d 3 xe ipm x φ(x) (2) d 3 x( φ(x)) 2 = 1 2 p n p 2 n φ(p n ) φ( p n ). (4) leads to a p n . (6) Finally, fill in the steps to show that = 1δpn,p m

Transcription

1 Physics 61 Homework Due Thurs 7 September 1 1 Commutation relations In class we found that φx, πy iδ 3 x y 1 and then defined φp m L 3/ d 3 xe ipm x φx and likewise for πp m. Show that it really follows from these definitions that φpn, πp m iδ pn, p m. 3 Then, show that 1 Next, show that the definition d 3 x φx 1 p n p n φp n φ p n. 4 a pn φp n +iπp n ωp leads to a p n φ p n iπ p n ωp 5 which together lead to apn, a p m 1δpn,p m. 6 Finally, fill in the steps to show that H 1 πp n π p n +p +m φp n φ p n 1 a p p n n a pn +a pn a p n. 7 p n That is, take the first expression for H to be shown, and derive the second expression from it, together with the definition of a and a. 1.1 Solution We have that φpn, πp m L 3 d 3 xd 3 ye ipn x ipm y φx, πy L 3 d 3 xd 3 ye ipn x ipm y iδ 3 x y il 3 d 3 xe ix pn+pm iδ pn, p m 8 1

2 Next, 1 d 3 x φx 1 d 3 xl 3 e ip n x φpn e ip m x φpm p n,p m 1 d 3 xl 3 ip n ip m e ix pn+pm φpn φp m p n,p m 1 δ pm, p n p n p m φp n φp m p n,p m 1 p n φp n φ p n 9 p n Use that φ x φx and the definition of φp n : φp n L 3/ d 3 xe ipn x φx L 3/ d 3 xe ipn x φx φ p n. 1 So a p n ωp φpn +iπp n ωp φ p n iπ p n ωp as desired. Carrying out the commutator, apn, a p m φ p n iπ p n ωp 11 1 ωpn m φpn, 4ωpn ω φ p m +im πpn, φ p m pm in φpn, π p m +i i πp n, π p m 1 4ωpn m +im iδ pn,p m in iδ pn,p m + iδ pn,p m. 1 Inverting the relation between a,a, φ and π, one finds φp n a p n +a p n ωp ia pn +ia p πp n ω n p 13 ωp and hence H 1 p n 1 ωp + p n apn a pn +a pn a p n +a p n a pn a p n a p n apn a pn +a pn a p ω n +a p n a pn +a p n a p n p apn a p n +a p n a pn. 14

3 Now just change the label from p n to p n in the last term note that p n is summed over and ω pn n. Momentum and field This is part of problem 3.4 from the book. Recall that the translation operator is Ta exp ip µ a µ and its action on the field is Ta 1 φxta φx a. 15 By considering infinitesimal a and the Taylor expansion of φx a about φx, find the expression for P µ, φx. Show that the time component µ of your relation is the Heisenberg equation of motion recall H P t φ i H, φ Solution At the infinitesimal level T 1 aφxta 1+iP µ a µ φx1 ip µ a µ φx+ia µ P µ, φx +Oa 17 while φx a φx a µ µ φ+oa. 18 The a terms are clearly the same. Equating the a 1 terms, ia µ P µ, φx a µ µ φ P µ, φx i µ φx. 19 Here I used that a µ is general so the quantities multiplying it must be equal. Now φ t φ, so for the case of P H the equation reads H, φx i t φx or t φx i H, φx. 3 Correlation functions Here we will study the values of several correlation functions in the free theory. First, use the expansion of the field in terms of creation and annihilation operators and their commutation 3

