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Background

As someone who has just finished my first year of undergrad, I don't really have any experience with QFT but have some experience with quantum mechanics and the math behind it. I have taken college courses on Newtonian mechanics, special relativity, and electromagnetism.

In the "Fields" episode of his "Biggest Ideas In The Universe" series on YouTube, Sean Carroll says that classical fields have a configuration $\Phi(\vec{x})$ which depends on the position $\vec{x}$.

He then goes on to say that quantum fields are described by a wave function $\Psi(\Phi(\vec{x}))$ which is the probability amplitude of the quantum field being in the configuration $\Phi(\vec{x})$.

Next, he considers a non-interacting (And also I guess he's assuming it's non-relativistic) "free" quantum field and says that you can use Fourier analysis to decompose the field configuration $\Phi(\vec{x})$ into its constituent frequencies or modes. Each mode is described by its wave vector $\vec{k}$ where the wavelength $\lambda = \frac{2\pi}{|\vec{k}|}$ So that the mode can be labeled as $\Phi_{\vec{k}}(\vec{x})$.

The kinetic energy of each mode is $K = \frac{1}{2}(\frac{\ d \Phi_{\vec{k}}}{\ dt})^2$ and the potential energy is $V = \frac{1}{2}m^2(\Phi_{\vec{k}})^2$ where $m$ is the mass of the particles described by the field.

In general, the energies of a mode are proportional to $h^2$ where $h$ is the amplitude of the mode and a wave function of the mode $\Psi(\Phi_{\vec{k}}(h))$ is the probability amplitude of the mode $\Phi_{\vec{k}}$ having the amplitude $h$. However, the energy levels of the mode are quantized (like in the quantum harmonic oscillator) so that there is a wave function for each energy level $n$ labeled $\Psi_n(\Phi_{\vec{k}}(h))$ for $n = 0$ to $\infty$.

Sean Carroll's Diagram: enter image description here

Finally, he says that an interpretation of the $n$th energy state $\Psi_n$ for the mode $\Phi_{\vec{k}}$ is as follows:

Since $\vec{k}$ represents the momentum of the mode, $\Psi_n$ is the probability amplitude of measuring $n$ particles with momentum/wave number $\vec{k}$.

The Question

It seems that this description basically allows you to find the probability of having $n$ particles with momentum/wave number $\vec{k}$. It tells you the probability amplitude of having $n$ particles with the same momentum but it doesn't tell you the probability amplitude of the positions of each of those $n$ particles.

My question is how can we find the probability of measuring a particle associated with the quantum field at some position $\vec{x}$ given that we know the probability of having $n$ particles with momentum / wave number $\vec{k}$. Would this require a different explanation than what Sean Carroll gave? Or can it be explained by continuing off of his explanation?

His lecture was a more casual explanation of quantum field theory with only some math and not going into too much detail. I was hoping for a similar casual explanation of my question with some math that builds off of his. Also, if there is anything in his explanation that is outright wrong and could be made more precise without overly complicating things please let me know of that as well.

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    $\begingroup$ Is the question specifically asking about relativistic (Lorentz-symmetric) QFT? That makes a big difference. $\endgroup$ Jul 25, 2020 at 19:00
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    $\begingroup$ @ChiralAnomaly Since it is an intro to QFT video I'm assuming it's non-relativistic. It wasn't really an advanced physics lecture. $\endgroup$
    – mihirb
    Jul 25, 2020 at 19:01
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    $\begingroup$ @ChiralAnomaly I don't really have much experience with QFT but have some experience with quantum mechanics $\endgroup$
    – mihirb
    Jul 25, 2020 at 19:04
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    $\begingroup$ "his explanation was meant to be more of a casual explanation of quantum field theory" casual explanations are casual. This sort of thing is clear in actual explanations. As @ChiralAnomaly says the context matters etc etc $\endgroup$
    – user21299
    Jul 25, 2020 at 23:11

2 Answers 2

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Always remember that the point of view of a single individual may contain certain biases that can skew one's understanding. As a general principle, always gather as much information from diverse sources to get a better picture.

