. trajectory, not trayectory), your English is remarkably good.

Ok, now i think i undertand what you mean. But, tou have to admitt that, in some fundamental level, the wavefunction and the classical trajectory are the same in the sense that both of them are mathematical entities (of totally different kind) that we use to describe some propieties.

Another question, which i consider interesting, is that if the wave function give us all the posible information about the particle, or, if we have a lack of information as a result of an incomplete theory.

 

[Shahin]

You can't actually say that. The particle in classical mechanics needs 6 quantities for a complete description. If you can describe it in less than 6, then you have done something quite remarkable.

3 quantities do not describe a particle in quantum mechanics. The value of the function at all possible co-ordinates (or all possible energy, or all possible momenta etc.) are required to fully describe just one single state.



Not only that, but you also need to have a complex-valued function of those 4 variables (i.e. an infinite set of complex numbers). However, in classical mechanics, all the information is contained in 6 numbers, and there is no need for a function. This is why a classical mechanical state space is 6-dimensional, whereas quantum mechanical state space is infinite dimensional.



I'm glad you agree that they are different, but you initially had a problem with QM because 2 particles require 6 variables, and I picked up on that. All I am saying is that, at any moment in time, in the single particle case, CM requires 6 numbers for a complete description, and QM requires an infinite set of numbers, but they can be indexed by 3 variables. Can you see why the two aren't even qualitatively equivalent?



Apart from the odd spelling mistake (e.g. trajectory, not trayectory), your English is remarkably good.

Ok, now i think i undertand what you mean. But, tou have to admitt that, in some fundamental level, the wavefunction and the classical trajectory are the same in the sense that both of them are mathematical entities (of totally different kind) that we use to describe some propieties.

Another question, which i consider interesting, is that if the wave function give us all the posible information about the particle, or, if we have a lack of information as a result of an incomplete theory.

 

[Shahin]

You can't actually say that. The particle in classical mechanics needs 6 quantities for a complete description. If you can describe it in less than 6, then you have done something quite remarkable.

3 quantities do not describe a particle in quantum mechanics. The value of the function at all possible co-ordinates (or all possible energy, or all possible momenta etc.) are required to fully describe just one single state.



Not only that, but you also need to have a complex-valued function of those 4 variables (i.e. an infinite set of complex numbers). However, in classical mechanics, all the information is contained in 6 numbers, and there is no need for a function. This is why a classical mechanical state space is 6-dimensional, whereas quantum mechanical state space is infinite dimensional.



I'm glad you agree that they are different, but you initially had a problem with QM because 2 particles require 6 variables, and I picked up on that. All I am saying is that, at any moment in time, in the single particle case, CM requires 6 numbers for a complete description, and QM requires an infinite set of numbers, but they can be indexed by 3 variables. Can you see why the two aren't even qualitatively equivalent?



Apart from the odd spelling mistake (e.g. trajectory, not trayectory), your English is remarkably good.

Ok, now i think i undertand what you mean. But, you have to admitt that, in some fundamental level, the wavefunction and the classical trajectory are the same in the sense that both of them are mathematical entities (of totally different kind) that we use to describe some propieties.

Another question, which i consider interesting, is that if the wave function give us all the posible information about the particle, or, if we have a lack of information as a result of an incomplete theory.

 

[Demystifier]

Another question, which i consider interesting, is that if the wave function give us all the posible information about the particle, or, if we have a lack of information as a result of an incomplete theory.

For me, this is THE MOST interesting question of current physics.

 

[gptejms]

Eugene:

I just saw your response.Please continue with your explanations--I'll ask my questions at the end.

 

[meopemuk]

This is exactly why it is said that particle localization in RELATIVISTIC QM does not make sense. But this is not really "my" approach, this is a standard definition of the wave function in relativistic QM/QFT. The problem, of course, is that you cannot interpret it in the usual way. You want to save the standard interpretation, but the price is that you must then crucify relativistic invariance.