4 relations Useful Formulae equations 13,14 φx ak, a p to derive that the Wightman correlation function takes the value d 3 k ak e ik x +a e ik x π 3 k k k k +m d 4 k +m Θk a π 4πδk k e ik x +a e ik x k 1 k π 3 δ 3 k p G > x G > x φxφ 3 d 4 k π 4eik x πδk +m Θk. 4 Hint: insert the explicit expressions for φx and φ in terms of creation and annihilation operators. Use that a and a and the commutation relation to evaluate the resulting expression. Next, use translation invariance to show that the lesser than Wightman correlator G < x φφx G > x It is also common to define the retarded Green function the advanced Green function d 4 k π 4eik x πδk +m Θ k. 5 G R x φx, φ Θx G > x G < x Θx 6 and the time-ordered Green function G A x G R x φx, φ Θ x 7 G T x G > xθx +G < xθ x. 8 Explain why G > x G < x for spacelike x, and how that ensures that G R vanishes unless x x, G A vanishes unless x x, and all three functions are Lorentz invariant despite the presence of the non-lorentz-invariant Θx in their expressions. Although we know the momentum-space formof G > x and G < x, the presence of Θx in the above expressions makes it tricky to get to the momentum-space expressions for these correlation functions. But we can do so by writing dω Θx lim ǫ + πi 4 e iωx ω iǫ. 9

5 Substitute this expression, as well as the Fourier-space expressions Eq. 4 and Eq. 5, into Eq. 6. Use the δk +m functions to perform the k integration, to show that G R x d 3 k k x dω π 3ei πi e iωx e iω k x ω iǫ ω k eiωkx ω k where ω k k +m. 3 This expression is the sum of two terms with different-appearing complex exponents. For each term, change variables from ω to k ω ±ω k such that the new variable appears in the exponent as e ik x. Re-organize the integrand in terms of this variable and show that it gives d 3 kdk ie ik x G R x π 4 k +m k +iǫ. 31 Therefore the momentum-space value of G R is G R k i/k +m k k +iǫ. You will take my word that one can similarly show that G T k i/k +m iǫω k. One last thing. The retarded function we found has poles in the complex k plane at the points k iǫ ± ω k, but it is analytic in the upper half-plane whenever k has positive imaginary part. Show that this is actually totally general. That is, defining the Fourier transform of the retarded function G R k d 4 xe ikµxµ G R x, 3 argue that so long as G > x and G < x are nonsingular for x >, that the integral is perfectly well defined if k has a positive imaginary part. Hint: write out the exponent e ikµxµ e ik x e i k x, and then write k kre + ik im. Does the exponent help or hurt convergence when x >? Does it help or hurt when x <? But what is the value of G R x when x <? Similarly, G A p is well defined for negative imaginary part. 3.1 Solution The problem has a lot of words but not that many things to do! The free theory Wightman function is φxφ d 3 k 1 d 3 k a π 6 k1k k1 e ik1 x +a k 1 e ik 1 x a k e ik +a k e ik d 3 k 1 d 3 k e ik1 x a π 6 k1k k1 a k d 3 k 1 d 3 k e ik1 x k π 6 k 1π 3 δ 3 k 1k 1 k d 3 k 1 e ik1 x d 4 k 4eik x πδk +m Θk. 33 π 3 k1 π 5

6 In the last line we changed names from k 1 to k and used what we learned in class about the relation between the integral and the explicitly Lorentz invariant integral over d 4 k. The lesser Wightman function is G < x φφx. 34 Now insert the identity in the form of TxT 1 x, several places: G < x TxT 1 xφtxt 1 xφxtxt 1 x. 35 The vacuum is translation invariant, which means that T 1 and Tx. And T 1 xφytx φy x. So G < x φ xφ G > x. 36 Equivalently we could just STATE that the correlation beteen and x should be the same as between -x and, since they are just shifted over in space and physics is invariant to such shifts. The position invariance of the vacuum is the rigorous way of showing that. Atequaltimes, φx, φ. ByLorentzinvariancethisshouldalsobetruewhenever x is spacelike. Since G > x and G < x differ by the order of the operators, they are equal at spacelike x, G > x G < x for x >. 37 Therefore G R x vanishes at spacelike x, since the commutator is zero. So it is only nonzero at lightlike or timelike separation, and the Θx simply picks out the positive light cone. The same is true of G A except it s the negative light cone; and there is no discontinuity in G T when x crosses zero because the two terms are the same; so it doesn t matter which is nonzero. Now we study G R k: G R x Θx G > x G < x Θx d 4 k 4eik x πδk +m Θk Θ k π dω d 4 k 4eik x 1 e iωx πi π ω iǫ πδk +m Θk Θ k d 3 k k x dk dω π 3ei π eix ω k i ω iǫ πδk +m Θk Θ k 38 Now perform the k integral, generating a 1/ω k and turning k ω k for the first step function and ω k for the second step function: G R x d 3 k k x dω i e ix ω ω k e ix ω+ω k 39 π 3ei π ω iǫω k 6