I don't want to say anything negative about Sean Carroll's video, but perhaps he is making the situation a bit more complicated than it needs to be in an attempt to be as general as possible.

Although the wave function of a multi-particle state can be rather complicated, the wave function for a single particle is essentially the same as that of the classical field. In other words, if the field $\Phi(x)$ in the classical case obeys the equations of motion, then in the quantum case for a single particle $\Phi(x)$ would be the wave function (barring some dimensional changes) that would obey the same equations of motion. The same equations of motion should govern both the classical field and the quantum wave function.

This can be confirmed experimentally. One can use a small detector to measure the classical optical field at different locations $x$. In the quantum case, one can use a single photon detector to measure the wave function as a function of $x$. If the initial conditions that produce these fields are the same, then the functions would be the same. In both cases, the measurements give us $|\Phi(x)|^2$. In the classical case, it is interpreted as a local intensity and, in the quantum case, it is a probability, but other than that they are the same.

For multiple particles, there are more possibilities. What makes it complicated is the fact that the different particles can have different wave functions and, on top of that, one can have superpositions of different such combinations. However, such complicated state are difficult to produce. Usually, all particles in a multi-particle state would have the same wave function. In that case, it is very much like the classical case. In fact, in the limit of a large number of particles with superpositions of different numbers of particles all having the same wave function, the state can essentially be considered a classical state. That is what one has with a coherent state.

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  • $\begingroup$ Thanks for the answer! I'll keep in mind to keep to not take Carroll's explanation as the only possible interpretation and look at other sources on QFT as well. $\endgroup$
    – mihirb
    Jul 26, 2020 at 5:22
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The question can be addressed and resolved in a way that spans the divide between classical and quantum physics, between relativity and non-relativistic theory, and applies across the board uniformly to them all; and this I will do here, now.

In the quantum case, the analysis will lead to a set of position-momentum Heisenberg triples - one for each non-interacting body in the system - and the usual method for resolving probability distributions under constraints on the momentum may then be applied. To get sensible results, you can't have a sharp constraint on the momentum, but rather something smeared, like with a Gaussian. Then, in that case, you are using coherent states with the position and momentum operators.

Here's the general answer.

A physical system has a state space that is acted on by the kinematic symmetry group associated with the underlying paradigm. In the case of non-relativistic physics, this is the central extension of the Galilei group, known as the Bargmann group. For Relativity, it is the PoincarΓ© group. However, to make it compatible with the correspondence limit, and to have the Bargmann group as its non-relativistic limit, it should actually be raised to a (trivially) central extended form of the PoincarΓ© group. We will see where that comes into play in a moment.

The kinematic groups define symmetry transforms for spatial translations, time translations, rotations and "boosts". The latter are what a transform from a stationary to a moving frame is called. By the Noether Theorem, for an isolated non-interacting system, associated with each type of symmetry transform is a conserved quantity - its Noether charge. The charges, themselves, are acted on by the symmetry transforms as the "co-adjoint" representation of the symmetry group. The Noether charges are conventionally identified as:

  • linear momentum $𝐏$, for spatial translations,
  • angular momentum $𝐉$, for rotations,
  • (negative) energy $-H$, for time translations,
  • adjusted mass moment $𝐊$, for boosts,
  • mass $ΞΌ$, for translations with respect to a fictitious fifth coordinate.

The last transform is needed, in the non-relativistic case, to provide a matching coordinate representation for the Bargmann group. This is exactly the extension that takes place when going from the Galilei group to the Bargmann group, and in the corresponding central extension, $ΞΌ$ is the central charge. I call it $ΞΌ$, instead of $m$, for reasons that will become shortly apparent.

