PX408 2016-2017 Lecture Notes 3 - Solving the Dirac equation PDF

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October 2016

PX408: Relativistic Quantum Mechanics Tim Gershon

([email protected])

Handout 3: Solving the Dirac Equation We would like to find, and investigate, the values of ψ which solve the Dirac equation: (iγ µ ∂µ − m)ψ = 0 .

(1)

We start by looking for (positive energy) plane wave solutions of the form ψ ∼ ωe−ip.x. It is convenient to use a form of the Dirac equation where space- and time-like parts have been separated (i ∂ψ = (−iα.∇ + βm)ψ – refer to earlier handouts for definitions of α, β and σ). We then find the ∂t momentum-space Dirac equation Eω = (α.p + βm)ω , (2)   ω+ into which we can insert ω = to give, in block form, ω− E



ω+ ω−



=



m σ.p σ.p −m



ω+ ω−



,

(3)

which we can rewrite as coupled equations (E − m)ω+ = σ.pω−

(E + m)ω− = σ.pω+ . Thus we can choose to substitute either ω+ or ω− in our working. Choosing the latter, we find   ω+ ω= . σ.p ω E+m +

(4) (5)

(6)

Note also that both choices give (E + m)(E − m)ω± = (σ.p)2 = p2 , i.e. we recover (again) the relativistic mass-energy relation E 2 = m2 +p2 . This implies the existence of negative energy solutions, which we will discuss in more detail below. From the above we can see that there are in fact two linearly independent solutions (corresponding to the fact that ω+ is a two-component spinor). For a particle at rest we can choose orthogonal (but unnormalised) solutions as     1 0 ω+ = (7) or 1 0 giving 

 1  0  −imt  ψ=  0 e 0

or



 0  1  −imt  e .  0  0

(8)

The Helicity Operator There must be some operator that distinguishes the two linearly independent solutions to the Dirac equation given above. Since they have the same energy, this operator must commute with the Hamiltonian. (Recall from Heisenberg’s equation of motion h i ˙ ˆ , ˆ H Aˆ = −i A, (9)

that any operator that commutes with the Hamiltonian represents a constant of motion.) The Hamiltonian is given by H = (α.p + βm), and choosing the particle’s rest frame for simplicity, this reduces to βm. We are therefore looking for an operator that commutes with β, while distinguishing     0 1  1   0      from (10)  0 .  0  0 0 It is fairly obvious that a possible choice is

Σz =



σz =



where we recall that

σz 0 0 σz 1 0 0 −1





.

(11)

(12)

It is worth noting that, based on the arguments above, we could have chosen the bottom right element of the block form matrix in Eq. (11) to take any value. Hopefully the choice we made appears sensible, but we will return to this point later. Q1 Show that Σz commutes with β . Q2 Show that Σz distinguishes the two solutions for ψ given in Eq. (8). Generalising, we might expect our operator in an arbitrary frame to be   σ 0 Σ = . 0 σ

(13)

This operator does not commute with (α.p + βm), and therefore is not the correct answer, but let us briefly examine its properties anyway. Using knowledge of the Pauli spin matrices, we see immediately    2 1 1 3 1 1 Σj , Σk = iǫjkl Σl (j 6= k) . Σ = , (14) 2 2 2 4 2 Q3 Show explicitly that

 1 2 Σ= 2

3 4

.

There are in fact many possible solutions for the operator that we are searching for, of which several can be found in the literature, but the most convenient is the so-called “helicity operator” given by ! σ.p 0 |p| . (15) h(p) = σ.p 0 |p|

