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Why is this true? Why i can' t use the operator X, P and S in QFT to describe particles as in the normal Q.M.?

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Why is this true? Why i can' t use the operator X, P and S in QFT to describe particles as in the normal Q.M.?

- #2

ShayanJ

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One of the purposes of QFT is to unify Quantum Mechanics and Special Relativity.

In QM position components are measurable quantities and there are operators associated to them. But the role of time is like in Classical Physics, as a parameter used in differential equations that evolve wave-functions or operators.

In SR, time and space are just different dimensions of the space-time manifold and are not as distinct as in CP. In fact SR says that time and space should be treated in a symmetric manner.

So to unify QM and SR, there are two options, 1) Promote time to an operator and make it the time component of a 4-position vector with other components being the position component operators and use proper time as the evolution parameter. 2) Demote the position component operators to labels on quantities along with time. Which means using fields on space-time.

The first option turns out to be much more complicated but equivalent to the second option. So what happens in QFT is that people use fields on Minkowski(flat) space-time.

The symmetry group of the Minkowski space-time is the group of the Poincare transformations. But not all kinds of fields transform the same way under these transformations. Different fields in terms of their transformation properties under Poincare transformations, constitute different representations of the Poincare group.

Now because in QFT particles are excitations of fields that arise because you give them enough energy to create them, these different representations of the Poincare group also constitute a classification of different particles according to their mass and spin angular momentum.

In QM position components are measurable quantities and there are operators associated to them. But the role of time is like in Classical Physics, as a parameter used in differential equations that evolve wave-functions or operators.

In SR, time and space are just different dimensions of the space-time manifold and are not as distinct as in CP. In fact SR says that time and space should be treated in a symmetric manner.

So to unify QM and SR, there are two options, 1) Promote time to an operator and make it the time component of a 4-position vector with other components being the position component operators and use proper time as the evolution parameter. 2) Demote the position component operators to labels on quantities along with time. Which means using fields on space-time.

The first option turns out to be much more complicated but equivalent to the second option. So what happens in QFT is that people use fields on Minkowski(flat) space-time.

The symmetry group of the Minkowski space-time is the group of the Poincare transformations. But not all kinds of fields transform the same way under these transformations. Different fields in terms of their transformation properties under Poincare transformations, constitute different representations of the Poincare group.

Now because in QFT particles are excitations of fields that arise because you give them enough energy to create them, these different representations of the Poincare group also constitute a classification of different particles according to their mass and spin angular momentum.

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- #3

strangerep

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Heh, have you obtained a copy of Ballentine's QM textbook yet (as I suggested in your other thread)?

Why is this true? Why i can' t use the operator X, P and S in QFT to describe particles as in the normal Q.M.?

There's stuff in there about nonrelativistic QM that one should master first.

The deeper reason is that quantization of a (class of) system essentially boils down to constructing a unitary representation of the dynamical symmetry group applicable to that system. In many case, this can be on the usual Hilbert space of square-integrable wavefunctions, and it's easy to get a position operator. But, more generally, in the non-relativistic case, we are constructing a unitary representation of the Galilei group -- actually the centrally-extended Galilei group (see Ballentine for more detail). In that case, it turns out that a position operator can be expressed in terms of the Galilean boost operator and the mass. I.e., the position operator arises from the Galilean boost operator, if one analyzes the associated group theory carefully.

However, the relativistic case is rather different. We have the invariant mass, and a Lorentz boost operator, but they don't play nicely together (in general) to yield something we would recognize as a position operator. Further, a closer analysis reveals that it's highly problematic to construct a position operator at all for some types of fields (e,g., spin-1, massless, i.e., the photon field). Another complication is that the Lorentz group is noncompact, and this implies that all its unitary representations are infinite-dimensional, i.e., field representations.

HTH.

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Actually, for massive particles one can do it, and it has been done in the literature; see, e.g.,Why i can' t use the operator X, P and S in QFT to describe particles as in the normal Q.M.?

L.L. Foldy, Synthesis of Covariant Particle Equations,

Physical Review 102 (1956), 568-581.