I think that our differences have deep roots in our different belief systems. In my opinion, one significant point of disagreement is this (let me see if you agree that we disagree here). I think you are saying that if $\psi(x)$ is a particle wave function in the reference frame at rest ($x \equiv (\mathbf{r},t)$), then from the point of view of a Lorentz-transformed observer the wave function should look like $\psi(\Lambda x)$, where $\Lambda$ is a $4 \times 4$ matrix of the Lorentz transformation. Is this what you call "relativistic invariance"?

 

[meopemuk]

Eugene:

I just saw your response.Please continue with your explanations--I'll ask my questions at the end.

In my previous post I explained how I understand n-particle wave functions in QFT. These definitions are not different from traditional definitions in ordinary QM. The probabilistic interpretation is clear, and probabilities are preserved in all reference frames, because transformations to different reference frames are represented by unitary operators.

Now, what about quantum fields? What is their role in QFT? Here I take the same position as in Weinberg's "The quantum theory of fields" vol. 1. His logic is the following. In order to construct a relativistic interacting theory in the Fock space we need to define a non-trivial unitary representation of the Poincare group there. This means that we cannot simply add some arbitrary interaction term $V$ to the free Hamiltonian $H_0$ to obtain the full interacting Hamiltonian $H = H_0 +V$. The interaction $V$ must satisfy some non-trivial conditions (commutation relations of the Poincare Lie algebra) in order to be consistent with the principle of relativity. There were a few proposals how to do that, but currently there is only one formalism that satisfies a number of additional conditions (e.g., cluster separability) and leads to a theory comparable with experiment (if we don't pay attention to renormalization difficulties). This formalism involves two steps:

1. For each type of particle define an operator function $\phi(x)$ on the Minkowski space-time, such that
1a. this function is built from creation and annihilation operators of particles,
1b. this function (anti) commutes with itself at space-like separations,
1c. this function transforms in a covariant way with respect to the non-interacting representation of the Poincare group in the Fock space.

In his book Weinberg shows how these three conditions uniquely determine the form of quantum fields for different types of particles. He also shows how these conditions imply that quantum fields satisfy relativistic wave equations (KG, Dirac, etc.)

2. Build operator $V$ as a polynomial in quantum fields with the additional condition that $V$ is a scalar with respect to the non-interacting representation of the Poincare group.

This is how interaction operators are constructed in QFT, and it should be clear that quantum fields $\phi(x)$ have relevance only to this construction. You may try to invent some probabilistic interpretation for quantum fields, but I think it is not necessary and, actually, goes against the logic of QFT, as I tried to explain above.

Eugene

 

[gptejms]

Eugene:

Thanks for your wonderful explanations.I now have the following questions:-

1.I had read that 'the wavefunction is a functional of the field and satisfies the Schrodinger equation'.Can you show that this is equivalent to what you said in your first post.

2.Suppose you have relativistic electron(s) fired into a double slit.Now how would you explain the interference pattern formed on the screen--in terms of interference of the above wavefunction or interference of the field(the field occurring in the wave equation of the relativistic electron(s)).

3.If you read Bjorken & Drell,you'll see that the K.G. equation and the Dirac equation can be reduced to the Schrodinger/Pauli equation.Now if I call what occurs in the relativistic equations as the field(and the \psi of Schrodinger equation as the wavefunction),a question arises.At what stage does the field decide to be a wavefunction?

 

[meopemuk]

1.I had read that 'the wavefunction is a functional of the field and satisfies the Schrodinger equation'.Can you show that this is equivalent to what you said in your first post.

I don't understand this phrase, and it doesn't make much sense, in my opinion.

2.Suppose you have relativistic electron(s) fired into a double slit.Now how would you explain the interference pattern formed on the screen--in terms of interference of the above wavefunction or interference of the field(the field occurring in the wave equation of the relativistic electron(s)).

By usual quantum-mechanical rules wavefunction is a set of coefficients in the decomposition of the state vector in a given orthonormal basis. This definition I used in my post
https://www.physicsforums.com/showpost.php?p=1379494&postcount=171
Two-slit interference should be described in terms of this wavefunction. Quantum fields have nothing to do with interference.