Q4 Show that Σdefined in Eq. (13) does not commute with (α.p + βm). Q5 Show that h(p) does commute with the Hamiltonian. Q6 Find the value of (h(p))2 , and state the eigenvalues of h(p). Q7 Consider the solutions to the Weyl equations discussed in Handout 2. Show that these are eigenstates of the helicity operators and give their eigenvalues. gives the projection of σ onto the direction of momentum p. Thus, h(p) can be Note that σ.p |p| understood, physically, as the projection of Σ onto the direction of motion. Helicity is not a Lorentz invariant quantity, except in the special case of massless particles, but is invariant under spatial rotations. (The spin matrix Σis Lorentz invariant, but is not a constant of the motion.) The Helicity Operator – Second Derivation It is fairly evident already that the operator we discovered above, called the helicity operator, is strongly related to the quantum property of spin that appears inevitably in the solution of the Dirac equation. To further illustrate this, an alternative derivation in which the connection is more apparent is useful. This approach, following Feynman (QED, p60) also makes use of some of the gamma matrix algebra we developed earlier. Let us start by restating the problem: we have two solutions to p/ω = mω ,

(16)

and therefore we are looking for an operator that commutes with p/. Firstly, recall that γ 5 anticommutes with the slashed version of any four vector. Therefore  5  γ , p/ = 0 . (17)

Let us now define an operator O such that Oµ pµ = 0. Then

{O/, p/} = 2Oµ pµ = 0 .

(18)

Now, define the eigenvalues of −γ 5 O/ as s, i.e. − γ 5 O/ω = sω

(19)

so that 

  −γ 5 O/ −γ 5 O/ ω = −γ 5 γ 5 O/O/ω µ

(20) 2

= −Oµ O ω = s ω

(21)

   2 (since γ 5 , O/ = 0, γ 5 = 1 and O/O/ = Oµ Oµ ). So with the choice Oµ Oµ = −1, we find eigenvalues s = ±1. (n.b. Although the choice −γ 5 O/ for the operator looks a bit unnatural, it is necessary to get the sign right in the end.) Note that the Lorentz invariant requirement Oµ pµ = 0, obliges us to put O0 = 0 in the particle’s rest frame. Therefore, in the rest frame, Oµ Oµ = −1 reduces to |O| = 1. Now, choosing the orientation to be along the z-axis, so that O/ = γ 3 , we find −γ 5 O/ω = −γ 5 γ 3 ω = −iγ 0 γ 1 γ 2 γ 3 γ 3 ω = iγ 0 γ 1 γ 2 ω 1 2 0

= iγ γ γ ω = iγ 1 γ 2 ω

(22) (23) (24) (25) (26)

where we have used the fact that γ 3 γ 3 = −1, used two anticommutation relations ((−1)2 = 1) to bring γ 0 through to the right, and, in the last step, used γ 0 ω = ω in the particle’s rest frame (recall that the two positive energy solutions in the rest frame are given by ω = (1 0 0 0)T , (0 1 0 0)T ). Now,        0 σx σz 0 −σx σy 0 0 σy 1 2 γ γ = . (27) = = −i 0 σz 0 −σx σy −σy 0 −σx 0 and thus 5

− γ 5 O/ω = i(−i)Σz ω = sω ,

(28)

i.e. −γ O/ is the Σz operator. The above derivation is still not very general, since we have chosen the particle to be in its rest frame and moreover specified an axis with which to align ourselves. Nonetheless, we can at this stage µ state that the Dirac spinor ψ ∼ ωe−ipµ x , where ω satisfies p/ω = mω and −γ 5 O/ω = sω, represents a particle moving with momentum pµ and with spin aligned either parallel with (s = +1) or antiparallel to (s = −1) the O axis in the co-moving frame. The generalisation to an arbitrary frame proceeds as before, to obtain the helicity operator of Eq. (15). Since (σ.p)2 = p2 , it is clear that (h(p))2 = 1, and so the eigenvalues of h(p) must be ±1 (in fact +1 twice and −1 twice). To find the eigenstates, consider !    σ.p 0 ω+ ω+ |p| h(p)ω = = ±ω = ± . (29) σ.p σ.p σ.p 0 ω ω |p| E+m + E+m + ω = ω+ , while for the negative solutions we find For the positive solution we find σ.p |p| + which as before can be solved for p aligned with the z-axis as     1 0 ω+ = . or 1 0

σ.p |p|

ω+ = −ω+ , (30)

