PHYS PS 4 Solutions

PHYS 610 - PS 4 Solutions

Sidan A

Fall 2020

Please email me if you have questions about these solutions :)

1

a) (6pt) When we quantize a free scalar fieldϕ(x) in 2+1D we get similar results as in 1+1D, from the KG eqn (+m2)ϕ = 0 where in 2+1D = ∂₀2 −∂₁2 −∂₂2 we have the solution

ϕ(x) ∼ e−ik0x0+ik1x1+ik2x2 _{with some factors that we dropped, and the energy is given by}

k0 =

p

k2

1 +k22+m2.

2D sheet: No periodic boundary condition, no restriction onk1, k2;

Cylinder: ϕ(x0, x1, x2) = ϕ(x0, x1+ 2πR, x2) where 2πR=L, which leads to k1 ∈2πZ/L,

no restriction on k2;

Torus: Scalar fields on a torus are periodic in both xand y direction, it leads tok1, k2 ∈

2π_Z/L.

b) (6pt) Compute the 2pt propagators on: 2D sheet: hϕ(y2)ϕ(y1)i= Z d2_kd2_k0 (2π)4 1 2pk0k0₀h(ak 0e−i~k 0_·y~ 2 _+a† k0ei~k 0_·y~ 2_)(a ke−i~k·y~1 +a † ke i~k·y~1_)i = Z d2kd2k0 (2π)4 1 2pk0k00 e−i~k0·y~2+i~k·y~1_ha k0a† ki = Z _d2_k (2π)2 1 2k0e i~k·(y~1−y~2) 1

Cylinder: hϕ(y2)ϕ(y1)i= 2π L 2 X k1,k10 Z _dk 2dk02 (2π)2 1 2pk0k00 e−i~k0·y~2+i~k·y~1_ha k0a† ki = 2π L X k1 Z _dk 2 2π 1 2k0 ei~k·(y~1−y~2) Torus: hϕ(y2)ϕ(y1)i= 2π L 4 X k1,k01 X k2,k02 1 2pk0k₀0e −i~k0·y~2+i~k·y~1_ha k0a† ki = 2π L 2 X k1,k2 1 2k0 ei~k·(y~1−y~2)

c) (6pt) Compute the 2pt amplitudes:

hx2|e−iHt|x1i= Z d2k (2π)2 hx2|e −iHt_{|ki hk|x} 1i = Z d2_k (2π)2e −ik2t 2m hx₂|ki hk|x₁i = Z _d2_k (2π)2e −ik2t 2mei~k·(~x2−~x1)

Perform the integral over ~k in radial coordinates (k, θ) by ~k·~x = kxcosθ, we see that the result doesn’t depend on momentum modes, thus it’s the same in all three spaces. In quantum mechanics, different spaces we have different results.

d) (6pt) Compute the configuration space wave-functions on: 2D sheet: φ(ϕ) =Y k exp(−k0 2 ak+a † −k √ 2k0 2 ) = exp − 1 4 Z _d2_k (2π)2|ak+a † −k| 2 Cylinder: φ(ϕ) =Y k exp(−k0 2 ak+a † −k √ 2k0 2 ) = exp − π 2L X k1 Z dk2 2π|ak+a † −k| 2

Torus: φ(ϕ) = Y k exp(−k0 2 ak+a † −k √ 2k0 2 ) = exp − π 2 L2 X k1,k2 |ak+a † −k| 2

The space-time wave-function for the fields are:

h0|ϕ(x)|ki= Z _d2_k0 (2π2₎ 1 p 2k0₀h0|(ak 0ei ~k 0_·~x +a†_k0e−i ~k0·~x)p2k0a † k|0i=e i~k·~x

For a 2D sheet,~k = (k1, k2) satisfies k1, k2 ∈_R; for a cylinder, k1 ∈2π_Z/L and k2 ∈_R; for a torus, k1, k2 ∈2πZ/L.

e) (6pt) Shrink the height of the cylinder to zero size, then quantize the fields, we get that ϕ(x) = Z _dk 2π 1 √ 2k0 (akeikx+a†_ke−ikx) with k ∈2π_Z/L

The 0pt energires are given by the following: i. On a circle: E0 = 1 2 X k∈2πZ/L √ k2_+m2 ii. On a 2D sheet: E0 = 1 2 Z d2k (2π)2 p |k|2_+m2 iii. On a cylinder: E0 = 1 2 X k1∈2πZ/L Z _dk 2 2π q k2 1+k22+m2 iv. On a torus: E0 = 1 2 X k1,k2∈2πZ/L q k2 1+k22+m2

On a circle, the configuration space wave-function is given by

φ(ϕ) = exp − 1 4 X k∈2πZ/L |ak+a † −k| 2

and the spacetime wavefcn is eikx with k ∈ 2π_Z/L. When taking the m → 0 limit, we see that the Klein-Gordon eqn becomes the usual wave equation ∂2

tϕ = ∂x2ϕ, and the particle

goes around the circle as a wave at the speed of light.