3.If you read Bjorken & Drell,you'll see that the K.G. equation and the Dirac equation can be reduced to the Schrodinger/Pauli equation.Now if I call what occurs in the relativistic equations as the field(and the \psi of Schrodinger equation as the wavefunction),a question arises.At what stage does the field decide to be a wavefunction?

Yes, Bjorken & Drell in the beginning of their book talk about K.G and Dirac equations as equations for relativistic wavefunctions. Then they list multiple problems with this intepretation (non-conserved probabilities, Klein paradox, zitterbewegung, etc.) and say that in fact, these are equations for quantum fields. I find this very confusing.

Personally, I was able to understand QFT only after reading Weinberg's works, especially vol. 1 of his book. This is not an easy read, but his approach is the only logical way to introduce QFT that I can understand.

Fields never "decide to be wavefunctions". Fields and wavefunctions are completely different objects. It may sound ridiculous, but relativistic quantum field theory (i.e., a quantum theory with variable number of particles) can be formulated without introducing quantum fields at all. One can build the Fock space as a direct sum of n-particle spaces. Then one can define orthonormal bases in each sector of the Fock space, and build particle creation and annihilation operators that move state vectors between sectors. Then one can construct a Hamiltonian as a function of these creation and annihilation operators. After we have the Hamiltonian, everything else (bound states, scattering, etc.) follows from simple rules of ordinary quantum mechanics. No fields needed. An example of such an approach is given in

H. Kita "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties" Progr. Theor. Phys. 35 (1966), 934

Eugene.

 

[Demystifier]

I think that our differences have deep roots in our different belief systems. In my opinion, one significant point of disagreement is this (let me see if you agree that we disagree here). I think you are saying that if $\psi(x)$ is a particle wave function in the reference frame at rest ($x \equiv (\mathbf{r},t)$), then from the point of view of a Lorentz-transformed observer the wave function should look like $\psi(\Lambda x)$, where $\Lambda$ is a $4 \times 4$ matrix of the Lorentz transformation. Is this what you call "relativistic invariance"?

Exactly!
(At least for spinless particles, with a spin there is an additional transformation of \psi itself).
Now, what is your way of thinking?

 

[Anonym]

1.I had read that 'the wavefunction is a functional of the field and satisfies the Schrodinger equation'.Can you show that this is equivalent to what you said in your first post.

2.Suppose you have relativistic electron(s) fired into a double slit.Now how would you explain the interference pattern formed on the screen--in terms of interference of the above wavefunction or interference of the field(the field occurring in the wave equation of the relativistic electron(s)).

3.If you read Bjorken & Drell,you'll see that the K.G. equation and the Dirac equation can be reduced to the Schrodinger/Pauli equation.Now if I call what occurs in the relativistic equations as the field(and the \psi of Schrodinger equation as the wavefunction),a question arises.At what stage does the field decide to be a wavefunction?

All that threat is the collection of the absurd statements made by ignorants. The numerous attempts of different posters (like you) to correct does not help. It starts so:

Post#1

So, in the double slit experiment, if a photon observes an electron, the interference pattern vanishes. Why is this so?

The photon do not observes an electron and the interference pattern do not vanish.

Post#2

a single electron does not create an interference pattern

No comment.

It finished with

Post#189

Personally, I was able to understand QFT only after reading Weinberg's works, especially vol. 1 of his book. This is not an easy read, but his approach is the only logical way to introduce QFT that I can understand…

Fields never "decide to be wavefunctions". Fields and wavefunctions are completely different objects. It may sound ridiculous, but relativistic quantum field theory (i.e., a quantum theory with variable number of particles) can be formulated without introducing quantum fields at all. One can build the Fock space as a direct sum of n-particle spaces.

S.Weinberg “The Quantum Theory of Fields” v.1 p.107 (introductory remark):
“The physical states before and after the collision consist of particles that are so far apart that they are effectively non-interacting, so they can be described as direct products of the one-particle states discussed in the previous chapter.

In addition,the writer of post # 189 does not know and does not understand the difference between the linear superposition and the direct sum.

