Physics question [QED]

[Surlethe]

I was thinking this morning about the Feynman-Stuckelberg interpretation of antiparticles as particles moving backward in time, and it occurred to me that time reversal (reflection across the xyz-space), given by

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[-1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
[ 0 0 0 1 ],

preserves the Minkowski norm, so it's a member of the general Lorentz group. So you should be able to produce the interpretation by transforming the Dirac equation according to that matrix. My problem is, how do you figure the spinor transformation associated with this Lorentz transformation? I've taken the following two matrices as ansatzes -

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    [ 0 0 1 0 ]  [ 0 0 0 1 ]
S = [ 0 0 0 1 ]  [ 0 0 1 0 ]
    [ 1 0 0 0 ]  [ 0 1 0 0 ]
    [ 0 1 0 0 ], [ 1 0 0 0 ]

- since, naively, they send the electron to the positron and vice-versa (the first preserving spin, the second reversing it, if I am understanding correctly), but neither works out with what the transformation ought to be. Maybe I'm just screwing up the basic math with the equation, but I was wondering if anybody here had some insight.

[Kuroneko]

I was thinking this morning about the Feynman-Stuckelberg interpretation of antiparticles as particles moving backward in time, and it occurred to me that time reversal (reflection across the xyz-space), given by ... preserves the Minkowski norm, so it's a member of the general Lorentz group. So you should be able to produce the interpretation by transforming the Dirac equation according to that matrix. My problem is, how do you figure the spinor transformation associated with this Lorentz transformation?

Let's unwind this a bit. The full Lorentz group O(1,3) is isomorphic to ℤ₂²×SO⁺(1,3), and has four connected components individually isomorphic to the proper orthochronous Lorentz group SO⁺(1,3) ≅ PSL(2,ℂ), the projective Möbius group, which is in turn double-covered by the simply connected SL(2,ℂ). Something is a bit fishy here--we're not getting a double-cover for O(1,3), which we need for half-integer spin, out of the spin group, but rather the pin group.

Look at parity first: xμ ↦ x'μ = (x0,-x).The Dirac equation is
[1] 0 = (p̸-m)ψ = (iγμ∂μ - m)ψ.
Hence γ0(p̸-m)ψ = (iγμ∂'μ-m)γ0ψ = 0. To keep the same form of the Dirac equation, ψ'(x') = γ0ψ(x), so that the parity exchange operator is
[2] P = γ0,
up to some overall phase factor.

That was simple enough, but time reversal is a bit tricky, so let's consider it in the nonrelativistic limit. For the Schrödinger equation i∂tψ = Hψ, you can see that whenever ψ(t,x) is a solution, the time-reversed ψ(-t,x) is generally not a solution, but the conjugate-reversed ψ*(-t,x) is. From this, we can get the ansatz for the time-reversal operator
[3] T = UK,
where K is the operator of complex conjugation and U is unitary.

There's your difficulty: time-reversal is antilinear!

Exercise: Following the same strategy as as for the parity operator, let ψ'(-t,x) = Tψ(t,x), and show that if [T,H] = 0, then keeping the same form of the Schrödinger equation forces T-1(-i)T = i. (The same conclusion as above, with less handwaving.)

Interesting note: for half-integer spins, T² = -1 causes Kramer degeneracy. For suppose toward a contradiction that ψ and Tψ represent the same state, i.e., they are identical up to overall phase φ. Then Tψ = exp(iφ)ψ, and hence T² = +1. Therefore, ψ and Tψ are different states of the same energy.

The parity trick worked so well, let's do it again, only now rearranging the Dirac equation into Schrödinger form after multiplying through by γ0:
[4] i∂tψ(t) = (-iγ0γi∂i + γ0m)ψ(t),
whereas we want
[5] i∂-tψ'(-t) = Hψ'(-t)
As before, T-1HT = H forces, with our ansatz T = UK, that (i) KU-1(iγ0γi)UK = iγ0γi and (ii) KU-1γ0UK = γ0. If you multiply through by K on both left and right, you'll see that in the Dirac basis γμ ↦ U-1γμU sign-reverses γ1,γ3, but keeps γ0,γ2 the same. Aha:
[6] T = γ1γ3K,
again up to overall phase.

Exercise: Verify that γ2 corresponds to charge conjugation.