Let ##M := R^3## be the manifold of 3-momenta ##p## (your P). On the Hilbert space ##H_m^d## obtained by completion of the space of all ##C^\infty## functions with compact support from ##M## to the space ##C^d## of d-component vectors with complex entries, with inner product defined by

$$<\phi|\psi> := \int dp \sqrt{p^2+m^2}\phi(p)^*\psi(p),$$

we define the position operator (your X)

$$q := i \hbar \partial_p,$$

which satisfies the standard commutation relations, the momentum in time direction,

$$p_0 := \sqrt{p^2+m^2},$$

where ##m>0## is a fixed mass, and the operators

$$J := q \times p + S,$$

$$K := (p_0 q + q p_0)/2 + p \times S/(m+p_0),$$

where ##S## is the spin vector in a unitary irreducible representation of ##so(3)## on the vector space ##C^d ##of complex vectors of length ##d##, with the same commutation relations as ##J.##

This is a unitary representation of the Poincare algebra; verification of the standard commutation relations (given, e.g., in Weinberg's Volume 1, p.61) is straightforward. It is not difficult to show that this representation is irreducible and extends to a representation of the Poincare group of mass ##m## and spin ##s=(d-1)/2##.

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This already holds for a single free nonrelativistic particle, since the Heisenberg group is also noncompact. Your argument only proves that the Hilbert space must be an infinite-dimensional space of wave functions, not that one needs a field theory to describe a single relativistic particle.the Lorentz group is noncompact, and this implies that all its unitary representations are infinite-dimensional, i.e., field representations.

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samalkhaiat

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The problem follows from the fact that QFT is not a single particle QM. So, in order to define a position operator one has to restrict oneself to the space of positive energy solutions of the equation of motion (as they are physically realizable by a free single particle). Even then, the usual definition of the position operator fails to be Hermitian and hence does not correspond to a measurable property, and when we define it to be Hermitian, the (localized) states, formed from its eigen-functions, fail to be Lorentz covariant.

Why is this true? Why i can' t use the operator X, P and S in QFT to describe particles as in the normal Q.M.?

To see this, consider the positive energy solutions of the KG equation. The space of such covariant solutions consists of all amplitudes [itex]\phi (x)[/itex] of the form [tex]\phi (x) = \frac{\sqrt{2}}{(2 \pi)^{3/2}} \int d^{4}p \ e^{- i p \cdot x} \theta (p_{0}) \delta (p^{2} - m^{2}) \Phi (p) ,[/tex] or, by doing the [itex]p_{0}[/itex] integration, [tex]\phi (x) = \frac{1}{\sqrt{2} (2 \pi)^{3/2}} \int \frac{d^{3}p}{p_{0}} \ e^{- i p \cdot x} \ \Phi (\mathbf{p}) ,[/tex] where [itex]p_{0} = + \sqrt{\mathbf{p}^{2} + m^{2}}[/itex] and [itex]\Phi(\mathbf{p})[/itex] is (like [itex]\phi(x)[/itex]) a scalar because [itex]d^{3}p/p_{0}[/itex] is the invariant measure over the hyperboloid [itex]p^{2} = m^{2}[/itex] and [itex]p \cdot x[/itex] is Lorentz invariant.