The Galilei group is what results from setting $ΞΌ = 0$. Its representations are "zero mass" systems, which can either be instantaneous: like an instantaneous action-at-a-distance impulse, at a specific time, acting over an interval in space in a specific direction, or homogeneous: uniform over space, with no center of mass. A special case of this is the homogeneous system that is isotropic and boost-invariant: the "vacuum".

Under the co-adjoint action, all the Noether charges are invariant with respect to translation on the extra coordinate, so that it acts as a gauge degree of freedom. By the Noether theorem, this makes the associated Noether charge, the mass $ΞΌ$ a conserved quantity.

For a non-relativistic system with these Noether charges, and with non-zero mass, it is possible to use the co-adjoint action to transform the momentum by a boost to $𝐏 β†’ 𝟬$. This lands in the "rest frame" of the system. Under this boost, the angular momentum transforms as $𝐉 β†’ 𝐒$ to the internal angular momentum of the system, and the energy $H β†’ U$ to the internal energy of the system. The mass and moment remain invariant: $ΞΌ β†’ ΞΌ$, $𝐊 β†’ 𝐊$ as they are both boost-invariant. In Relativity, matters are a little more complicated: the moment in that case is not quite boost invariant.

In the rest frame, $𝐊$ is a bona fide mass moment. Under a spatial translation, it may be transformed $𝐊 β†’ 𝟬$. All other Noether charges remain unchanged. This lands you at the center of mass of the system.

Applying these processes in reverse with the following transforms leads to the following results for the step-by-step transform of $(𝐉,𝐊,𝐏,H,ΞΌ)$:

  • Spatial translation by $𝐫$:
    $(𝐒,𝟬,𝟬,U,ΞΌ) β†’ (𝐒,μ𝐫,𝟬,U,ΞΌ)$,
  • Boost by $-𝐯$:
    $(𝐒,μ𝐫,𝟬,U,ΞΌ) β†’ \left(μ𝐫×𝐯 + 𝐒, μ𝐫, μ𝐯, Β½ΞΌ|𝐯|^2 + U, ΞΌ\right)$,
  • Time translation by $t$:
    $\left(μ𝐫×𝐯 + 𝐒, μ𝐫, μ𝐯, Β½ΞΌ|𝐯|^2 + U, ΞΌ\right) β†’ \left(μ𝐫×𝐯 + 𝐒, ΞΌ(𝐫 - 𝐯t), μ𝐯, Β½ΞΌ|𝐯|^2 + U, ΞΌ\right)$,
  • Final result:
    $\left(μ𝐫×𝐯 + 𝐒, ΞΌ(𝐫 - 𝐯t), μ𝐯, Β½ΞΌ|𝐯|^2 + U, ΞΌ\right) = (𝐉,𝐊,𝐏,H,ΞΌ),$

which leads to the following decomposition: $$𝐉 = m𝐫×𝐯 + 𝐒, \quad 𝐊 = m(𝐫 - 𝐯t), \quad 𝐏 = m𝐯, \quad H = \frac{m|𝐯|^2}{2} + U, \quad ΞΌ = m.$$ This describes a system of mass $m = ΞΌ$, a center of mass position $𝐫$ and associated velocity $𝐯$. You will also notice that the mass moment is projected back to time $t = 0$. It has an explicit dependency on $t$ and - by this account - a degree of ambiguity. The locus of all positions, as you vary time $t$ is the trajectory of the system's center of mass.