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In the Dirac basis, γ^1 and γ^3 are
[  0  0  0  1  ]   [  0  0  1  0  ]
[  0  0  1  0  ]   [  0  0  0 -1  ]
[  0 -1  0  0  ] , [ -1  0  0  0  ]
[ -1  0  0  0  ]   [  0  1  0  0  ]

[Surlethe]

Let's unwind this a bit. The full Lorentz group O(1,3) is isomorphic to ℤ₂²×SO⁺(1,3), and has four connected components individually isomorphic to the proper orthochronous Lorentz group SO⁺(1,3) ≅ PSL(2,ℂ), the projective Möbius group, which is in turn double-covered by the simply connected SL(2,ℂ). Something is a bit fishy here--we're not getting a double-cover for O(1,3), which we need for half-integer spin, out of the spin group, but rather the pin group.

Why do we need a double-cover for O(1,3) to get half-integer spin? And what's the pin group?

(Note to myself: ℏ = c = 1.)

Look at parity first: xμ ↦ x'μ = (x0,-x).The Dirac equation is
[1] 0 = (p̸-m)ψ = (iγμ∂μ - m)ψ.
Hence γ0(p̸-m)ψ = (iγμ∂'μ-m)γ0ψ = 0.

Not following this step. Two questions. First, under xμ ↦ x'μ = (x0,-x), we should have ∂μ ↦ ∂'μ by ∂0 = ∂0, ∂μ = - ∂μ, which changes the form of the equation, right? Or is the new coordinate system going to take derivatives by the new coordinates? Second, why γ0 and not γ1, γ2, or γ3?

Exercise: Following the same strategy as as for the parity operator, let ψ'(-t,x) = Tψ(t,x), and show that if [T,H] = 0, then keeping the same form of the Schrödinger equation forces T-1(-i)T = i. (The same conclusion as above, with less handwaving.)

For clarity, denote the "new" time τ = -t. Then ∂τ = [∂t/∂τ]∂t = -∂t. Put ψ'(τ,x) = ψ'(-t,x) = Tψ(t,x). Then the Schrödinger equation in the new coordinates is i∂τψ'(τ,x) = Hψ'(τ,x), which leads to ...
i∂τTψ(t,x) = HTψ(t,x)
-i∂tTψ(t,x) = HTψ(t,x)
i∂tψ(t,x) = (-T-1HT)ψ(t,x),
leading to the requirement that H = -T-1HT, or that H and T are anticommutative. Therefore, if we also have [H,T] = 0, we get T = 0.

I'm screwing up something with the derivatives, aren't I?

Since the rest of the presented argument relies on these two points, I'll end my response here.

[Kuroneko]

Why do we need a double-cover for O(1,3) to get half-integer spin?

Hmm... my statement was too ambiguous. I did not mean that we need to double-cover of O(1,3) in particular to get half-integer spin, but that (i) double-covers are the reason for half-integer spins, in the sense described below, and (ii) Spin(1,3) ≅ SL(2,ℂ) is doing this to SO(1,3) rather than O(1,3). Those two statements ran together somehow and produced something rather misleading.

Let's consider just the spatial part. In quantum mechanics, our states live in Hilbert space H (or close to it, anyway), so since we can rotate our physical system any way we like, we need a unitary representation of SO(3). The Lie algebra of SO(3) has the following generators:
[1] y∂x-x∂y, z∂x-x∂z, z∂y-y∂z,
which should look familiar (cf. other thread), and correspond to angular velocities around the x,y,z axes, respectively. But this Lie algebra is identical to that of SU(2), from which we get extra irreducible representations for each dimension n: n = 1 is trivial (identity map), giving scalars (spin-0), n = 2 is the fundamental representation, giving spinors, and so on. Without this, SO(3) proper only has irreducible representations of odd dimensionality, which corresponds to integer spins. The relativistic case is similar, except that unlike SU(2), SL(2,ℂ) has two distinct n = 2 representations, producing two different types of spinors: left-handed and right-handed.

And what's the pin group?

The group double-covering O(1,3) in the same way that Spin(1,3) double-covers SO(1,3) that admits a Clifford algebra representation (actually, there are two). The point is that your assumption that a general Lorentz transformation is representable with Spin(1,3) is false; only for the proper Lorentz transformations SO(1,3) is this the case. Since time reversal is an improper Lorentz transformation, something needs to give. And give it does--the time reversal operator is not even linear, although it's close.

Not following this step. Two questions. First, under xμ ↦ x'μ = (x0,-x), we should have ∂μ ↦ ∂'μ by ∂0 = ∂0, ∂μ = - ∂μ, which changes the form of the equation, right? Or is the new coordinate system going to take derivatives by the new coordinates?