Clearly, the space of positive energy solution is a linear vector space. So, in order to turn it into a Hilbert space, we need to define a suitable (i.e. Lorentz-invariant and time-independent) scalar product. Let us define the scalar product of two positive energy amplitudes by [tex]\left( \phi , \psi \right)_{\sigma} \equiv i \int_{\sigma} d \sigma^{\mu}(x) \ \phi^{*}(x) \overleftrightarrow{\partial_{\mu}} \psi (x) , \ \ \ \ (1)[/tex] where [itex]\sigma[/itex] is an arbitrary space-like hypersurface. The 4-vector surface element [itex]d\sigma^{\mu} (x) = n^{\mu}(x) d\sigma[/itex], (where [itex]n^{\mu}(x)[/itex] is the unit normal to [itex]\sigma[/itex] at [itex]x[/itex]) has the components [tex]d\sigma^{\mu} = \{ dx^{1} dx^{2} dx^{3} , dt dx^{2}dx^{3}, dt dx^{1}dx^{3}, dt dx^{1}dx^{2} \} .[/tex] For the hyperplane [itex]\sigma = t = \mbox{const.}[/itex] equation (1) reduces to [tex]\left( \phi , \psi \right)_{t} \equiv i \int_{t} d^{3}x \ \phi^{*}(x) \overleftrightarrow{\partial_{t}} \psi (x) . \ \ \ \ (2)[/tex] Going over to momentum space, the scalar product, (2), becomes [tex]\left( \phi , \psi \right) = \int_{+} \frac{d^{3}p}{p_{0}} \ \Phi^{*}(\mathbf{p}) \Psi (\mathbf{p}) . \ \ \ \ \ (3)[/tex] This shows that [itex](\phi , \phi)[/itex] is positive definite if [itex]\phi \neq 0[/itex]. It is also evident that [itex]( \phi , \psi ) = ( \psi , \phi )^{*}[/itex] and [itex]( \phi + \chi , \psi ) = ( \phi , \psi ) + ( \chi , \psi )[/itex]. Thus, our scalar product satisfies all the properties required of a scalar product.

Clearly the scalar product, eq(1), is Lorentz invariant. Also, it is easy to show that [tex]\frac{\delta}{\delta \sigma (x)} \left( \phi , \psi \right)_{\sigma} = 0 ,[/tex] if [itex]\phi[/itex] and [itex]\psi[/itex] satisfy the KG equation. This means that the scalar product does not depend on the space-like surface [itex]\sigma[/itex] used to calculate it. Therefore, it is time-independent.

Now, it is easy to show that the usual quantum mechanical expression for the position operator [itex]\hat{X} = i \vec{\nabla}_{\mathbf{p}}[/itex] is not Hermitian. Indeed, integrating by parts and ignoring a surface term gives

[tex]\int_{+} \frac{d^{3}p}{p_{0}} \ \Phi^{*}(\mathbf{p}) ( i \vec{\nabla}) \Psi (\mathbf{p}) = \int_{+} \frac{d^{3}p}{p_{0}} \left(- i \vec{\nabla} \Phi^{*}(\mathbf{p}) \right) \Psi (\mathbf{p}) + \int_{+} \frac{d^{3}p}{p_{0}} \Phi^{*}(\mathbf{p}) \frac{ i \mathbf{p}}{p_{0}^{2}} \Psi (\mathbf{p}) . \ \ (4)[/tex] Thus, [itex](\phi , \hat{X}\psi) \neq (\hat{X}\phi , \psi)[/itex]. It follows that the wave function [itex]\phi (x)[/itex] cannot be a probability amplitude for finding the particle at [itex](t , \mathbf{x})[/itex].

Equation (4), allows us to define the following Hermitian operator [tex]\hat{Q} = i \vec{\nabla}_{\mathbf{p}} - \frac{i \mathbf{p}}{2 p_{0}^{2}} ,[/tex] which agrees with the definition of the centre of mass in special relativity. It is the so-called Newton-Wigner position operator: they derived it from certain physical conditions on localized states. It turns out that the localized states do not form a Lorentz covariant manifold. They only have the symmetry properties in the hyperplane [itex]t = \mbox{constant}[/itex] in spacetime.

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From the equations above, it looks as if one can also take m=0. What goes wrong (if anything) in the massless case?Actually, for massive particles one can do it, and it has been done in the literature; see, e.g.,

L.L. Foldy, Synthesis of Covariant Particle Equations,

Physical Review 102 (1956), 568-581.

