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single-particle wavepackets in QFT and position measurement

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Consider a scalar field $\phi$ described by the Klein-Gordon Lagrangian density $L = \frac{1}{2}\partial_\mu \phi^\ast\partial^\mu \phi - \frac{1}{2} m^2 \phi^\ast\phi$.

As written in every graduate QM textbook, the corresponding conserved 4-current $j^\mu = \phi^\ast i \overset{\leftrightarrow}{\partial^\mu} \phi$ gives non-positive-definite $\rho=j^0$. If we are to interpret $\phi$ as a wave function of a relativistic particle, this is a big problem because we would want to interpret $\rho$ as a probability density to find the particle.

The standard argument to save KG equation is that KG equation describes both particle and its antiparticle: $j^\mu$ is actually the charge current rather than the particle current, and negative value of $\rho$ just expresses the presence of antiparticle.

However, it seems that this negative probability density problem appears in QFT as well. After quantization, we get a (free) quantum field theory describing charged spin 0 particles. We normalize one particle states $\left|k\right>=a_k^\dagger\left|0\right>$ relativistically:

$$ \langle k\left|p\right>=(2\pi)^3 2E_k \delta^3(\vec{p}-\vec{k}), E_k=\sqrt{m^2+\vec{k}^2} $$

Antiparticle states $\left|\bar{k}\right>=b_k^\dagger \left|0\right>$ are similarly normalized.

Consider a localized wave packet of one particle $\left| \psi \right>=\int{\frac{d^3 k}{(2\pi)^3 2E_k} f(k) \left| k \right>}$, which is assumed to be normalized. The associated wave function is given by

$$ \psi(x) = \langle 0|\phi(x)\left|\psi\right> = \int{\frac{d^3 k}{(2\pi)^3 2E_k} f(k) e^{-ik\cdot x}} $$

$$ 1 = \langle\psi\left|\psi\right> = \int{\frac{d^3 k}{(2\pi)^3 2E_k} |f(k)|^2 } = \int{d^3x \psi^\ast (x) i \overset{\leftrightarrow}{\partial^0} \psi (x)}$$.

I want to get the probability distribution over space. The two possible choices are:

1) $\rho(x) = |\psi(x)|^2$ : this does not have desired Lorentz-covariant properties and is not compatible with the normalization condition above either.

2) $\rho(x) = \psi^\ast (x) i \overset{\leftrightarrow}{\partial^0} \psi(x)$ : In non-relativistic limit, This reduces to 1) apart from the normalization factor. However, in general, this might be negative at some point x, even if we have only a particle from the outset, excluding antiparticles.

How should I interpret this result? Is it related to the fact that we cannot localize a particle with the length scale smaller than Compton wavelength ~ $1/m$ ? (Even so, I believe that, to reduce QFT into QM in some suitable limit, there should be something that reduces to the probability distribution over space when we average it over the length $1/m$ ... )

This post imported from StackExchange Physics at 2015-05-16 18:38 (UTC), posted by SE-user pdfs
asked Nov 3, 2013 in Theoretical Physics by pdfs (35 points) [ no revision ]
retagged May 16, 2015
Most voted comments show all comments
In your mind, what does mean the notation $\psi^\ast i \overset{\leftrightarrow}{\partial^0} \psi$ ? If you think that , here, $\psi$ is a state $|\psi\rangle$, this is a nonsense, because $|\psi\rangle$ does not depends on $t$. If you think that $\psi$ is an operator, this is a nonsense too, because t the integration on $x$ gives an operator too, and it cannot be equals to $1$, which is a real number, but not an operator.

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Trimok
Yes, I do. By substituting $ \psi(x) = \int{\frac{d^3 k}{(2\pi)^3 2\omega_k} f(k) exp(-ik\cdot x)}$ , you can directly verify the equation you pointed out.. By the way, how can I use LaTeX in comment lines?

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user pdfs
Circle the formulae with \$,for instance \$ \psi \$ gives $\psi$

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Trimok
I see. I edited what you have commented in order to avoid possible confusion. Thank you.

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user pdfs
First your equation is valid only for $t=0$, remember that $kx = \vec k.\vec x - k_0t$, and you are using properties of a 3-dimensional Fourier transform. But the main problem is that you are mixing operators ($\Phi$), states ($|\psi\rangle$), and, in the very particular case of a one-particle state, a pseudo - "wavefunction" $\psi(x)$. So, you cannot start from a possible negative eingenvalue of an operator $j^0$, and conclude that you have a possible negative value for $\rho(x)$ for some $x$, because these are 2 different mathematical entities.