The situation is clearly explaned by Jimmysnyder:

“The OP wasn't really interested in knowing how many people understand the physics, they were asking how many can understand the math. I'm pretty sure they meant the math in an introductory text. When Feynman said no one understands QM, he didn't mean to imply that the authors couldn't do the math in their own books. In my opinion, anyone who is not mentally handicapped can understand an introductory QM text like Shankar's by the simple act of applying themselves to the task. However, lack of interest will prevent most people from commiting themselves for the amount of time it would take.”

I guess that the existence of that collection of nonsense is the consequence that Zz took a long vacation.

Regards, Dany.

 

[gptejms]

I don't understand this phrase, and it doesn't make much sense, in my opinion.

If the hamiltonian involves the field,won't the wavefunction be a functional of the field?

Two-slit interference should be described in terms of this wavefunction. Quantum fields have nothing to do with interference.

If I do a Kirchoff integral kind of analysis with the KG equation(given a \psi at the slits,calculate the \psi at the screen--where \psi satisfies KG equation), I guess an interference pattern will result.So you can get an interference pattern with fields.

Yes, Bjorken & Drell in the beginning of their book talk about K.G and Dirac equations as equations for relativistic wavefunctions. Then they list multiple problems with this intepretation (non-conserved probabilities, Klein paradox, zitterbewegung, etc.) and say that in fact, these are equations for quantum fields. I find this very confusing.

Fields never "decide to be wavefunctions". Fields and wavefunctions are completely different objects.

i) do you agree that the KG equation describes a quantum field(and not a relativistic wavefunction)?ii)do you agree that the KG equation reduces to the Schrodinger equation in the non-relativistic limit?

If your answers to both the questions are yes,then it is important to answer how the field transforms into a wavefunction.

Finally a question:-

If I take your definition of the wavefunction in QFT and consider two slit interference experiment,then we would have vacuum interfering with itself,n=1 state interfering with itself,n=2 state interfering with itself etc.--would this lead to an interference pattern?

 

[meopemuk]

Exactly!
(At least for spinless particles, with a spin there is an additional transformation of \psi itself).
Now, what is your way of thinking?

Before I tell you my way of thinking, could you please explain why do you think this $\psi(x) \to \psi(\Lambda x)$ is a good transformation formula for the wavefunction? Where this formula comes from? Let's consider spinless particles only to keep things simple.

Eugene.

 

[meopemuk]

S.Weinberg “The Quantum Theory of Fields” v.1 p.107 (introductory remark):
“The physical states before and after the collision consist of particles that are so far apart that they are effectively non-interacting, so they can be described as direct products of the one-particle states discussed in the previous chapter.

Dany,

you are right that n-particle states in quantum mechanics are described as direct (tensor) products of 1-particle states. However, in quantum field theory, one cannot fix the number of particles, so the full Hilbert space of the theory is built as a direct sum of such n-particle direct products, with n varying from 0 to infinity. This is called Fock space.

Eugene.

 

[meopemuk]

If the hamiltonian involves the field,won't the wavefunction be a functional of the field?

I don't understand what you mean by that. Hamilltonian and wavefunction are two different objects. Quantum field is a mathematical tool for constructing the Hamiltonian.

If I do a Kirchoff integral kind of analysis with the KG equation(given a \psi at the slits,calculate the \psi at the screen--where \psi satisfies KG equation), I guess an interference pattern will result.So you can get an interference pattern with fields.

Yes, if you interpret thing like $\langle 0 | \phi(x) | 1 \rangle$ as wave functions (i.e., probability amplitudes), then using K.G. wave equation you may "explain" interference patterns. But this interpretation and explanation would be not correct.

i) do you agree that the KG equation describes a quantum field(and not a relativistic wavefunction)?

Yes, of course.

ii)do you agree that the KG equation reduces to the Schrodinger equation in the non-relativistic limit?

No. KG equation has 2nd time derivative, but Schroedinger equation should have only 1st time derivative. Even in the (improbable) case when wave equations for quantum fields coincided with Schroedinger equations for wavefunctions, I wouldn't say that "quantum field = wavefunction". Two completely different objects may satisfy the same equation. So what?