Let ##M := R^3## be the manifold of 3-momenta ##p## (your P). On the Hilbert space ##H_m^d## obtained by completion of the space of all ##C^\infty## functions with compact support from ##M## to the space ##C^d## of d-component vectors with complex entries, with inner product defined by

$$<\phi|\psi> := \int dp \sqrt{p^2+m^2}\phi(p)^*\psi(p),$$

we define the position operator (your X)

$$q := i \hbar \partial_p,$$

which satisfies the standard commutation relations, the momentum in time direction,

$$p_0 := \sqrt(m^2+|p|^2),$$

where ##m>0## is a fixed mass, and the operators

$$J := q \times p + S,$$

$$K := (p_0 q + q p_0)/2 + p \times S/(m+p_0),$$

where ##S## is the spin vector in a unitary irreducible representation of ##so(3)## on the vector space ##C^d ##of complex vectors of length ##d##, with the same commutation relations as ##J.##

This is a unitary representation of the Poincare algebra; verification of the standard commutation relations (given, e.g., in Weinberg's Volume 1, p.61) is straightforward. It is not difficult to show that this representation is irreducible and extends to a representation of the Poincare group of mass ##m## and spin ##s=(d-1)/2##.

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The question is - does it really matter? Does the position operator really needs to be Lorentz covariant?It is the so-called Newton-Wigner position operator: they derived it from certain physical conditions on localized states. It turns out that the localized states do not form a Lorentz covariant manifold. They only have the symmetry properties in the hyperplanet=constantt = \mbox{constant} in spacetime.

The answer critically depends on interpretation of quantum theory. If the position operator is supposed to describe a property which the particle

- #9

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In the massless case, the resulting representation is no longer irreducible but splits into irreps of helicity ##s,s-1,\ldots,-s##. This (and much more) is described in my chapter on the Poincare group inFrom the equations above, it looks as if one can also take m=0. What goes wrong (if anything) in the massless case?

my theoretical physics FAQ.

- #10

samalkhaiat

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No, it has nothing to do the interpretation problem in QM: I would not have written a single word if it does. The transformation laws of states and operators are independent of any experimental set up or interpretation issues: the relative motion of two laboratories is not (and should not be) part of the measurement set up. Other wise you wouldn’t have any theoretical framework for a physically acceptable theory.The question is - does it really matter? Does the position operator really needs to be Lorentz covariant?

The answer critically depends on interpretation of quantum theory. If the position operator is supposed to describe a property which the particlehasirrespective of the measurement, then yes, it is very desirable that it is Lorentz covariant. By contrast, if the position operator is interpreted as an observable that only makes sense in a context of a given measurement setup, then it does not need to be Lorentz covariant.

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I disagree. According to the general principles of quantum theory,No, it has nothing to do the interpretation problem in QM: I would not have written a single word if it does. The transformation laws of states and operators are independent of any experimental set up or interpretation issues: the relative motion of two laboratories is not (and should not be) part of the measurement set up. Other wise you wouldn’t have any theoretical framework for a physically acceptable theory.

- #12

DrDu

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Ok, but irreducibility is rather mathematical convenience than a physical requirement. So e.g. it might be possible then to localize linearly polarized photons but not helically polarized ones?In the massless case, the resulting representation is no longer irreducible but splits into irreps of helicity ##s,s-1,\ldots,-s##. This (and much more) is described in my chapter on the Poincare group in

my theoretical physics FAQ.

- #13

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The massless case of the Foldy representation of spin 1 does not describe photons since it has longitudinal (helicity 0) modes in addition to the transversal (helicity ##\pm1##) ones.Ok, but irreducibility is rather mathematical convenience than a physical requirement. So e.g. it might be possible then to localize linearly polarized photons but not helically polarized ones?

Without adding the unphysical longitudinal modes one cannot construct a position operator for a massless vector representation.

- #14

DrDu

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But the longitudinal modes do not interact and are therefore unobservable. So they are basically gauge degrees and you can choose them as seems convenient.The massless case of the Foldy representation of spin 1 does not describe photons since it has longitudinal (helicity 0) modes in addition to the transversal (helicity ##\pm1##) ones.

Strangely enough there seem to be no other massless particles which transform as irreducible representations of the Lorenz group, since we know that neutrinos have mass.

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Unfortunately, the position operator depends on how one chooses the gauge - which is unacceptable in a physically valid interpretation.But the longitudinal modes do not interact and are therefore unobservable. So they are basically gauge degrees and you can choose them as seems convenient.

Strangely enough there seem to be no other massless particles which transform as irreducible representations of the Lorenz group, since we know that neutrinos have mass.

