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PHY489 Lecture 14 the Dirac Equation

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PHY489 Lecture 14 the Dirac Equation PHY489 Lecture 14 The Dirac Equation First a reminder (hopefully) from non-relativistic quantum mechanics: 2 p E = + V 2m To express this in the form of a wave equation we make the operator substitutions ∂ p → −i∇ E → i ∂t and allow these to act on a wavefunction ψ 2 ∂ − ∇2ψ + Vψ = i ψ (Schrödinger equation) 2m ∂t One can do the same thing for the relativistic case, starting with the usual relationship between the relativistic energy and momentum: 2 2 2 2 4 µ 2 2 E = p c + m c or p µ p − m c = 0 (in relativistic notation) 1 ∂ p i Starting with this, we make the operator substitution µ → ∂ µ with ∂µ ≡ µ ∂x and allow the resulting expression to act on a wavefunction ψ : 2 1 ∂2ψ ⎛ mc⎞ −2∂µ∂ ψ − m2c2ψ = 0 ⇒ − + ∇2ψ = ψ (Klein-Gordon equation) µ c2 ∂t 2 ⎝⎜ ⎠⎟ xµ = ct, x, y, z x = g xν = ct, −x, − y, −z ( ) µ µν ( ) ∂ ⎛ 1 ∂ ∂ ∂ ∂ ⎞ ⎛ 1 ∂ ⎞ , , , = , µ = ⎜ ⎟ ⎜ ∇ ⎟ ∂x ⎝ c ∂t ∂x ∂y ∂z ⎠ ⎝ c ∂t ⎠ ∂ ⎛ 1 ∂ ∂ ∂ ∂ ⎞ ⎛ 1 ∂ ⎞ = ⎜ , − , − , − ⎟ = , −∇ x c t x y z ⎜ ⎟ ∂ µ ⎝ ∂ ∂ ∂ ∂ ⎠ ⎝ c ∂t ⎠ 2 µ µ 1 ∂ 2 2 ∂µ∂ = ∂ ∂µ = 2 2 − ∇ ≡ c ∂t 2 This equation describes free particles without spin (e.g. spin 0). We didn’t need it for the ABC model because we had the Feynman rules. For historical reasons (see pg 227… or other texts) Dirac was looking for a relativistic wave equation that was linear in the time derivative (like the Schrödinger equation). His strategy was to try to factorize the expression for the energy/momentum relation that we began with: p pµ − m2c2 = 0 µ To see how this might work, consider first the simple case of a particle at rest: pµ = ( p0 , 0 ) 2 p pµ − m2c2 = 0 ⇒ p0 − m2c2 = 0 ⇒ p0 + m p0 − m = 0 µ ( ) ( )( ) This equation is easily satisfied by p0 mc = ± 3 For p ≠ 0 the factorization is less straightforward. Can try something like pµ p − m2c2 = βκ p + mc γ λ p − mc ( µ ) ( κ )( λ ) βκ γ λ p p − m2c2 − mc βκ p − γ λ p κ λ ( κ λ ) Since there are no terms linear in p (or m) we require that γk =βk, so we get µ κ λ p pµ = γ γ pκ pλ or, more explicitly: 2 2 2 2 2 2 2 2 2 2 2 2 p p p p 0 p 1 p 2 p 3 p ( 0 ) − ( 1 ) − ( 2 ) − ( 3 ) = (γ ) ( 0 ) + (γ ) ( 1 ) + (γ ) ( 2 ) + (γ ) ( 3 ) + 0 1 1 0 p p (γ γ + γ γ )( 0 1 ) + 0 2 2 0 p p (γ γ + γ γ )( 0 2 ) + 0 3 3 0 p p (γ γ + γ γ )( 0 3 ) +............ 4 The Gamma Matrices Could chose γ0=1, γ1= γ2 =γ3=i but that leaves cross-terms that do NOT vanish if γi is simply a number. What we need is for the γi factors to anti- commute, e.g. 2 2 2 2 γ 0 = 1 γ 1 = γ 2 = γ 3 = −1 ⎫ ( ) ( ) ( ) ( ) ⎪ γ µ ,γ ν = 2g µν ⎬ { } µ ν ν µ ⎪ γ γ + γ γ = 0 for µ ≠ ν ⎭ Can’t do this with numbers since they commute (AB=BA always) but we can do it with matrices (which do not, in general, commute). The smallest matrices that will work are 4x4. There are various conventions: we will follow that of the text (which follows the text by Bjorken & Drell). ⎫ Here, by convention, each element is ⎛ 0 ⎞ ⎪ a 2x2 matrix, and the σ ’s are the Pauli 0 ⎛ 1 0 ⎞ i σ i i γ = ⎜ ⎟ γ = ⎜ ⎟ ⎬ spin matrices. 