If I take your definition of the wavefunction in QFT and consider two slit interference experiment,then we would have vacuum interfering with itself,n=1 state interfering with itself,n=2 state interfering with itself etc.--would this lead to an interference pattern?

Could you please clarify your question. What do you mean by "vacuum interfering with itself"?

 

[gptejms]

I don't understand what you mean by that. Hamilltonian and wavefunction are two different objects. Quantum field is a mathematical tool for constructing the Hamiltonian.

If H and \psi(wavefunction) are connected by an equation(Schrodinger)(and H is constructed from fields),then the wavefunction would be a functional of the field.Anyway,I am not a QFT specialist,but I read this somewhere--may be others can add their comments.

Even in the (improbable) case when wave equations for quantum fields coincided with Schroedinger equations for wavefunctions, I wouldn't say that "quantum field = wavefunction". Two completely different objects may satisfy the same equation. So what?

Now that's a kind of statement about which I don't know what to say--it is 'escapist' to say the least.

Could you please clarify your question. What do you mean by "vacuum interfering with itself"?

You said in your 'first' post:-
'In each of these sectors you can have an orthonormal basis (e.g. a basis of position eigenvectors). So, in each of these sectors you can have an n-particle wave function. Then the total wave function corresponding to the state is a suporposition of all these wave functions with complex coefficients.'

So in your grand wavefunction you have a number of wavefunctions with different n(n=0,n=1,n=2 etc.) i.e. the particle number is not fixed.Now can you work out the interference pattern with this combination--please do the calculation and you'll see what I mean.

 

[Anonym]

in quantum field theory, one cannot fix the number of particles, so the full Hilbert space of the theory is built as a direct sum of such n-particle direct products, with n varying from 0 to infinity. This is called Fock space.

What is called Fock space is defined by V.A.Fock and not by Eugene. It is clearly presented in every standard textbook on QT. The difference between the linear superposition and the direct sum also is clearly presented in every standard textbook on QT.

If you know Russian (I do not believe that it is translated to English) I suggest reading 28 pages book written by V.A.Fock,”Quantum Physics and Structure of Matter”, Leningrad Univ. Press (1965) that definitely will help you to understand QF.

V.A. Fock was the head of the theoretical physics department where I was educated. His spirit and his pupils taught us from the very first lecture in the physics and the mathematics (the Dedekind theorem was presented after 15min from the beginning of Classical Analysis).

I consider the continuation of this discussion surrealistic and tasteless.

Regards, Dany.

 

[meopemuk]

If H and \psi(wavefunction) are connected by an equation(Schrodinger)(and H is constructed from fields),then the wavefunction would be a functional of the field.

Your premises are correct: wavefunction and Hamiltonian are related by the Schroedinger equation, the Hamiltonian of QFT is (usually) constructed from fields. However, I don't understand how you reach your conclusion about the wavefunction being a "functional of fields". What does "functional of fields" mean, anyway?

You said in your 'first' post:-
'In each of these sectors you can have an orthonormal basis (e.g. a basis of position eigenvectors). So, in each of these sectors you can have an n-particle wave function. Then the total wave function corresponding to the state is a suporposition of all these wave functions with complex coefficients.'

So in your grand wavefunction you have a number of wavefunctions with different n(n=0,n=1,n=2 etc.) i.e. the particle number is not fixed.Now can you work out the interference pattern with this combination--please do the calculation and you'll see what I mean.

Sure, the interference pattern can be described. Take, for example, Feynman's double slit experiment. In this case we are dealing with an 1-electron system moving in a potential (the screen with two slits can be described by a potential $V(\mathbf{r})$). The number of particles is definitely conserved in this case. So, our system can be fully described in the 1-particle sector of the Fock space. Then the relativistic Schroedinger equation in this sector can be written as

$$-i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r},t) = \Bigl(\sqrt{-\hbar^2c^2 \nabla^2 + m^2c^4} + V(\mathbf{r}) \Bigr) \psi(\mathbf{r},t)$$

from which you can obtain the time evolution of the wave function, interference, and everything else.

 

[meopemuk]

What is called Fock space is defined by V.A.Fock and not by Eugene. It is clearly presented in every standard textbook on QT. The difference between the linear superposition and the direct sum also is clearly presented in every standard textbook on QT.

