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HW8_soln

# HW8_soln - PHY 5667 Problem Set no 8 solution Problem 1 In...

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Unformatted text preview: PHY 5667 Problem Set no. 8 solution Problem 1 In a frame k = (E, 0, 0, p) the solutions for right-handed spinors read √ E+p √0 √0 uk,1 = , vk,1 = , , uk,2 = E−p E−p 0 vk,2 = √ − E+p 0 (i) = √ E+p 0 = E+p 0 2 s=1 uk,s u† k,s ( 0 0 0 0 + = E · 1 + p · σ3 = k · σ ¯ 2 = √0 E−p = s=1 vk,s v † k,s 0 0 0 E−p √0 E−p E + p, 0) + (0, 0 E−p 0 E+p E+p 0 = √ − E+p 0 E − p) + + (0, 0 0 E − p) 0 E−p (− E + p, 0) E+p 0 0 E−p = = E · 1 + p · σ3 = k · σ ¯ (ii) The transformation that brings k = (E, 0, 0, p) to k ′ = (E, p sin θ cos φ, p sin θ sin φ, p cos θ) is a rotation by −θ around the y -axis followed by a rotation by −φ around the ˆ z -axis. This means that the spinors get rotated to ˆ uk′ ,s = e−iφ σ3 2 e −i θ σ2 2 and similar for vk′ ,s = e−iφ uk,s σ3 2 e −i θ σ2 2 vk,s This means 2 s=1 uk′ ,s u† ′ ,s = e−iφ k σ3 2 e −i θ s=1 2 3 =e 2 σ2 2 −i φ σ 2 e −i θ σ 2 uk,s u† eiφ k,s σ3 2 e iθ (E · 1 + p · σ 3 )eiθ = E · 1 + p · e −i φ = E · 1 + p · e −i φ 2 σ3 2 σ2 2 σ2 2 e iφ σ3 2 σ3 2 e−iθσ eiφ σ3 2 (cos θ − i sin θσ 2 )eiφ σ3 = E · 1 + p · (cos θ − i sin θe−iφσ3 σ 2 )σ 3 σ3 2 σ3 = E · 1 + p · (cos θ − i sin θ(cos φ − i sin φσ 3 )σ 2 )σ 3 1 (1) . = E · 1 + p · cos θσ 3 − p sin θ cos φ iσ 2 σ 3 −p sin θ sin φ σ 3 σ 2 σ 3 −σ 1 1 −σ 2 3 2 = E · 1 + p · sin θ cos φ · σ + p sin θ sin φ · σ + p cos θ · σ = k′ · σ ¯ It works exactly the same way for vk,s , which means that the identity 2 2 uk,s u† k,s s=1 = s=1 vk,s v † = k · σ ¯ k,s holds independent of the frame of reference. Problem 2 In a frame k = (E, 0, 0, p) the solutions for left-handed spinors read √ E−p √0 √0 , uk,2 = uk,1 = , vk,1 = , 0 E+p E+p vk,2 = √ − E−p 0 . (i) = √ E−p 0 = 0m 00 2 T uk,s vk,s s=1 0 0 −m 0 + = imσ2 √0 E+p E + p) + (0, = (− E − p, 0) 0 m −m 0 This property is valid in all frames, which can be shown by performing the same rotation as in problem (1) (rotation transformation is the same for leftand right-handed spinors) 2 T uk′ ,s vk′ ,s = e−iφ σ3 2 e −i θ σ2 2 s=1 2 T uk,s vk,s eiθ σ2 2 s=1 2 3 =e −i φ σ 2 = e −i φ σ3 2 e −i θ σ 2 e −i θ σ2 2 imσ2 eiθ e iθ σ2 2 e iφ σ2 2 σ3 2 e −i φ 2 σ3 2 because σ 2T = −σ 2 and σ 3T = σ 3 σ3 2 imσ2 = imσ2 e −i φ b/c σ 2 commutes w/ itself & anticommutes w/ σ 3 (ii) = √ E−p 0 = E−p 0 2 s=1 uk,s u† k,s ( √0 E+p E − p, 0) + (0, E + p) 0 E+p =k·σ 2 s=1 † vk,s vk,s = = √0 E+p E−p 0 (0, E + p) + √ − E−p 0 (− E − p, 0) 0 E+p =k·σ The proof that this independent of the frame of reference goes through exactly as in problem 1. Problem 3 In a frame k = (E, 0, 0, p) the solutions for √ E−p √0 E+p 0 uk,1 = √ E + p , uk,2 = √0 E−p 0 (i) 2 uk,s u† s=1 k,s Dirac spinors read √ E−p , v = √0 k,1 − E+p 0 √ E−p 0 = √ E + p ( E − p, 0, E + p, 0) + 0 0 0 E−p 0 m 0 0 0 0 0 0 E+p + = m 0 0 E+p 0 0 0 m 0 0 0 0 = k·σ m1 2 m1 2 k·σ ¯ = kµ · 0 σµ ¯ σµ 0 , √0 E+p (0, √0 E−p 0 0 0 m 0 0 0 E−p + m1 0 12 vk,2 E + p, 0, 12 0 = (kµ γ µ + m1)γ 0 We must show that this result in fact holds in all frames. Therefore, we need to perform rotations of the Dirac spinors (which are 4-vectors, thus we need 4 × 4transformation matrices. Knowing that a Dirac spinor consists of a left-handed 3 √0 E+p . = √0 − E−p E − p) and a right-handed Weyl spinor and that these both transform under rotations as speciﬁed in (1), we get the transformation of Dirac spinors as uk′ ,s = e−iφ σ3 2 e −i θ 0 1 1 0 σ2 2 and similar for vk′ ,s = e−iφ uk,s σ3 2 e −i θ 1 0 σ2 2 0 1 vk,s (2) From this point on, it is straighforward to show that this holds independent of the frame of reference. (ii) √ E−p √0 2 0 E+p (0, E + p, 0, − E − p) vk,s v † = √ − E + p ( E − p, 0, − E + p, 0) + k,s √0 s=1 − E−p 0 E − p 0 −m 0 0 0 0 0 0 0 0 0 0 E + p 0 −m = −m 0 E + p 0 + 0 0 0 0 0 0 0 0 0 −m 0 E − p = k·σ − m1 2 − m1 2 k·σ ¯ = kµ · 0 σµ ¯ σµ 0 − m1 0 12 12 0 = (kµ γ µ − m1)γ 0 Again using the transformations (2), it is straighforward to show that this holds independent of the frame of reference. Problem 4 {ψα (x), ψβ (y )} = = dk dk ′ √ (2π )3 2 k 0 k ′0 dk dk ′ √ (2π )3 2 k 0 k ′0 2 ′ ′ uk,sα ak,s e−ik·x + vk,sα a† eik·x , uk′ ,s′ β ak′ ,s′ e−ik ·y + vk′ ,s′ β a† ′ ′ eik ·y ¯ ¯ k,s s,s′ =1 k ,s 2 ′ s,s′ =1 ′ uk,sα uk′ ,s′ β e−i(k·x−k ·y) {ak,s , ak′ ,s′ } + uk,sα vk′ ,s′ β e−i(k·x−k ·y) {ak,s , a† ′ ′ } ¯ k ,s ′ ′ +vk,sα uk′ ,s′ β ei(k·x−k ·y) {a† , ak′ ,s′ } + vk,sα vk′ ,s′ β ei(k·x+k ·y) {a† , a† ′ ,s′ } ¯k,s ¯k,s ¯k =0 k0 ,k′0 ﬁxed since all the anticommutators vanish 4 k0 ,k′0 ﬁxed † ψα (x), ψβ (y ) = = dk dk ′ √ (2π )3 2 k 0 k ′0 dk dk ′ √ (2π )3 2 k 0 k ′0 2 ′ ′ ∗ uk,sα ak,s e−ik·x + vk,sα a† eik·x , u∗′ ,s′ β a† ′ ′ eik ·y + vk′ ,s′ β ak′ ,s′ e−ik ·y ¯ ¯ k k,s s,s′ =1 k ,s 2 ′ s,s′ =1 ′ ∗ uk,sα u∗′ ,s′ β e−i(k·x−k ·y) {ak,s , a† ′ ′ } + uk,sα vk′ ,s′ β e−i(k·x+k ·y) {ak,s , ak′ ,s′ } ¯ k k ,s ′ ′ ∗ +vk,sα u∗′ ,s′ β ei(k·x+k ·y) {a† , a† ′ ′ } + vk,sα vk′ ,s′ β ei(k·x−k ·y) {a† , ak′ ,s′ } ¯ ¯ ¯ k k,s = = dk (2π )3 2k 0 dk (2π )3 2k 0 k0 ,k′0 ﬁxed k ,s k,s k0 ,k′0 ﬁxed 2 s=1 (uk,s u† )αβ · e−ik·(x−y) − (vk,s v † )αβ · eik·(x−y) k,s k,s k0 ﬁxed / / ((k + m1)γ 0 )αβ e−ik·(x−y) − ((k − m1)γ 0 )αβ · eik·(x−y) k0 ﬁxed where we changed k in the exponent to k since (x0 − y 0 ) = 0 anyway now in the second term replace k → −k = dk (2π )3 2k 0 ((k0 γ 0 − ki γ i + m1)γ 0 )αβ e−ik·(x−y) + ((k0 γ 0 + ki γ i − m1)γ 0 )αβ e−ik·(x−y) dk 2k0 ((γ 0 )2 )αβ e−ik·(x−y) (2π )3 2k 0 = δαβ δ (x − y) = k0 ﬁxed where in the last line we used (γ 0 )2 = 1. 5 k0 ﬁxed Problem 5 (i) † † 0|ψRα (x)ψRβ (y )|0 † † − 0|ψRβ (y )ψRα (x)|0 † † 0|T ψRα (x)ψRβ (y ) |0 = if x0 > y 0 if x0 < y 0 Let’s start with the x0 > y 0 case: †(a† ) †(a) ψRβ (y ) + ψRβ (y ) |0 †(a† ) †(a) = 0|ψRα (x)ψRβ (y )|0 dk dk ′ √ (2π )3/2 2k 0 2k ′0 = †(a† ) †(a) † † 0|ψRα (x)ψRβ (y )|0 = 0| ψRα (x) + ψRα (x) 2 ′ ∗ (vk,s )α (u∗′ ,s′ )β 0|ak,s a† ′ ,s′ |0 e−ik·x eik ·y k k s,s′ =1 δs,s′ δ (k−k′ ) dk (2π )3/2 2k 0 = = −im(σ2 )αβ 2 s=1 ∗ (vk,s )α (u∗ )β · e−ik·(x−y) k,s dk e −i k · (x −y ) (2π )3/2 2k 0 where in the last step we used 2 2 2 ∗ (vk,s u† )αβ = ∗ (vk,s )α (u∗ )β = k,s s=1 s=1 k,s s=1 T ((uk,s vk,s )† )αβ = ((miσ2 )† )αβ = −im(σ2 )αβ . For y 0 > x0 : †(a† ) †(a) † † − 0|ψRα (y )ψRβ (x)|0 = − 0| ψRβ (y ) + ψRβ (y ) †(a) †(a† ) ψRα (x) + ψRα (x) |0 †(a† ) †(a) = − 0|ψRβ (y )ψRα (x)|0 = im(σ2 )βα dk e −i k · (y −x ) (2π )3/2 2k 0 = −im(σ2 )αβ dk e −i k · (y −x ) (2π )3/2 2k 0 We got from the second to the third line by realizing that †(a) †(a† ) 0|ψRβ (y )ψRα (x)|0 = †(a) †(a† ) 0|ψRα (x)ψRβ (y )|0 , exchanging α ↔ β and x ↔ y . Finally using what we showed in class when we talked about scalar ﬁelds dk −i k · (x −y ) (2π )3 2k0 e dk −i k · (y −x ) (2π )3 2k0 e d4 k i e −i k · (x −y ) = (2π )4 k 2 − m2 + iε 6 if x0 > y 0 if x0 < y 0 (3) we ﬁnd d4 k i e −i k · (x −y ) 4 k 2 − m2 + iε (2π ) † † 0|T ψRα (x)ψRβ (y ) |0 = −im(σ2 )αβ † 0|ψRα (x)ψRβ (y )|0 † − 0|ψRβ (y )ψRα (x)|0 † 0|T ψRα (x)ψRβ (y ) |0 = if x0 > y 0 if x0 < y 0 Let’s start with the x0 > y 0 case: (a† ) (a) ψRβ (y ) + ψRβ (y ) |0 †(a† ) (a) = 0|ψRα (x)ψRβ (y )|0 = †(a† ) †(a) † 0|ψRα (x)ψRβ (y )|0 = 0| ψRα (x) + ψRα (x) dk dk ′ √ (2π )3/2 2k 0 2k ′0 2 ′ (uk,s )α (u∗′ ,s′ )β 0|ak,s a† ′ ′ |0 e−ik·x eik ·y k k ,s s,s′ =1 δs,s′ δ (k−k′ ) = = dk (2π )3/2 2k 0 2 s=1 (uk,s )α (u∗ )β · e−ik·(x−y) k,s dk (k · σ )αβ e−ik·(x−y) ¯ (2π )3/2 2k 0 where in the last step we used 2 2 (uk,s )α (u∗ )β = k,s s=1 0 s=1 (uk,s u† ′ )αβ = (k · σ )αβ . ¯ k,s 0 For y > x : †(a† ) †(a) (a) † − 0|ψRα (y )ψRβ (x)|0 = − 0| ψRβ (y ) + ψRβ (y ) (a† ) †(a) = − 0|ψRβ (y )ψRα (x)|0 =− (a† ) ψRα (x) + ψRα (x) |0 dk dk ′ √ (2π )3/2 2k 0 2k ′0 2 ′ ∗ (vk,s )β (vk′ ,s′ )α 0|ak,s a† ′ ′ |0 e−ik·y eik ·x s,s′ =1 k ,s δs,s′ δ (k−k′ ) 2 =− dk (2π )3/2 2k 0 =− dk (k · σ )αβ e−ik·(y−x) ¯ (2π )3/2 2k 0 s=1 ∗ (vk,s )β (vk,s )α · e−ik·(y−x) where in the last step we used 2 2 ∗ (vk,s )β (vk,s )α = s=1 s=1 (vk,s v † )αβ = (k · σ )αβ . ¯ 7 k,s Since there is a k under the integral, we cannot use (3) directly. However by diﬀerentiation w.r.t. (x − y ), we can derive: i∂µ d4 k i e −i k · (x −y ) = (2π )4 k 2 − m2 + iε i∂µ dk e−ik·(x−y) + θ(y 0 − x0 ) (2π )3 2k 0 θ(x0 − y 0 ) d4 k ikµ e −i k · (x −y ) = (2π )4 k 2 − m2 + iε dk kµ e−ik·(x−y) + θ(y 0 − x0 ) (2π )3 2k 0 θ(x0 − y 0 ) + δ (x0 − y 0 ) dk e−ik·(x−y) − δ (x0 − y 0 ) (2π )3 2k 0 dk e −i k · (y −x ) (2π )3 2k 0 dk (−kµ )e−ik·(y−x) (2π )3 2k 0 dk e −i k · (y −x ) (2π )3 2k 0 The last line, stemming from ∂0 acting on the θ-functions can be rewritten as dk e−ik·(x−y) − δ (x0 − y 0 ) (2π )3 2k 0 δ (x0 − y 0 ) dk e i k · (x −y ) (2π )3 2k 0 = δ (x0 − y 0 ) 0 0 0 dk e−ik (x −y ) eik·(x−y) − (2π )3 2k 0 = δ (x0 − y 0 ) dk eik·(x−y) − (2π )3 2k 0 0 0 0 dk e i k (x − y ) e − i k · (x − y ) (2π )3 2k 0 dk e−ik·(x−y) (2π )3 2k 0 now take k → −k in the second term (this does not aﬀect k 0 = = δ (x0 − y 0 ) dk eik·(x−y) − (2π )3 2k 0 |k |2 + m2 ) dk e i k · (x − y ) (2π )3 2k 0 =0 d4 k ikµ e −i k · (x −y ) (2π )4 k 2 − m2 + iε = θ(x0 − y 0 ) = dk kµ e−ik·(x−y) + θ(y 0 − x0 ) (2π )3 2k 0 dk −i k · (x −y ) (2π )3 2k0 (kµ )e dk −i k · (y −x ) (2π )3 2k0 (−kµ )e if x0 > y 0 if x0 < y 0 8 dk (−kµ )e−ik·(y−x) (2π )3 2k 0 (4) Using (4) we can combine the results for x0 > y 0 and y 0 > x0 to ¯ d4 k i(k · σ )αβ −ik·(x−y) e 4 k 2 − m2 + iε (2π ) † 0|T ψRα (x)ψRβ (y ) |0 = T (ii) Since 2=1 uk,s vk,s = imσ2 for left-handed as well as for right-handed Mas jorana spinors, it follows, 0|T {ψLα (x)ψLβ (y )} |0 = 0|T {ψRα (x)ψRβ (y )} |0 † † † † and 0|T ψLα (x)ψLβ (y ) |0 = 0|T ψRα (x)ψRβ (y ) |0 and we can directly copy the results from what was derived in class and above: d4 k i e −i k · (x −y ) 4 k 2 − m2 + iε (2π ) 0|T {ψLα (x)ψLβ (y )} |0 = im(σ2 )αβ d4 k i e −i k · (x −y ) (2π )4 k 2 − m2 + iε † † 0|T ψLα (x)ψLβ (y ) |0 = −im(σ2 )αβ † For 0|T ψLα (x)ψLβ (y ) |0 , we need 2 s=1 (instead of = k · σ for right-handed spinors), so ¯ 2 s=1 uk,s u† k,s = † 0|T ψLα (x)ψLβ (y ) |0 = 2 s=1 vk,s v † k,s uk,s u† k,s = 2 s=1 vk,s v † k,s = k·σ † 0|T ψRα (x)ψRβ (y ) |0 under k · σ → k · σ ¯ so we get d4 k i(k · σ )αβ −ik·(x−y) e (2π )4 k 2 − m2 + iε † 0|T ψLα (x)ψLβ (y ) |0 = (iii) 0|T {ψα (x)ψβ (y )} |0 = 0|ψα (x)ψβ (y )|0 − 0|ψRβ (y )ψRα (x)|0 Since (¯ † ) a ( 0|ψα (x)ψβ (y )|0 = 0|ψαa) (x)ψβ and similarly (a) if x0 > y 0 if x0 < y 0 (y )|0 ∼ 0|aa† |0 = 0 ¯ † † (¯ 0|ψRβ (y )ψRα (x)|0 = 0|ψβ (y )ψαa ) (x)|0 ∼ 0|aa† |0 = 0, ¯ it follows that 0|T {ψα (x)ψβ (y )} |0 = 0 By the same argument ¯ ¯ 0|T ψα (x)ψβ (y ) |0 = if x0 > y 0 if x0 < y 0 † a ¯ ¯ ¯(a ) ¯(¯ ) 0|ψα (x)ψβ (y )|0 = 0|ψα (x)ψβ (y )|0 ∼ 0|aa† |0 = 0 ¯ † ¯β (y )ψα (x)|0 = − 0|ψ (a ) (y )ψ (¯ ) (x)|0 ∼ − 0|aa† |0 = 0 ¯ ¯ ¯a ¯ − 0|ψ Rα Rβ 9 So ¯ ¯ 0|T ψα (x)ψβ (y ) |0 = 0 ¯ 0|ψα (x)ψβ (y )|0 ¯ − 0|ψRβ (y )ψRα (x)|0 ¯ 0|T ψα (x)ψβ (y ) |0 = if x0 > y 0 if x0 < y 0 Let’s start with x0 > y 0 : (a† ) ( ¯ ¯ 0|ψα (x)ψβ (y )|0 = 0|ψαa) (x)ψβ (y )|0 dk dk ′ √ (2π )3/2 2k 0 2k ′0 = 2 ′ (uk,s )α (¯k′ ,s′ )β 0|ak,s a† ′ ′ |0 e−ik·x eik ·y u k ,s s,s′ =1 δs,s′ δ (k−k′ ) dk (2π )3/2 2k 0 = 2 (uk,s )α (¯k,s )β e−ik·(x−y) u s=1 dk / (k + m1)αβ e−ik·(x−y) (2π )3/2 2k 0 = Where in the last line we used 2 2 / / (uk,s u† γ 0 )αβ = ((k + m1)γ 0 γ 0 )αβ = (k + m1)αβ (uk,s )α (¯k,s )β = u s=1 k,s s=1 For y 0 > x0 : † (¯ ) a ¯ ¯(¯ − 0|ψRβ (y )ψRα (x)|0 = − 0|ψαa) (x)ψβ (y )|0 =− dk dk ′ √ (2π )3/2 2k 0 2k ′0 2 ′ (¯k,s )β (vk′ ,s′ )α 0|ak,s a† ′ ′ |0 e−ik·y eik ·y v k ,s s,s′ =1 δs,s′ δ (k−k′ ) =− = dk (2π )3/2 2k 0 2 (¯k,s )β (vk,s )α e−ik·(y−x) v s=1 dk / (−k + m1)αβ e−ik·(x−y) (2π )3/2 2k 0 Where in the last line we used 2 2 (¯k,s )β (vk,s )α = v s=1 2 (vk,s )α (¯k,s )β = v s=1 s=1 / / (vk,s v † γ 0 )αβ = ((k −m1)γ 0 γ 0 )αβ = (k −m1)αβ k,s Combining the two cases x0 > y 0 and y 0 > x0 using (3) and (4) we get ¯ 0|T ψα (x)ψβ (y ) |0 = / d4 k i(k + m1)αβ −ik·(x−y) e 4 k 2 − m2 + iε (2π ) 10 ...
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