∂'0 = ∂0 and ∂'μ = -∂μ, yes. What we want is to have a completely identical Dirac equation with primed coordinates over some transformed state, but what we get upon substitution is the wrong sign of the spatial components of the momentum term. So we need to fix this somehow, and multiplying by γ0 from the left does the trick.

Second, why γ0 and not γ1, γ2, or γ3?

Notice that we first multiplied through by it from the left, and then switched the order to the right. In doing so, γ0γ0 trivially doesn't change sign, but γ0γi = -γiγ0. Flipping the sign of only the spatial parts of p̸ is exactly what we want, and the result is (iγμ∂'μ-m)(γ0ψ) = 0. This is exactly the Dirac equation with the transformed state (γ0ψ), from which we conclude that ψ'(x') = γ0ψ(x)

For clarity, denote the "new" time τ = -t. Then ∂τ = [∂t/∂τ]∂t = -∂t. Put ψ'(τ,x) = ψ'(-t,x) = Tψ(t,x). Then the Schrödinger equation in the new coordinates is i∂τψ'(τ,x) = Hψ'(τ,x), which leads to ...
i∂τTψ(t,x) = HTψ(t,x)
-i∂tTψ(t,x) = HTψ(t,x)

Exactly right.

i∂tψ(t,x) = (-T-1HT)ψ(t,x),

That doesn't follow. Don't assume that T is linear. Well, you did prove that if T were linear, then either it's not invertible or H = 0 (not T = 0), so in a sense you proved that it can't be linear, but the exercise's conclusion is stronger.

Let's explore what would happen if T were linear. The time evolution of a state ψ to t = τ is, to first order, (1-iHτ)ψ. If the system is invariant under time-reversal, then evolving a reversed state should be the same as reversing the earlier t = -τ state: (1-iHτ)Tψ = T(1+iHτ)ψ. Thus -iHT = TiH. If T were linear, then for any energy eigenstate φ (Hφ = Eφ), Tφ is an eigenstate of energy -E (HTφ= -THφ = -ETφ); in other words, every state of every system whatsoever would have a negative-energy twin. (This is actually mathematically impossible for free particles, and in general any reasonable particles under a potential bounded from below.)

[Surlethe]

∂'0 = ∂0 and ∂'μ = -∂μ, yes. What we want is to have a completely identical Dirac equation with primed coordinates over some transformed state, but what we get upon substitution is the wrong sign of the spatial components of the momentum term. So we need to fix this somehow, and multiplying by γ0 from the left does the trick.

Okay, so use the properties of the Dirac γs to build an ansatz. Does this hold generally? That is, since γiγj = -γjγi, any time you want to reverse any three of the four coordinates (e.g, (x0, x1, s2, x3) ↦ (-x0, -x1, s2, -x3)) you can get a transformation based on the appropriate γ (γ2 in the example).

This is exactly the Dirac equation with the transformed state (γ0ψ), from which we conclude that ψ'(x') = γ0ψ(x)

Uniqueness doesn't follow -- any matrix S that anticommutes with the last three γ and commutes with the first γ will do the trick.

That doesn't follow. Don't assume that T is linear. Well, you did prove that if T were linear, then either it's not invertible or H = 0 (not T = 0), so in a sense you proved that it can't be linear, but the exercise's conclusion is stronger.

I'm not sure where I'm using the assumption that T is linear. Is it in inverting it or in pulling it out from behind the ∂t?

... in other words, every state of every system whatsoever would have a negative-energy twin. (This is actually mathematically impossible for free particles, and in general any reasonable particles under a potential bounded from below.)

I don't think I'm following. Wouldn't this be exactly the interpretation of antiparticles?

[Kuroneko]

Okay, so use the properties of the Dirac γs to build an ansatz.

The ansatz has a complex-conjugate operator, which we guessed from the structure of the Schrödinger equation: just flipping the time-coordinate gives us the wrong sign, but doing that and conjugating flips it back because of the i.

Does this hold generally? That is, since γiγj = -γjγi, any time you want to reverse any three of the four coordinates (e.g, (x0, x1, s2, x3) ↦ (-x0, -x1, s2, -x3)) you can get a transformation based on the appropriate γ (γ2 in the example).

Yes, but note that the charge conjugation operator is C = γ2γ0 (I left the last γ by accident in the previous post).