7 Think of this as two terms, and do a shift of variables in each: dω i e ix ω ω k e ix ω+ω k π ω iǫω k dω i ω ω k dω i ω+ω k π ω iǫω k eix π ω iǫω k eix dω i dω i π ω +ω k iǫω k eixω ω π ω ω k iǫω k eix add them back together and replace ω k : dω i e ix ω ω k e ix ω+ω k π ω iǫω k dω ω i π eix ω +ω k iǫω k i ω ω k iǫω k dk k i k ω k iǫ+k ω k +iǫ e ix π ω k k +ω k iǫ k ω k iǫ dk iω k k e ix π ω k k +ω k iǫk +ω k +iǫ dk i k e ix π ωk k +iǫ 4 41 and turning ω k k +m again, sticking back into the expression for G R, we find In general G R x d 4 k π 4eik x i k +m k k +iǫ. 4 G R k d 4 xe ikµxµ G R x 43 can be defined for complex k ; the time part of the integral is dte ikt G > G < 44 which will converge provided that e ikt is well behaved. In terms of the complex k, this is e ik re t e k im t ; since t due to the Θt function, there is no problem with convergence unless G > or G < diverges, which would indicate severe problems with the theory. 4 Energy associated with the vacuum We saw that the Hamiltonian for a free field theory is H p π L n a pn a pn +a pn a p n 45 7

8 where a pn, a p m δ n, m the Kroneker delta. Show that the energy of the vacuum H is E vac H p. 46 Suppose that, because space is somehow discrete on some scale a, there is a maximum size which the momentum p is allowed to take, call it p max. Show that the energy of the vacuum is extensive that is, that it grows as we increase the size of our box as L 3. This is to be expected. Then show that it also depends on p max with proportionality p 4 max, that is, E vac L 3 p 4 max. Do not try to evaluate the actual coefficient, which depends on exactly what we mean by momenta cannot exceed some scale p max. Show that, if we include a constant term C in the Lagrangian density, the Hamiltonian is shifted by L 3 C. This energy associated with space is of no relevance in particle physics because it is not observable. Particle physicists who also worry about gravity do worry about it, though, because it should behave as a cosmological constant, and we know observationally that the cosmological constant is very small, whereas presumably p 4 max is very large. In principle it might be that C is just the right value to balance the energy associated with all the SHO zero-points, but Solution The vacuum value of the energy is H p p p ω p ap a p +a pa p ap a p + a pa p 1+ p. 47 The spacing of terms in the sum is p n π/l n. If we make L twice as large, there are twice as many n x values needed to get to a particular value of p x, and similarly for p y,p z. Therefore the number of terms in the sum grows as L 3. Alternatively, if we think about space as a lattice of points, we know there is one term in the sum for each lattice point; at fixed spacing, the number of points is clearly volume-extensive. The sum is roughly p pmax L3 d 3 p L3 π 3p 8π p4 max 48 8

9 which scales as L 3 p 4 max, as we could guess from the fact that we are computing an energy and it is extensive. Since H P Q L, a constant in L becomes a negative constant in H; a constant C in L becomes d 3 x C in H, evaluating to L 3 C. 9