What is your definition of the Fock space and how it is different from mine? I always thought I was following the standard definition. I could be wrong, though.

Eugene.

 

[meopemuk]

If you know Russian (I do not believe that it is translated to English) I suggest reading 28 pages book written by V.A.Fock,”Quantum Physics and Structure of Matter”, Leningrad Univ. Press (1965) that definitely will help you to understand QF.

Dany,

Thank you for the reference. I would love to read this book. Yes, I can read Russian, as you can guess from my user name.

Eugene.

 

[gptejms]

Your premises are correct: wavefunction and Hamiltonian are related by the Schroedinger equation, the Hamiltonian of QFT is (usually) constructed from fields. However, I don't understand how you reach your conclusion about the wavefunction being a "functional of fields". What does "functional of fields" mean, anyway?

You know what a functional is.The field is changing all the time and,starting with a particular field configuration, different field configurations have different probability amplitudes--calculated from the wavefunction satisfying the Schodinger equation.This is my guess(a reasonable one I hope).

Sure, the interference pattern can be described. Take, for example, Feynman's double slit experiment. In this case we are dealing with an 1-electron system moving in a potential (the screen with two slits can be described by a potential $V(\mathbf{r})$). The number of particles is definitely conserved in this case. So, our system can be fully described in the 1-particle sector of the Fock space. Then the relativistic Schroedinger equation in this sector can be written as

$$-i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r},t) = \Bigl(\sqrt{-\hbar^2c^2 \nabla^2 + m^2c^4} + V(\mathbf{r}) \Bigr) \psi(\mathbf{r},t)$$

from which you can obtain the time evolution of the wave function, interference, and everything else.

When you want to describe the interference pattern you jump to a 1-particle system--of course there is no problem here!Though there would be no problem with an n-particle system also if n is fixed(which remains fixed if there are no interactions).It would be interesting to see what happens when the interactions are switched on--interference pattern should come crumbling down.

 

[meopemuk]

You know what a functional is.The field is changing all the time and different field configurations have different probability amplitudes(given by the wavefunction corresponding to the field configuration)--the wavefunction satisfies the Schodinger equation.This is my guess(a reasonable one I hope).

OK, now I am lost completely. I understand particles and their wavefunctions. I can even understand why one can confuse particle wavefunction with the expression

$$\langle 0 | \phi(x) | 1 \rangle$$ (1)


where $\phi(x)$ is quantum field (this expression was mentioned already in this thread). This expression is, indeed, very similar to 1-particle wavefunction in the position representation, apart from the factor $E_p$ in the denominator, as we discussed earlier.

Note also that in eq (1), the field $\phi(x)$ is a fixed object. It does not depend on the state of the particle. All changes related to different states of the particle are contained in the particle state vector [itex] | 1 \rangle [/tex].

Now you moved to a completely different territory. You declared that fields are some physical objects that can exist in different states and have certain observables (what are they?). The wavefunction now describes not probabilities of different particle properties, but probabilities of different field properties. This theory is completely different from QFT I know. I have no idea how your theory can be related to the non-relativistic limit of QM, or to physical measurements in few-particle systems.

When you want to describe the interference pattern you jump to a 1-particle system--of course there is no problem here!Though there would be no problem with an n-particle system also if n is fixed(which remains fixed if there are no interactions).It would be interesting to see what happens when the interactions are switched on--interference pattern should come crumbling down.

What interaction do you want to switch on? The operator $V(\mathbf{r})$ describing interaction of the electron with the screen is always on.

 

[gptejms]

Now you moved to a completely different territory. The wavefunction now describes not probabilities of different particle properties, but probabilities of different field properties. This theory is completely different from QFT I know. I have no idea how your theory can be related to the non-relativistic limit of QM, or to physical measurements in few-particle systems.

It's a new QFT you know!I was only trying to give meaning to the statement 'wavefunction is a functional of the field'( a statement I picked up from somewhere).I think the onus of explaining is on one who made/endorses this statement.As I wrote earlier,I am no QFTist(in fact,I've never had a formal course in QFT)--my knowledge of this subject is confined to bits and pieces that I've picked up from here and there.