Uniqueness doesn't follow -- any matrix S that anticommutes with the last three γ and commutes with the first γ will do the trick.

Simply commuting with γ0 gets rid of eight degrees of freedom, and similarly anticommuting with the rest halves the number each time. The space of such S's thus only one (complex) degree of freedom, and preserving the norm means we're left with just the phase.

Well, alright, while individually it's obvious that (anti)commutation with a gamma matrix gets us eight independent constraints on the components of S, it's not immediately clear that there are no overlap if we constrain them one after another (there isn't, but it's too bothersome to check). So instead we can make a basis for all the matrices out of the gamma matrices: γμ, γμγν, γμγνγξ, γ5, with all different indices (or just one product of four matrices with indices allowed to be the same), and it is obvious that (say) every γμγν fails to achieve what we want.

I'm not sure where I'm using the assumption that T is linear. Is it in inverting it or in pulling it out from behind the ∂t?

Neither; it's in pulling it across the i. You flipped the signs of both sides, which is fine, but then you also multiplied by T-1 from the left, and canceled it with the T on the left-hand side that was already there.

I don't think I'm following. Wouldn't this be exactly the interpretation of antiparticles?

Oh, no. That this is the conclusion of Dirac equation in describing the electron is an indicator that the Dirac equation is fundamentally broken for such a person. Instead, what it describes is a field rather than a single particle, and the antiparticles do not have negative energy.

If you interpret my above handwaving of 'reasonable particle' as a particle that's localized enough in the sense of having a wavefunction that goes to zero at infinity sufficiently fast, then it's easy to verify that for any potential V≥Vmin, there are no energy eigenstates with E < Vmin. For example, the energy specturm of a free particle (V = 0 everywhere) is positive-semidefinite. More physically, this is just the requirement that the kinetic energy of the particle is never negative.

Try solving Hψ = Eψ with the usual H = p²/2m + V(x) with E<Vmin. (The boundary term being zero is becomes important via integration by parts.)

[Kuroneko]

Hmm... the successive-constraint counting isn't actually all that terrible. In the Dirac basis, if Sγ0 = γ0S, then S is block-diagonal. Anticommutation with γi then gives:

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    S       γ                                -γ       S
[ A  0 ][  0  s ] = [ 0  As ] = [ 0  sB ] = [ 0 -s ][ A  0 ]
[ 0 -B ][ -s  0 ] = [ Bs  0 ] = [ sA  0 ] = [ s  0 ][ 0 -B ]

where s = σi is the corresponding Pauli spin matrix, and in particular σ3 forces A = B, so that [A,H] = 0 for every traceless Hermitian H. This condition for σ1 makes the diagonal elements equal (for both diagonals), leaving two degrees of freedom, and so on. The end result is that A = aI, where I is the identity matrix, i.e., we just get multiples of γ0.

But every sign-flip {++++,+++-,...,----}, of which there are sixteen, can be built by composing successive 3-flips (which γμ perform, as we saw before). So we simply don't have any room for more than one degree of freedom for each flip operation, because the space of matrices only 16-dimensional. Thus, we can restrict ourselves to consideration of {I,γμ,γμγν,γμγνγξ,γ5}, which is a basis for the matrices.

[Kuroneko]

Exercise: Verify that γ2 corresponds to charge conjugation.

Alright, as I noted above, I left off a factor here. What I should have done is to have checked it beforehand. It's actually somewhat inconsistent in the literature--and one of the peculiarities is that for obscure reasons (probably because of the convenience of ψ̅ = ψ†γ0), C's effect on ψ is sometimes defined with a factor of γ0, which perversely enough requires C itself to have another γ0 just to cancel it the extra padding factor. So let's instead work through the Dirac equation directly without any such shenanigans.

Before that, though, perhaps some reminders would be in order. The Lorentz force law is F/q = E + (1/c)v×B. For the nonrelativistic case, the Lagrangian responsible for this is L = (1/2)mv² - qΦ + (q/c)v·A, where Φ,A are the electric and magnetic potentials. What's interesting about this is that the conjugate momentum is not mv, but P = (∂L/∂ẋi) = mv + (q/c)A. The relativistic case is very similar, with Pμ = muμ + (e/c)Aμ, where A = (Φ/c,A).