The appearance of internal values for angular momentum and energy are justified, after the fact, by the requirement that a composite system, having such elementary systems as components, should also have a set of attributes of its own that are obtained by requiring all the Noether charges to be additive. Here, we can see that for two bodies, with the following rule of composition: $$ ΞΌ = ΞΌ_0 + ΞΌ_1, \quad 𝐫 = \frac{ΞΌ_0𝐫_0 + ΞΌ_1𝐫_1}{ΞΌ}, \quad 𝐯 = \frac{ΞΌ_0𝐯_0 + ΞΌ_1𝐯_1}{ΞΌ},\\ 𝐒 = 𝐒_0 + 𝐒_1 + \bar{ΞΌ}\bar{𝐫}Γ—\bar{𝐯}, \quad U = U_0 + U_1 + \frac{\bar{ΞΌ}|{\bar{𝐯}}|^2}{2},\\ \bar{ΞΌ} = \frac{ΞΌ_0ΞΌ_1}{ΞΌ}, \quad \bar{𝐫} = 𝐫_0 - 𝐫_1, \quad \bar{𝐯} = 𝐯_0 - 𝐯_1. $$ The internal dynamics of the two component bodies making up the larger body is captured by a fictitious body with effective mass $\bar{ΞΌ}$, position $\bar{𝐫}$ and velocity $\bar{𝐯}$. The non-linearity of the expressions for $𝐉$ and $H$ is compensated for by extra additions made to $𝐒$ and $U$, respectively.

There is nothing in principle that requires either of these internal values to vanish for non-composite systems. Therefore, it is meaningful to speak of "spin" and not just "internal energy" - even for classical physics, and even for non-relativistic theory. Also, the decomposition helps bolster the description of $𝐫$ and $𝐯$, respectively, as center of mass position and velocity.

For any system, be it elementary or composite, the relations may be inverted, and position may be defined in terms of the Noether charges as $$𝐫 = \frac{𝐊 + 𝐏t}{μ}.$$

All of what was described here applies to both classical and quantum theory. However, when passing over to quantum theory, the Noether charges become "q-numbers" and may have non-zero commutators, which can lead to operator ordering ambiguity. Fortunately, for the decomposition just presented, there is no non-commutativity between the factors at all. In Relativity, however, there will be a minor exception in the case of $𝐊$.

How are the commutators determined? In the classical theory, Poisson brackets may be associated with the Noether charges by replicating the Lie brackets on the Noether charges, themselves. For instance: $$\left\{K_i, P_j\right\} = δ_{ij} μ, \quad \left\{J_x, J_y\right\} = J_z,$$ with cyclic permutations of $(x,y,z)$ in the latter case. In quantized form, this becomes, respectively: $$\left[K_i, P_j\right] = iħ δ_{ij} μ, \quad \left[J_x, J_y\right] = iħJ_z.$$ The parameter $t$ is treated as a classical coordinate and does not directly enter any of the relations.

From this follows the usual Heisenberg relations, e.g. $$\left[𝐫, P_x\right] = \left[\frac{𝐊 + 𝐏t}{ΞΌ}, P_x\right] = \frac{\left[𝐊, P_x\right]}{ΞΌ} = \frac{iΔ§ΞΌ(1,0,0)}{ΞΌ} = iΔ§(1,0,0),$$ where $𝐫 = (x,y,z)$, as well as those for internal angular momentum, e.g.: $$\left[S_x, S_y\right] = iΔ§S_z.$$ The two sets of operators for $𝐫$ and $𝐒$ arise from quantities that have zero Poisson brackets with each other, so that they are mutually commuting.

Your question restricted its focus to free systems - many-body systems whose component bodies are mutually non-interacting. In such a system, each body possesses bona fide Noether charges of their own as conserved quantities, so that all of what was described applies to each body individually, and not just to the whole system.

This is true in the classical case and in the quantum case of a free field described as a many-body system with a Fock space.

If the bodies are interacting, then you lose the ability to make a clear description of the individual bodies, since their "Noether charges" are no longer conserved quantities and are not actually Noether charges at all anymore. In the non-relativistic case, you can still use them as time-varying quantities and describe the system dynamics in terms of them. That would be a lot more difficult to do in the Relativistic case, since you have a road-block known as the no-interaction theorem (or Leutwyler Theorem) in the classical case, and the Haag Theorem in the quantum case.