What interaction do you want to switch on? The operator $V(\mathbf{r})$ describing interaction of the electron with the screen is always on.

interaction between the electrons,I meant.

 

[meopemuk]

meopemuk: What interaction do you want to switch on? The operator describing interaction of the electron with the screen is always on.

interaction between the electrons,I meant.

But in the double-slit experiment electrons are released one-by one, so it is, effectively, a one-electron problem. If electrons were released in pairs, then, of course, we would need to take the electron interaction into account, and the interference picture would have changed.

Eugene

 

[Fra]

I think I lost track here.

You can construct a QFT either from a supposedly "classical field" - ie a field that has meaning in the classical sense, like the potential of the free EM radiation field. This QFT deals with our information (probabilistic) of the classical field.

But one can also try to do QFT, on a wavefunction, which is why one can call it "second quantization". This QFT deals with our information (probabilistic) of the wavefunction.

This is what I previously loosely referred to as "probability of probability", which implicitly acknowledges that there is an uncertainty in the "observations of a frequency" in the frequentist interpretation.

But there are some deep problems to a completely consistent understand of this, at least for me, but may it's just me. Of course this is related to a many particle interpretation, because one interpretation of the non-unitarity of one-particle probabilities is that it's many particles. If we mistake two particles for one, we are obviously going to conclude violation of probability conservations (P > 1), or we mistake one particle for two, we can get (P < 0). This is exactly why I think of probabilities as expectations. An example of this is the "dirac wavefuncton/field", which is hardly a classical field. I think this also boils down to the issue of fermions vs bosons, I'm not sure how you'd explain a fermion in classical mechanics. The KG <-> Dirac transformation is interesting and I think the key. I have planned to resume this later, but until he spacetime concept is finished this would not make sense, because all of the above is conditional on that. Of course the time evolutions are different, but then I think it's all entangled up. The way I see it in ordinary QM the wave function really doesn't contain all information. A significant amount of information is also contained in the Hamiltonian, this I view as a sort of broken symmetry. I personally don't understand how you can consistently separate the Information about dynamics from the information about what is the subject of dynamics.

This is why I am trying to find a new framework, that updates the overly idealistic probability concept of QM, and also adds a better relations between dynamics and the subject of the dynamics, which I think will ultimately make the integration of relativity(GR+SR) more natural... but I don't know if this is what we are talking about in this thread? or if we are talking about sometihng else?

/Fredrik

 

[gptejms]

But in the double-slit experiment electrons are released one-by one, so it is, effectively, a one-electron problem. If electrons were released in pairs, then, of course, we would need to take the electron interaction into account, and the interference picture would have changed.

Eugene

Depends on how electrons are released:-it could be one by one,in pairs,or more than two at a time.

I wish someone here clarifies the statement 'wavefunction is a functional of the field'.

 

[Demystifier]

Before I tell you my way of thinking, could you please explain why do you think this $\psi(x) \to \psi(\Lambda x)$ is a good transformation formula for the wavefunction? Where this formula comes from? Let's consider spinless particles only to keep things simple.

It seems so obvious to me that I can hardly imagine what kind or argument you would like to see. But let me try: Because this provides that the equation of motion for psi (the Klein Gordon equation) will have the same form in all Lorentz frames.
As it is the standard way of viewing this, I think it is YOU who should explain the reason for disagreement.

 

[Demystifier]

I wish someone here clarifies the statement 'wavefunction is a functional of the field'.

See, e.g., eqs. (87)-(89) in
http://xxx.lanl.gov/abs/quant-ph/0609163
Compare it also with eq. (80).

 

[gptejms]

See, e.g., eqs. (87)-(89) in
http://xxx.lanl.gov/abs/quant-ph/0609163
Compare it also with eq. (80).

So I was right with my guess(post #201)--why don't you answer Eugene's objection to this.

 

[Demystifier]

So I was right with my guess(post #201)--why don't you answer Eugene's objection to this.

He is allowed to read what I suggested above as well. That would be my answer to his objection.