0 −1 ⎜ −σ 0 ⎟ ⎝ ⎠ ⎝ i ⎠ ⎪ ⎭ NB: these four objects are NOT the elements of a four-vector 5 Now we have p µ p − m 2 c 2 = γ κ p + m c γ λ p − m c = 0 as desired. µ ( κ )( λ ) Can choose either factor to define the Dirac equation. By convention γ µ p − m c = 0 . µ Making the usual operator substitution p → i ∂ and allowing the µ µ resulting expression to operate on the (now necessarily four component) wavefunction ψ yields the Dirac equation: ⎛ ψ ⎞ ⎜ 0 ⎟ µ ⎜ ψ 1 ⎟ iγ ∂ ψ − mcψ = 0 ψ = ⎜ ⎟ is called a Dirac spinor (more on this later) µ ⎜ ψ 2 ⎟ ⎜ ψ ⎟ ⎝ 3 ⎠ It is important to note that while the Dirac spinor ψ is a four component object, is it NOT a four-vector. It’s transformation properties are important however, and we will discuss these later on. 6 ∂ So we have what we wanted: a relativistic wave equation linear in : ∂t iγ µ∂ − mc ψ = 0 ( µ ) Where ψ is a four component object called a Dirac spinor. 7 Transformation Properties of Dirac Spinor ψ We will state the transformation properties without proof. Under a Lorentz transformation (along x) we have: Note that this is a 4x4 matrix. ⎛ a a σ ⎞ The σ are the 2x2 Pauli spin ψ → ψ ′ = Sψ S = ⎜ + − 1 ⎟ i ⎜ a a ⎟ matrices (here we use σ1 for the ⎝ −σ1 + ⎠ case of a boost along x). −1/2 with a = ± γ ± 1 / 2 and γ = 1 − β 2 ± ( ) ( ) Note that the wavefunction ψ is not an observable. What we want to construct (from this) are quantities that do have well defined transformation properties. For example, try to form a Lorentz scalar: ⎛ ⎞ ψ 1 ⎜ ⎟ This is more or less the same ⎜ ψ ⎟ 2 2 2 2 ψ ✝ψ = ψ ∗ ψ ∗ ψ ∗ ψ ∗ 2 = ψ + ψ + ψ + ψ thing we tried when we tried ( 1 2 3 4 )⎜ ⎟ 1 2 3 4 ⎜ ψ 3 ⎟ to define an invariant based ⎜ ⎟ µ ⎝ ψ 4 ⎠ on the four-vector x . Actually, didn’t do this this year, but you can try it and see that it doesn’t work 8 Transformation Properties of ψ✝ψ How does ψ✝ψ transform under a Lorentz transformation (again along x)? ✝ ✝ ′ ✝ ✝ ✝ ✝ ✝ ψ ψ → ψ ψ = ψ ′ ψ ′ = ψ S Sψ ≠ ψ ψ since S S ≠ 1 ( ) ( ) ( ) ⎛ a 0 0 a ⎞ ⎛ a 0 0 a ⎞ ⎛ a2 + a2 0 0 2a a ⎞ + − + − ⎜ + − + − ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 ✝ ⎜ 0 a+ a− 0 ⎟ ⎜ 0 a+ a− 0 ⎟ ⎜ 0 a+ + a− 2a+ a− 0 ⎟ S S = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ 0 a a 0 0 a a 0 0 2a a a2 a2 0 ⎜ − + ⎟ ⎜ − + ⎟ ⎜ + − + + − ⎟ ⎜ ⎟ ⎜ a 0 0 a ⎟ ⎜ a 0 0 a ⎟ 2a a 0 0 a2 a2 ⎝ − + ⎠ ⎝ − + ⎠ ⎜ + − + + − ⎟ ⎝ ⎠ ⎛ a2 + a2 2a a σ ⎞ ⎛ 1 −βσ ⎞ = ⎜ + − − + 1 ⎟ = γ ⎜ 1 ⎟ [e.g. in 2x2 form] ⎜ 2a a a2 a2 ⎟ ⎜ −βσ 1 ⎟ − +σ1 + + − ⎝ 1 ⎠ ⎝ ⎠ 1 1 ⎧ a2 a2 1 1 + + − = (γ + ) + (γ − ) = γ ⎪ 2 2 ⎪ since ⎨ ⎪ 1 1− 1+ β 2 2a a = − γ − 1 γ + 1 = − γ 2 − 1 = − − 1 = − = − β 2γ 2 = −βγ ⎩⎪ + − 1− β 2 1− β 2 9 Transformation Properties ψ✝ψ So ψ✝ψ does not form an invariant (under Lorentz transformations). [Just as the sum of the squares of the components of a four vector did not, when we were looking at four-vector invariants. In that case, needed a relative negative sign on the spatial µ components, for which we introduced the covariant four vector to define the invariant a aµ.] Need to do something similar here. Note that the elements of S✝S are either a2 + a2 or 2a a (from a a + a a ) + − + − + − − + If the sign were reversed in each case, we would get 1 1 a2 a2 1 1 1 a a a a 0 + − − = (γ + ) − (γ − ) = + − − − + = 2 2 So we can get what we need if we introduce a matrix that effects the required change of sign: (γ0 will do the trick). 10 The Adjoint Spinor ψ ≡ ψ ✝γ 0 Define the adjoint spinor as ⎛ 1 0 0 0 ⎞ ⎜ 0 1 0 0 ⎟ ∗ ∗ ∗ ∗ ⎜ ⎟ ∗ ∗ ∗ ∗ e.g. ψ 1 ψ 2 ψ 3 ψ 4 = ψ 1 ψ 2 −ψ 3 −ψ 4 ( )⎜ 0 0 −1 0 ⎟ ( ) ⎝⎜ 0 0 0 −1 ⎠⎟ ✝ 0 The quantity ψ ψ = ψ γ ψ is a Lorentz invariant scalar quantity, since ⎡ ⎛ 1 0 0 0 ⎞ ⎤ ⎢ ⎜ ⎟ ⎥ ′ ✝ ′ 0 ✝ ✝ 0 ✝ 0 0 1 0 0 0 ψψ → ψψ = ψ γ ψ ′ = ψ S γ Sψ = ψψ ⎢S γ S =⎜ ⎟ = γ ⎥ ( ) ( ) ( ) ⎢ ⎜ 0 0 −1 0 ⎟ ⎥ ⎢ ⎥ ⎝⎜ 0 0 0 −1 ⎠⎟ ⎣ ⎦ So we know that this quantity is a Lorentz scalar. We can also ask whether is it a regular scalar or a pseudoscalar. That is, how does it transform under parity? 11 To determine how the quantity ψ ψ behaves under parity, first need to know how ψ transforms: 0 Quote without proof: under parity we have ψ → ψ ′ = γ ψ ✝ ✝ ′ 0 ✝ 0 0 0 ✝ 0 ψψ → ψψ ′ = ψ γ ψ ′ = ψ γ γ γ ψ = ψ γ ψ = ψψ ( ) ( ) ( ) ( ) 2 since γ 0✝ = γ 0 γ 0 = 1 [e.g. the 4x4 identity matrix] ( ) Can also make a pseudoscalar combination: ψ γ 5 ψ with γ 5 ≡ iγ 0γ 1γ 2γ 3 5 ⎛ 0 1 ⎞ γ = ⎜ ⎟ in the usual 2x2 form ⎝ 1 0 ⎠ 12 5 Parity Transformation of ψ γ ψ Under parity, we have 5 5 ′ ✝ ′ 0 5 ✝ 0 0 5 0 ✝ 5 0 ψγ ψ → ψγ ψ = ψ γ γ ψ ′ = ψ γ γ γ γ ψ = ψ γ γ ψ ( ) ( ) ( ) ✝ 0 5 5 0 5 5 0 = − ψ γ γ ψ = −ψγ ψ since γ γ = −γ γ γ0 anti-commutes with γ1, γ2, γ3 and commutes with itself, so 3 γ 5γ 0 = iγ 0γ 1γ 2γ 3γ 0 = −1 γ 0iγ 0γ 1γ 2γ 3 = −γ 0γ 5 ( ) γ5 anti-commutes with all other γ matrices 2 ✝ γ µ ,γ 5 = 0 γ 5 = 1 γ 5 = γ 5 { } ( ) ( ) ψγ 5ψ Can also show that is Lorentz invariant (Problem 7.12) 13 5 Lorentz Invariance of ψ γ ψ Under Lorentz transformation, we have 5 5 ′ ✝ ′ 0 5 ✝ ✝ 0 5 ψγ ψ → ψγ ψ = ψ γ γ ψ ′ = ψ S γ γ Sψ ( ) ( ) ( ) Can show that S, γ 5 commute: ⎛ a a ⎞ ⎛ a a ⎞ 5 + −σ1 ⎛ 0 1 ⎞ −σ1 + Sγ = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ a σ a ⎟ ⎝ 1 0 ⎠ ⎜ a a σ ⎟ ⎝ − 1 + ⎠ ⎝ + − 1 ⎠ ⎛ a a ⎞ ⎛ a a ⎞ 5 ⎛ 0 1 ⎞ + −σ1 −σ1 + γ S = ⎜ ⎟ = ⎜ ⎟ ⎜ 1 0 ⎟ ⎜ a a ⎟ ⎜ a a ⎟ ⎝ ⎠ ⎝ −σ1 + ⎠ ⎝ + −σ1 ⎠ ✝ ✝ 0 5 ✝ ✝ 0 5 ✝ 0 5 5 So ψ S γ γ Sψ =ψ S γ Sγ ψ = ψ γ γ ψ = ψγ ψ Reminder: the form of S used is for a Lorentz transformation along x.
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