2

(10pt) By Noether’s thm we have Tµν _=∂µ_φ∂ν_φ−ηµν_{L. When coupling a free scalar theory}

to gravity, we get Tµν =−√2 −g ∂(√−gL) ∂gµν =−√2 −g 1 2 ∂√−g ∂gµν (∂µφ∂µφ+m2φ2) + √ −g ∂ ∂gµν (∂µφgµν∂νφ) +m2φ2 =−√2 −g _√ −ggµνL −√−g∂µφ∂νφ =∂µφ∂νφ−ηµνL

The minimal change to do is to replace ηµν bygµν.

3

a) (5pt) Define an operator ϕ(~x) = R d3_k

(2π)3a~_kei~k·~x. For a relativistic theory, the simple

nor-malization h~k|~k0_{i = (2π)}3_δ(3)_{(~k −~k}0_{) is NOT invariant under boosts, and we need a factor}

of ωk so ωkδ(3)(~k−~k0) is Lorentz invariant. Thus we define |~ki= √

2ωka

†

k|0i and this is the

origin of this factor of√2ωk. However, for non-relativistic theories, the simple normalization

b) (5pt) To see thatXcan be viewed as a natural position operator, actX on a state |~yi X|~yi= Z d3x~xϕ†(~x)ϕ(~x)ϕ†(~y)|0i = Z d3x~xϕ†(~x)[ϕ(~x), ϕ†(~y)]|0i = Z d3x~xδ(3)(~x−~y)|~xi =~y|~yi c) (5pt) For a state|ψi ≡R d3_xψ(~x)|~xi X|ψi= Z d3xψ(~x)X|~xi= Z d3x~xψ(~x)|~xi P|ψi= Z d3pd3x~pa†_~pa~pψ(~x)|~xi = Z d3pd3x~pψ(~x)a_p~†a~pϕ†(~x)|0i = Z d3pd3x~pψ(~x)a_p~†[a~p, ϕ†(~x)]|0i = Z d3pd3x~pe−i~p·~xψ(~x)ϕ†(~p)|0i = Z d3pd3xi∇ei~p·x~ψ(~x)a†_~p|0i = Z d3pd3x−i∇ψ(~x)ei~p·~xa†_~p|0i = Z d3x−i∇ψ(~x)ϕ†(~x)|0i = Z d3x−i∇ψ(~x)|~xi [Xi,Pj]|ψi= [xi,−i∂j]|ψi=iδij|ψi

d) (5pt) For a static free field we have Hψ(~x, t) = 1 2m Z d3x∂µϕ†∂µϕψ(~x, t) = 1 2m Z _d3_k (2π)2k 2 a†_~ ka~kψ(~x, t) = P 2 2mψ(~x, t) =− 1 2m∇ 2 ψ(~x, t)

which leads to non-relativistic the Schrodinger eqn i∂ψ_∂t =− 1 2m∇

2_ψ.

4

a) (5pt) As we saw in a previous problem set that a static solution for ϕi to the Lagrangian

satisfies _∂ϕ∂U

i =∇

2_ϕ

i, and this leads to ∂W_∂ϕ

i = 0, thus it is the extrema of W[ϕ].

b) (5pt) Consider a one-parameter family of solutions, lety=λx, then we have

V1[ϕλ] +V2[ϕλ] = Z dDx 1 2(∇ϕi(λx)) 2_+U(ϕ(λx)) = Z dDyλ−D λ21 2(∇ϕi(y)) 2_+U(ϕ(y)) =λ2−DV1[ϕ] +λ−DV2[ϕ]

Evaluate the following

∂W[ϕλ] ∂λ λ=1 = (2−D)V1[ϕ]−DV2[ϕ] ∂2_W[ϕ λ] ∂λ2 λ=1 = (1−D)(2−D)V1[ϕ] + (1 +D)DV2[ϕ]

A static solution requires the first derivative to be 0. Since V1, V2 are non-negative, we see

that for D ≥ 3 a stable static solution requires V1 = V2 = 0 so the second derivative is non-negative. Thus ϕi must be time-independent and it’s the abs min of U(ϕ). The only

solution that satisfies these requirements is the constant solution. So there’s no non-trivial static solution.