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Trimok
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There is a (short) example of use of the wave function in QFT in this paper, paragraphs $5.3.1,5.3.2$, pages $133-138$

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Trimok
Also relevant to the discussion of wavefunctions in QFT is the question of the causal behaviour of the "naive" wavefunction (not the Schroedinger wavefunctional that Trimok is referring to). This is discussed here

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user twistor59

3 Answers

+ 5 like - 0 dislike

The following does not completely answer OP's question, rather, this is going to be a clarification of subtleties and difficulties on the issue. Notice that sometimes I am going to use the same notations as OP used, but not necessarily having the exact same meanings and I will make them clear in the context.

The usage of "wavefunction"

It has two possible meanings when people refer to something as wavefunctions :

(1). The collection of some functions $g(x)$ furnishes a positive-energy, unitary representation of the underlying symmetry group(in our case just the Poincare group).

(2). In addition to (1) being satisfied, we should be able to interpret $|g(x)|^2$ as the probability distribution of finding the particle in $(\mathbf{x},\mathbf{x}+d\mathbf{x})$. It really has to be of the form $|g(x)|^2$ according to the standard axioms of quantum mechanics, provided you interpret $g(x)$ as an inner product $\langle x|g\rangle$ where $\langle x|$ is an eigen bra of the position operator. I will discuss what position operator means later.

The 2nd meaning is of course much stronger and OP is searching for a wavefunction in this sense. However, the $\psi(x)$ written by OP is only a wavefunction in the 1st sense, because clearly $\langle 0|\phi(x)$ cannot be eigen bras of any Hermitian position operator, as easily seen from the fact that they are not even mutually orthogonal, i.e. $\langle 0|\phi(x)\phi^\dagger(y)|0\rangle\neq0$ even when $x$ and $y$ are spacelike separated. As a consequence, $\int d^3\mathbf{x}|\psi(x)|^2\neq1$ as OP has already noted. Moreover, $|\psi(x)|^2$ is invariant under Lorentz transformation, but a density distribution should transform like the 0th component of a 4-vector in relativistic space-time, as OP has also noticed.

The localized states and position operator(Newton-Wigner)

In this section I will mostly rephrase(for conceptual clarity in sacrifice of technical clarity) what is written in this paper.

What is a sensible definition of position operator of single-particle states? First we need to think about what the most spatially-localized states are, and then it would be natural to call these states $|\mathbf{x} \rangle$, then it is also natural to call the operator having these states as the eigenstates the position operator. It seems pretty reasonable and not too much to ask to require localized states to have the following properties:

(a). The superposition of two localized states localized at the same position in space should again be a state localized at the same position.

(b). Localized states transform correctly under spatial rotation, that is, $|\mathbf{x} \rangle \to |R\mathbf{x} \rangle$ under a rotation $R$.

(c). Any spatial translation on a localized state generates another localized state that is orthogonal to the original, that is, $\langle\mathbf{x}+\mathbf{y}|\mathbf{x} \rangle=0$ if $\mathbf{y}\neq 0$.

(d). Some technical regularity condition.

It turns out these conditions are restrictive enough to uniquely define localized states $|\mathbf{x}\rangle$. It can be worked out that, borrowing OP's notations and normalization convention, if $|\psi\rangle=\int\frac{d^3 k}{(2\pi)^3 2E_k} f(k) |k\rangle$, then(including time dependence) $$\langle x|\psi\rangle=\int\frac{d^3 k}{(2\pi)^3 \sqrt{2E_k}} f(k) e^{-ik\cdot x},$$

and this (unsurprisingly) gives $\int d^3\mathbf{x}|\langle x|\psi\rangle|^2=1$. However, this is not the full solution to OP's problem, because we can show that, although not as bad as transforming invariantly, $|\langle x|\psi\rangle|^2$ is not as good as transforming like a 0th component, either. The underlying reason is, as already realized by Newton and Wigner, that a boost on a localized state will generate a delocalized state, so the interpretation is really frame dependent.

As I disclaimed, I do not know if there is a complete satisfactory solution, or if it is even possible, but I hope it helps to clarify the issue.