Q8 Find the eigenvectors of h(p) in the cases that (i) p is aligned along the x-axis; (ii) p is aligned along the y-axis; (iii) p has a general direction. Helicity and Chirality As discussed above, the helicity operator h(p) distinguishes the solutions of the Dirac equation (i.e. h(p) commutes with the Hamiltonian). The term “chirality” is often confused with helicity, and therefore it is worthwhile to briefly discuss its meaning at this stage. Hopefully this will reduce the possibility for potential confusion. The chirality operator is actually just γ 5 . Recall that in block form, and in the standard representation, we write   0 1 γ 5 = iγ 0 γ 1 γ 2 γ 3 = . (31) 1 0

It is common to encounter the following operators that project out right- and left-handed chiral states, respectively:   1 1 1 − γ5 (32) 1 + γ5 PL = PR = 2 2 and we use these to define right- and left-handed spinors ψR = PR ψ

ψL = PL ψ

(33)

The importance of the chirality operators will be apparent to those who have studied particle physics since the weak interaction operates only on the left-handed spinors. (Some of the profound consequences of this fact will be discussed in the Gauge Theories module.) Note that the chirality operator is Lorentz invariant, as are the PR and PL projection operators.

Q9 Show that ψR and ψL are eigenstates of the chirality operator γ 5 , and give their eigenvalues. Q10 Show that ψ = ψR + ψL . Q11 Find expressions for PR ψR , PR ψL , PL ψR and PL ψL . Q12 Using the standard convention, give matrix forms for PR and PL . Q13 Using matrix representation, show that (PR )2 = PR , (PL )2 = PL and PR PL = PL PR = 0. Now, the expressions for the helicity operator of Eq. (15) and the chirality operator of Eq. (31) appear completely different, so it is not obvious how any confusion could possibly arise. However, let us recall that a spinor that is a solution of the Dirac equation can be written   ω+ (E − m) ω+ = σ.p ω− −ipµ xµ ψ ∝ ωe where ω= and (34) (E + m) ω− = σ.p ω+ ω− Thus we can write   0 1 γ 5ω = 1 0

σ.p ω E−m − σ.p E+m ω+



=



σ.p ω E+m + σ.p ω E−m −



=



σ.p E+m

0

0 σ.p E−m



ω+ ω−



.

(35)

Consequently, when acting on solutions of the Dirac equation, the chirality operator can in fact be represented as   σ.p   0 0 1 5 E+m ω = ω. (36) γ ω= σ.p 1 0 0 E−m It should now be apparent how helicity and chirality operators can be confused. In particular, in the highly relativistic limit (E ≫ m such that E − m ∼ E + m ∼ |p|), they tend to become equivalent. In particular, for massless particles helicity and chirality are equivalent. Correspondingly, it is for the nearly massless neutrinos where the largest confusion between the two operators tends to arise. Nonetheless, in a general frame helicity and chirality are distinct quantities represented by different operators. A nice summary of their similarities and differences can be found at http://www.quantumfieldtheory.info/Chiralityvshelicitychart.htm. Q14 Find the commutator of the chirality operator with the Hamiltonian. The Dirac Current As before, we want to find something that satisfies the continuity equation ∂ρDirac = −∇.j Dirac ∂t

or

∂µ j µDirac = 0 .

(37)

It is tempting to try ρ = ψ † ψ, where ψ † is the Hermitian conjugate (or adjoint) of ψ. That is   ψa  ψb   then ψ † = (ψ ∗a ψ b∗ ψc∗ ψ ∗d ) . (38) if ψ =  ψc  ψd

Then ρ = |ψa|2 + |ψb |2 + |ψc |2 + |ψd |2 is automatically positive definite, and looks like a good candidate for the probability density. (n.b. We will discuss the normalisation of the Dirac spinors later, which

will cover possible concerns that the density should transform as the time-like component of a fourvector.) To find j Dirac, we use a similar approach to that taken for the Klein-Gordon case, and start by taking the conjugate of the Dirac equation. It is easiest to see how this works by explicitly writing out the Lorentz invariant term which gives (iγ µ ∂µ − m)ψ = 0 iγ 0 ∂0 ψ + iγ 1 ∂1 ψ + iγ 2 ∂2 ψ + iγ 3 ∂3 ψ − mψ = 0