So here is the Dirac equation of a charge e coupled to an electromagnetic field:
[1] 0 = [iγμ(∂μ-ieAμ) - m]ψ.
We want to flip the sign of e but leave the ∂μ alone, and the only thing that formally different between them is an extra factor of i. Hence, charge conjugation should involve complex conjugation: C = UK. There's also a very good reason why U must be unitary--conjugating the above equation changes γμ ↦ -γμ*, but {-γμ*} have the exact same algebra, so they should just be {γμ} expressed in some other basis.

So: we're looking for a solution to a change of basis problem, U-1γμU = -γμ*. After solving it, we have the all the relevant operators:
[2] C = γ2K, P = γ0, T = γ1γ3K,
again up to some overall phase factor.

[Surlethe]

Neither; it's in pulling it across the i. You flipped the signs of both sides, which is fine, but then you also multiplied by T-1 from the left, and canceled it with the T on the left-hand side that was already there.

Ah-hah.
i∂τTψ(t,x) = HTψ(t,x)
-i∂tTψ(t,x) = HTψ(t,x)
∂tψ(t,x) = (T-1iHT)ψ(t,x),
i∂tψ(t,x) = (iT-1iHT)ψ(t,x)
so we must have iT-1iHT = H. Then
T-1iHT = (-i)H
iHT = T(-i)H
i = T(-i)H(HT)-1
= T(-i)H(TH)-1 (employing [H,T] = 0)
= T(-i)T-1.

What sort of operator on a scalar function would obey this? Edit: Stupid question. Conjugate reversal, like you said.

[Kuroneko]

Ah-hah.

Oh-hoh! That works, but methinks you're making it harder than it has to be. In particular, we don't need to know whether H is invertible for this problem.

What sort of operator on a scalar function would obey this?

Well, you answered this question already, but antilinearity, A(αx+βy) = α*Ax+β*Bx, actually occurs in another interesting way in quantum mechanics--a way that's usually hidden by furious handwaving.

Side note: To clear up a potential confusion, the prior discussion does mean that the charge conjugacy operator is also antilinear--in the Dirac theory of the electron. However, it is not antilinear in QED, the corresponding field theory (but time reversal still is). Also, you may see the action of the operator defined through their square roots in that context, as in Pψ(t,x)P = ηγ0ψ(t,-x) and Tψ(t,x)T = ηγ1γ3ψ(-t,x), etc., with some phase η, but overall the situation is analogous.

In quantum mechanics, our states live in Hilbert space H (or close to it, anyway), ... .

You've probably learned the probabilistic interpretation of quantum mechanics, which involves interpreting a particle state at a particular time as a wavefunction ψ in in ℋ = L², the space of square-integrable functions, and that |ψ|² as a probability density of finding the particle at a given point. This is all a lie, albeit usually not a very important one.

First, we trivially can't take all of L², since we some Hamiltonian like H = -(ℏ²/2m)∇² + V doesn't make sense for most square-integrable functions. That's fine; we can just take some dense subspace Φ where the position, momentum, and Hamiltonian operators Q,P,H are give well-defined (square-integrables) results at all powers. Second, the energy eigenfunctions of the free particle, the plane wave, is not in L² (and neither are a position eigenfunctions, etc.), but we can deal with this through some Fourier trickery and the Dirac delta. In other words, now we have to consider Φ*, the dual space of linear functionals over Φ (1).

This makes a lot of sense: if we have a bra <ψ| acting on some linear combinations of states |0> and |1> (that we can get via correspondence from Φ), then
[1] <ψ|a0+b1> = a<ψ|0> + b<ψ|1>,
so bras live in the dual space Φ*. What of kets? Since
[2] <a0+b1|ψ> = a*<0|ψ> + b*<1|ψ>,
kets live in the antidual space Φ× of antilinear functionals over Φ.

Antilinearity was there all along. It just looked unobtrusive because we pretended that the correspondence between bras and kets was between L² and its dual, rather than between the dual and the antidual. As a matter of terminology, (Φ,ℋ,Φ*,×) are called Gel'fand triples. Gel'fand also explained why there is a Hilbert space in the first place, but that's another (more interesting) story.

(1) More generally, we need Φ* whenever we're dealing with continuous spectra. Also note that dual space Φ* is the space of all linear functionals f:Φ→ℂ, f(αx+βy) = αf(x)+βf(y), rather than just the continuous ones (which wouldn't improve anything, since the continuous dual space Φ' is isomorphic to Φ by the Riesz representation theorem). For example, the Dirac delta is a discontinuous linear functional (why?). The antidual space is defined in the same manner, just with complex conjugates.