Gravitons are postulated to be massless particles which transform as irreducible spin 2 representations of the Poincare group

- #16

samalkhaiat

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Why do you state things backward? The postulate says: every observable is represented by Hermitian operator. The converse is not true for we can construct an infinite number of hermitian operators which do not correspond to any observable.According to the general principles of quantum theory,anyhermitian operator (acting in the physical Hilbert space) can, in principle, be an observable.

Physical Hilbert space is

Name to me one non-covariant observable in QFT.And even if action of the theory is Lorentz invariant, most hermitian operators do not transform as tensors under Lorentz transformations.

This would contradict the principle of relativity:And it is certainly possible to arange measurement procedures the results of which do depend on motion of the measuring apparatus.

Yeah, one can measure all sort of things: I can do experiment and find that I have £20 in my pocket. What are you going to gain from my experiment?A non-covariant measurable quantity is probably not fundamental, but it doesn't mean that it isn't measurable.

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Electric field! (Or magnetic field.)Name to me one non-covariant observable in QFT.

It is a 3-vector, but does not transform as a space-part of a 4-vector.

There is a way to introduce a 4-vector for the electric (or the magnetic) field, but this explicitly depends on the 4-velocity of the observer:

http://arxiv.org/abs/1302.5338

And don't tell me that the "true" observable is the electromagnetic tensor ##F_{\mu\nu}##. Yes, it is also an observable, but electric field E and magnetic field B are observables too because there are well defined measurement procedures that measure only electric (or only magnetic) field.

So yes, some observables do depend on the observer. I see nothing very surprising about that.

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I can gain a lot from it. For instance, it increases my subjective estimate of probability that at the time of measurement you were in UK.Yeah, one can measure all sort of things: I can do experiment and find that I have £20 in my pocket. What are you going to gain from my experiment?

- #19

DrDu

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This is just a mathematical game without physical content. The lack of vector properties under rotations disqualifies it as a physical position operator, as we observe this rotation covariance in very ordinary situations. Hawton's position operator does not give orientation-independent probabilities for observing a photon in a spherical region of space. See also this discussion.

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- #21

samalkhaiat

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[itex](\vec{E} , \vec{B})[/itex] have well defined Lorentz transformations. You can not have consistent physical theory based on [itex]\vec{E}[/itex] alone or [itex]\vec{B}[/itex] alone.Electric field! (Or magnetic field.)

It is a 3-vector, but does not transform as a space-part of a 4-vector.

QFT of photons is based on [itex]A_{\mu}[/itex] not on [itex]\vec{E}[/itex] alone or [itex]\vec{B}[/itex] alone.And don't tell me that the "true" observable is the electromagnetic tensor ##F_{\mu\nu}##. Yes, it is also an observable, but electric field E and magnetic field B are observables too because there are well defined measurement procedures that measure only electric (or only magnetic) field.

Who said they don’t? The field tensor [itex]F_{\mu\nu}[/itex], the energy-momentum 4-vector [itex]P_{\mu}[/itex] and all other covariant objects are observer (i.e. frame) dependent quantities. Being covariant means that you can unambiguously communicate (i.e. transform) the values you get, say for [itex]P_{\mu}[/itex], to other people.So yes, some observables do depend on the observer. I see nothing very surprising about that.

Look, I am not going to argue about an issue which was resolved 50 years ago. The last nail in the coffin was hammered by the following excellent paper (with long title):

R. A. Coleman (1978), “

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samalkhaiat

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No, you don’t gain anything because I don’t have any rule to transform (or communicate) the outcome of my experiment to you. It is my money, why should I let you know this piece of information about me?I can gain a lot from it. For instance, it increases my subjective estimate of probability that at the time of measurement you were in UK.

- #23

DrDu

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Interesting, thanks!See also this discussion.

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Actually, I have to take it back.And don't tell me that the "true" observable is the electromagnetic tensorFμν. Yes, it is also an observable

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Can you name one hermitian operator in non-relativistic QM which is not an observable?Why do you state things backward? The postulate says: every observable is represented by Hermitian operator. The converse is not true for we can construct an infinite number of hermitian operators which do not correspond to any observable.

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