−i∂0 ψ † γ 0 + i∂1 ψ † γ 1 + i∂2 ψ † γ 2 + i∂3 ψ † γ 3 − mψ † = 0 ¯ 0 + i∂1 ψγ ¯ 1 + i∂2 ψγ ¯ 2 + i∂3 ψγ ¯ 3 + m ψ¯ = 0 i∂0ψγ ¯ µ + mψ¯ = 0 i∂µ ψγ

(39) (40) (41) (42) (43)

where between the second and third lines we have taken the Hermitian conjugate, and exploited the fact that (γ 0 )† = γ 0 while (γ i )† = −γ i (i = 1, 2, 3), and between the third and fourth lines we have right-multiplied by −γ 0 , exploited the anticommutation relations of the gamma matrices, and introduced the conjugate spinor ψ¯ = ψ † γ 0 . Q15 Work carefully through the steps above to derive Eq. (43). ¯ right-multiply the conjugate Dirac Now, we left-multiply the Dirac equation of Eq. (39 by ψ, equation of Eq. (43) by ψ, and add ¯ µ ∂µ − m)ψ ψ(iγ ¯ µ + mψ)ψ ¯ (i∂µ ψγ ¯ µ (∂µ ψ) + i(∂µ ψ)γ ¯ µψ i ψγ ¯ µ ψ) i∂µ (ψγ

= 0

(44)

= 0 = 0 = 0

(45) (46) (47)

µ ¯ µ ψ, with components ρDirac = j 0 = ψγ ¯ 0ψ = ψ†ψ allowing us to identify the Dirac current j Dirac = ψγ i i ¯ and j Dirac = ψγ ψ.

Normalisation of Dirac Spinors µ Since ρDirac = ψ † ψ is the time-like component of j Dirac, and must behave accordingly under Lorentz transformations, an appropriate choice of normalisation is ψ † ψ = 2E (or often 2E particles per unit volume, making the connection with density explicit). It should be noted that this is merely a convention (albeit a very useful one), and other choices appear in the literature. Let us write our solutions for the Dirac spinors as     ω+ ω+ µ −ipµ xµ −ipµ xµ ψ = N ωe =N e =N e−ipµ x , (48) σ.p ω ω− E+m +

where N is the normalisation factor to be found. Without loss of generality, we can choose ω+ to be † normalised such that ω + ω+ = 1 (any other choice will just scale the resulting value of N ). We then find 2 !  2E (E + m)2 + p2 σ.p † = N2 = 2E , (49) = N2 ψ † ψ = N 2 ω + ω+ 1 + 2 ( E + m ) E +m E +m where we have used (σ.p)2 = p2 and the √ relativistic mass energy relation E 2 = p2 + m2 . Clearly, to † normalise to ψ ψ = 2E, we require N = E + m. Finally, the (positive energy) solutions of the Dirac equation are ψ = u(p, s)e−ipµ x

µ

(50)

where u(p, s) = (E + m)1/2



φs

σ.p φs (E+m)



φ↑ =

and



1 0



φ↓ =



0 1



.

Until now we have only considered solutions of the Dirac equation of the form     ω+ µ ω+ −ipµ xµ −ipµ xµ e =N e−ipµ x . ψ = N ωe =N σ.p ω ω− E+m +

(51)

(52)

However, as can be seen from Eq. (4), we could equally well write   σ.p   µ µ µ ω+ ω E−m − e−ipµ x = N ψ = N ωe−ipµ x = N e−ipµ x . ω− ω−

(53)

The reason that we have avoided this solution is that its behaviour in the case that |p| → 0 (i.e. E → m) is not immediately apparent. Q16 By giving the non-relativistic expression for E, find the limit of

σ.p E−m

as |p| → 0.