Appendix: Some interesting properties of Newton-Wigner(NW) states and operator

I decide to make it an appendix since I think this is not directly relevant yet very interesting(all the following are for scalar field, and NW also discusses spinor field in their paper):

(1). A state localized at the origin, projected to the bras $\langle 0|\phi(x)$, has the form

$$\langle 0|\phi(x)|\mathbf{x}=0\rangle=\left(\frac{m}{r}\right)^{\frac{5}{4}}H_{\frac{5}{4}}^{(1)}(imr),$$ where $r=(x_1^2+x_2^2+x_3^2)^{\frac{1}{2}}$ and $H_{5/4}^{(1)}$ is a Hankel function of the first kind. So it is not a delta function under such bras.

(2)The NW position operator $q_i (i=1,2,3)$ acting on momentum space wavefunction(defined as $f(k)$ in OP's notation) is $$q_if(k)=-i\left(\frac{\partial}{\partial k_i}+\frac{k_i}{2E_k}\right)f(k),$$ and in nonrelativistic limit the second term approaches 0, giving the familiar expression of a position operator. We can also get, if $\Psi(x)=\langle 0|\phi(x)|\Psi\rangle$, then $$q_i\Psi(x)=x_i\Psi(x)+\frac{1}{8\pi}\int\frac{\exp(-m|\mathbf{x-y}|)}{|\mathbf{x-y}|}\frac{\partial \Psi(y)}{\partial y_i}d^3\mathbf{y},$$ and here the nonrelativistic limit is hidden in the unit of $m$, when converting back to SI units, the $m$ on the exponent is really the inverse of Compton wavelength, which can be taken as $\infty$ for low energy physics, so again the 2nd term vanishes.

(3)$[q_i,p_j]=i\delta_{ij}$.


This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Jia Yiyang

answered Nov 4, 2013 by Jia Yiyang (2,465 points) [ revision history ]
edited May 18, 2015 by Jia Yiyang
+ 3 like - 0 dislike

One textbook explanation of this problem is to interpret the probability density as a charge density. However this explanation is meaningless in the case of a real Klein-Gordon field.

There is an other solution in which one works with the following Hamiltonian for a spinless particle:

$H(p, x) = + \sqrt{p^2+m^2}+ V(x)$

This Hamiltonian is positive definite by construction, but it is a fractional pseudo-differential operator. As a consequence, its "wave equation" (named the Salpeter equation) after quantization is nonlocal. (The parentheses are added because it is not a hyperbolic equation).

However, its nonlocality is consistent with the relativistic causality (light-cone structure), therefore, it is a viable relativistic field equation for a spinless particle.

This Hamiltonian allows a probabilistic interpretation of its "wave mechanics". Its probability amplitude returns to be of the form $\bar{\psi}\psi$ at the expense of nonlocal spatial components of the probability current. Please see the following article by Kowalski and Rembielinśki in which explicit constructions of the positive definite probability amplitudes in many particular cases.

The space of solutions of the Salpeter equation is a Hilbert space, thus it can be second quantized to form a quantum free field, please see the following article by: J.R. Smith. The relativistic causality is also manifested in the second quantized version.

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user David Bar Moshe
answered Nov 4, 2013 by David Bar Moshe (3,875 points) [ no revision ]
+1, interesting references.

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Jia Yiyang
+ 1 like - 0 dislike

I am a tyro here, but I suspect you are finding that you cannot exclude anti-particles in a relativistic quantum theory.

This line of reasoning is presented by Feynman in his 1986 Dirac Memorial Lecture, "The reason for antiparticles", published in Elementary Particles and the Laws of Physics.

He considers the amplitude for a process in which an initial state $\phi_0$ is scattered by a "disturbance" at time $t_1$ into intermediate states, and then back to $\phi_0$ after a second disturbance at a later time $t_2$. The insight: if the intermediate states are restricted to only positive energies, there will be non-zero contributions from intermediate states traveling faster-than-light.

This result follows from a property of the Fourier transform: if

$$f(t) = \int_0^\infty e^{-i \omega t} F(\omega) \, d \omega $$

(i.e. only positive energies) "then $f$ cannot be zero for any finite range of $t$, unless trivially it is zero everywhere".

The sum over intermediate states has exactly this form, so no matter how far apart in distance, the amplitude will be non-zero if $t_2>t_1$. The argument completes by noting that these faster-than-light solutions will appear, in some Lorentz frames, to be time-order-reversed, and therefore as anti-particles.

For the same reason (I think), the positive-energy wave packet you constructed has a non-zero amplitude to travel faster-than-light, and hence cannot be a relativistic single-particle state. Multiple particles (including anti-particles) to the rescue...

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Art Brown
answered Nov 7, 2013 by Art Brown (40 points) [ no revision ]

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