Now, for the relativistic case, if we use the central charge, then the brackets for the moment and momentum become the following: $$\left\{K_i, P_j\right\} = Ξ΄_{ij} \left(ΞΌ + \frac{H}{c^2}\right) \quad (i,j = x, y, z),$$ while for the moment, the brackets become $$\left\{K_x, K_y\right\} = -\frac{J_z}{c^2},$$ with similar brackets obtained by cyclically permuting $(x,y,z)$. These are actually the only differences between the extended PoincarΓ© group and Bargmann group. So, we can treat it as a deformation of the Bargmann group with an extra "deformation" parameter that goes from $0$ to $1/c^2$.

What makes this an extension of PoincarΓ© is that it has 11 Noether charges, instead of 10, and the respective Lie algebra has 11 dimensions, instead of 10. The reduction to 10 is obtained by combining the energy $H$ - which is just kinetic energy plus internal energy - and the mass $ΞΌ$ into a single parameter $E$ for "total energy". Equivalently we can also express this as the "relativistic mass" $M = ΞΌ + H/c^2$. The relation between the two, of course, is $E = Mc^2$.

Both $H$ and $ΞΌ$ can then be replaced by $E$ or by $M$. Conventionally, the choice is with $E$, though $M$ actually gives rise to more intuitive descriptions and - more importantly - has a correspondence limit to non-relativistic theory. There is no non-relativistic limit for $E = Mc^2$, since it blows up to infinity as $1/cΒ² β†’ 0$.

In non-relativistic theory, there are two invariants that can be formed from the central charge, energy and momentum: $$ΞΌ, \quad |𝐏|^2 - 2ΞΌH.$$ Under deformation to the extended PoincarΓ© group, these respectively become: $$ΞΌ = M - \frac{H}{c^2}, \quad |𝐏|^2 - 2ΞΌH - \frac{H^2}{c^2} = |𝐏|^2 - 2MH + \frac{H^2}{c^2}.$$ The only combination (up to functional dependence) that can be formed from them that removes the explicit dependency on $(ΞΌ, H)$ is: $$ΞΌ^2 - \frac{1}{c^2}\left(|𝐏|^2 - 2ΞΌH - \frac{H^2}{c^2}\right) = M^2 - \frac{|𝐏|^2}{c^2} = m^2,$$ which - for systems satisfying the condition $M^2 c^2 β‰₯ |𝐏|$ - may be identified as the square of the rest mass $m$. The term "rest" is justified - at least in the case $|m| > 0$ by the ability to transform to a rest frame, where $M β†’ m$.

If we keep the central charge intact and stay within the extended PoincarΓ© group, then it is possible to carry out a set of transforms similar to the non-relativistic case, but slightly more complicated, that lands at: $$(𝐉,𝐊,𝐏,H,ΞΌ) β†’ (𝐒,𝟬,𝟬,U,ΞΌ),$$ just like before. Finding the right combination of transforms to achieve this, however, turns into an exercise much like solving the Rubik's Cube.

Going in the opposite direction, upon transform back to a general state, what we get is the following decomposition that is a deformation of the non-relativistic case: $$ 𝐉 = M𝐫×𝐯 + 𝐒, \quad 𝐊 = M(𝐫 - 𝐯t) + \frac{1}{c^2}\frac{M𝐯×𝐒}{m+M},\\𝐏 = M𝐯, \quad H = \frac{M|𝐯|^2}{m + M} + U, \quad ΞΌ = m - \frac{U}{c^2}, $$ where $$M = \frac{m}{\sqrt{1 - |𝐯|^2/c^2}}.$$

The reduction from 11 charges to 10 that arises from replacing $H$ and $ΞΌ$ with $E$ or $M$ corresponds to the elimination of $U$, resulting in the following relations in their place $$ 𝐉 = M𝐫×𝐯 + 𝐒, \quad 𝐊 = M(𝐫 - 𝐯t) + \frac{1}{c^2}\frac{M𝐯×𝐒}{m+M},\\ 𝐏 = M𝐯, \quad H = \frac{M|𝐯|^2}{m + M} = E - mc^2, \quad ΞΌ = m\\ M = \frac{m}{\sqrt{1 - |𝐯|^2/c^2}} = \frac{E}{c^2}. $$ This makes both $ΞΌ$ and $H$ redundant and unnecessary, so that they drop out.