To avoid such problems, we can “fix” the energy by writing Efix = −E. We then obtain   σ.p fix   σ.p ω− ω− − Efix E +m +m fix = , ω= ω− ω−

(54)

where pfix = −p. We can write the solutions as ψ = N ωe−ipµ x = N µ



σ.p fix ω Efix+m −



ω−

µ

e+ipfixxµ ,

(55)

√ where the normalisation factor N can be found to be Efix + m. We can interpret these as negative energy solutions. Alternatively, yet equivalently, we can drop the subscript “fix”, and consider these solutions as corresponding to antiparticles with positive energy. We will discuss the interpretation of these solutions further below (and more in the next handout). The complete expression for the negative energy solutions is then ψ = v(p, s)e+ipµ x

µ

where v(p, s) = (E + m)

1/2



σ.p χs (E+m) s χ



and

(56)

↑

χ =



0 1



χ↓ =



1 0



.

(57)

The two-component vectors φs and χs describe the spin-states of particles and antiparticles respectively. As noted after Eq. (11), we have some freedom to choose how to define the relative helicity of particles and antiparticles. (Specifically γ 0 h(p)  would be a valid choice  the   for  alternative 1 0 0 helicity operator.) It is conventional to choose φ↑ = and , with χ↑ = and φ↓ = 1 1 0   1 . With this choice the absence of a “spin-up” negative-energy electron is equivalent to χ↓ = 0 the presence of a “spin-down” positive-energy positron. The utility of this equivalence will become clear when we discuss the negative energy solutions in more detail below. Q17 What are the values of u ¯(p, s)u(p, s) and v¯(p, s)v(p, s)? Q18 What are the values of u ¯(p, s)γ 5 u(p, s) and v¯(p, s)γ 5 v (p, s)?

Lorentz Transformations of Dirac Spinors We discussed the Lorentz transformation properties of gamma matrices earlier, showing that they could be treated as invariant since the transformation is an equivalence transformation. Effectively, we can choose to implement the Lorentz transformation in one of two ways: • transform γ µ like a 4-vector, leaving ψ unchanged ֒→ γ µ′ differs by an equivalence transformation • leave γ µ unchanged ֒→ ψ transforms according to an equivalence transformation The question we would like to answer here is, exactly which equivalence transformation is required? µ That is, what is the form of S such that S −1 γ µ S = Λ ν γ ν , where Λνµ is a Lorentz transformation? To solve this, we start by considering a simpler transformation: that of a rotation in spatial dimensions. For a rotation by θ about the x axis, we can write  ′       x 1 0 0 x x 7→ x′ =  y ′  =  0 cos θ sin θ   y  . (58) x= y  z z′ 0 − sin θ cos θ z

Note also that p transforms in the same way. Let us consider how a two component spinor φ transforms under this rotation. This question is not trivial since the internal space described by the two components of the spinor are not related to our spatial coordinates in any straightforward way. However, we can still progress by noting that ′ . the helicity will be unchanged under the transformation, i.e. if h(p)φ+ = +φ+ then h(p′ )φ′ + = +φ+   1 Then, if we start with a positive helicity spinor φ+ = , we require 0 σ.p′ ′ φ = φ′+ , |p′ | +

(59)

i.e. that φ′+ is an eigenvector of the helicity operator in the transformed space, with eigenvector +1. Without loss of generality we can take p to be normalised to unity, and since rotation preserves normalisation, p′ is also normalised to unity. Then σ.p′ = σx px + σy (py cos θ + pz sin θ) + σz (−py sin θ + pz cos θ) .

(60)

Choosing p to be aligned with the z-axis, we then find σ.p′ = σy sin θ + σz cos θ . |p′ |

(61)

[Note that the choice of aligning p with the z-axis here is arbitrary, but simplifies the maths. Since we are looking at a rotation about the x-axis it would be silly to align in that direction – nothing would happen! One can choose to align along the y-axis, and the same solutions are obtained. If you are feeling brave you can try to prove this, but it is not recommended.] So, using explicit forms for σy and σz , Eq. (59) tells us        a a cos θ −i sin θ a ′ . (62) with φ+ = = b b b i sin θ − cos θ Q19 Find the solution to Eq. (62). hint: recall that cos 2θ = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1.

The solution to Eq. (62) is ′ φ+ =



a b



=



cos(θ/2) i sin(θ/2)



,

(63)

† where we have chosen normalisation such that (φ′+ )† φ′+ = φ+ φ+ = 1.

Q20 Find a similar description for the...