As a result, the internal energy $U$ drops out of the picture. I don't know if retaining it would provide a means to overcome the no-interaction theorem, but without its presence, there is nothing to absorb the non-additivity for the energy, and this creates part of the problem that leads to the no-interaction theorem. But the theorem may be robust enough to continue to block non-trivial interacting many-body systems, even if internal energy is included for each body.

You will notice that the moment has acquired a dependence on the spin or internal angular momentum $𝐒$. This arises from the $𝐊-𝐊$ Poisson brackets, now being a non-zero multiple of $𝐉$. This is directly connected to a feature unique to Relativity: that two successive boosts in non-collinear directions results also in a small rotation.

Another feature that arises is that the components of the Noether charge don't all commute (or have non-zero Poisson brackets) anymore. Though they continue to do so for the Noether charges $𝐉$, $𝐏$, $H$, $ΞΌ$ (and for $M$ and $E$), they don't for $𝐊$. The Poisson bracket for the mass and moment deforms from the non-relativistic case to the relativistic case as: $$0 = \{ΞΌ, 𝐊\} β‡’ \{M, 𝐊\} = \frac{1}{c^2}\{E, 𝐊\} = \frac{M}{c^2}\{E, 𝐫\} = \frac{M}{c^2}𝐯 = \frac{𝐏}{c^2}.$$ Thus, $$\{M, 𝐫\} = \frac{\{M, 𝐊\}}{M} = \frac{𝐯}{c^2}.$$ Correspondingly, for the commutators, we have: $$[M, 𝐫] = iΔ§\frac{𝐯}{c^2}.$$ The operator ordering ambiguity may be absorbed into a shift on $t$ in the expression $M(𝐫 - 𝐯t)$, which is already a free-running parameter (though the shift is by an imaginary value). This results in a time-uncertainty in the position of the body on its center of mass trajectory.

The prevailing convention is to adopt Weyl operator ordering, i.e. $$\widehat{M𝐫} = \frac{1}{2}(\hat{M}\hat{𝐫} + \hat{𝐫}\hat{M}).$$ Similarly, when inverting the relation between $𝐊$ and $𝐫$ to obtain an expression for $𝐫$, Weyl operator ordering may be used in the quantum case. The resulting expression is known as the Newton-Wigner position operator. $$𝐫 = \frac{𝐊 + 𝐏t}{M} + \frac{1}{Mc^2}\frac{𝐒×𝐏}{m+M} $$ As before, the Poisson brackets on the Noether charges reduce to the usual Poisson and Heisenberg brackets on $𝐫$ and $𝐒$, as a result. Upon quantization, with Weyl ordering, the part of $𝐊$ with the non-commuting factors is set as: $$\widehat{\frac{𝐊}{M}} = \frac{1}{2}\left(\frac{1}{\hat{M}}\hat{𝐊} + \hat{𝐊}\frac{1}{\hat{M}}\right).$$

There is a dependence on the internal angular momentum or (for elementary systems: spin), $𝐒$. But the extra term commutes with $𝐏$, so it has momentum eigenstates as its own eigenstates.

For a non-interacting many-body system, this may be applied to each body's set of Noether charges, as well as to the system as a whole. The way seems to be blocked, however, to making this work - in both the classical and quantum cases - for interacting bodies, where we allow the "Noether charges" to be time-varying. The Haag Theorem blocks non-trivial many-body dynamics on Fock space in the quantum case, while the Leutwyler Theorem does the same for the classical case.

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