University of California at San Diego – Department of Physics – Prof. John McGreevy
Quantum Field Theory C (215C) Spring 2013
Assignment 1 – Solutions
Posted April 3, 2013
Due 11am, Thursday, April 11, 2013
Please remember to put your name at the top of your homework.
Announcements
• The 215C web site is:
http://physics.ucsd.edu/∼mcgreevy/s13.
• Please accept my apologies for the typos in the problem statement. They will make
you stronger!
Problem Set 1
1. Scale invariant quantum mechanics
Consider the action for one quantum variable r ∈ IR+
Z
λ
1 2
S[r] = dt
mṙ − V (r) , V (r) = 2 .
2
r
[Note: relative to earlier versions of this problem set, I’ve restored the (non-relativistic)
mass parameter m. It can be eliminated by a multiplicative redefinition of the field r
or of the time t. You should convince yourself that the physics of interest here should
only depend on mλ.]
(a) Show that the coupling λ is dimensionless: [λ] = 0.
As
demanding that
R λ
√ before 1in 0 + 1 dimensions,
√ [Skinetic ] = 0 = 1 − 2 + 2[r] implies
[ mr] = 2 . So 0 = [ dt r2 ] = 1 + [λm] − 2[ mr] – the more accurate statement
is that [λm] = 0. A funny thing about QM is that the mass can be scaled away.
The physics will only depend on λm.
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(b) Show that this action is scale invariant, i.e. show that the transformation
r(t) → sα · r(st)
(1)
(for some α which you must determine), (with s ∈ IR+ ) is a symmetry. Find the
associated Noether charge D. For this last step, it will be useful to note that the
infinitesimal version of (1) is (s = ea , a 1)
d
δr(t) = a α + t
r(t).
dt
With the finite form of the transformation is easier to check that the action is
invariant. Especially with a symmetry that acts on spacetime, we must be careful
about active and passive issues – this problem is actually a bit of a nightmare of
signs because of that. Then
Z
Z
λ ts ≡st
1 2 2α
1
2
2
−2α −2
−1
S[rs ] = dt
m (∂t rs (st)) − 2
ms s (∂ts r(ts )) − λs r
= S[r]
=
dts s
2
rs
2
if 2α + 2 − 1 = 0 and −2α − 1 = 0 which both require α = − 21 . Note that this
agrees with our naive dimensional analysis.
In field theory the way to find the Noether current is the following. If we know
that under a transformation φ → φ with parameter constant in spacetime, the
action does not change: S[φ] = S[φ ] then if we allow = (x) (infinitesimal) then
the variation must be proportional to derivatives of :
Z
δS ≡ S[φ(x) ] − S[φ] = dD x∂µ j µ (x)
R
for some functional of the fields j µ . The RHS is δS = − dD x∂µ j µ by integration
by parts (we assume no boundary); but this must vanish for constant epsilon!
Therefore ∂µ j µ = 0. j µ is the conserved noether current. Its time component,
integrated over space, is time independent:
Z
Z
d
d
D−1
0
d
~xj = − ∇ · j = 0
Q≡
dt
dt
(again we ignore boundary terms, in space). This method is superior to formulae
you remember from classical mechanics (like Q = ∂L
δr +Lδt) because it makes no
∂ ṙ
assumptions about the dependence of the Lagrangian on q̇ and it doesn’t require
remembering anything.
In this example there is no space and we have j 0 = Q. The infinitesimal variation
of r is δr(t) = a(− 21 +Rt∂t )r(t). Under the transformation (1) with constant s, the
lagrangian (L in S = dtL) rescales
R
Rby L(t) → sL(st); this is necessary to cancel
the variation of the measure dt → d(ts ). The infinitesimal statement is
δL(t) = a (1 + t∂t ) L(t) =
2
d
(atL) .
dt
But under a variation with a = a(t), we would acquire an extra term from the
variation of the kinetic term:
1
1 2
mṙ = mṙδ ṙ = mṙȧ − + t∂t r + terms that are there when ȧ = 0.
δ
2
2
So the variation of L is
1
δL = ... + ȧ tL − mṙr + mtṙ2
2
where again the ... is terms that would be there if ȧ = 0. So by the method
described above we have:
1
1 2 λ
1
2
D = tL − mṙr + mtṙ = t
mṙ + 2 − mṙr.
2
2
r
2
We conclude that the Noether charge is
1
1
D = +Ht − rmṙ = Ht − rp.
2
2
We can check that on a solution of the EoM ṗ = −∂r H, ṙ = +∂p H :
d
1
1
1
1
1 p2 1 2λ
D = H − rṗ − pṙ = +H − p∂p H + r∂r H = +H −
−
= 0.
dt
2
2
2
2
2 m 2 r2
(c) Find the position-space Hamiltonian H governing the dynamics of r. Show that
the Schrödinger equation is Bessel’s equation
λ
∂r2
+
ψE (r) = EψE (r).
−
2m r2
Show that the Noether charge associated D with scale transformations (≡ dilatations) satisfies: [D, H] = iH. This equation says that the Hamiltonian has a definite scaling dimension, i.e. that its scale tranformation is δH = ia[D, H] = −aH.
2
p
+ rλ2 . The quantum dilatation operator is then
The Hamiltonian is H = 2m
1
D = − 2 rp + tH (plus a possible constant term related to ordering issues of x and
p which we can neglect). We can check that this generates the correct variation
of r by commutators:
1
1
1
δr = −ia[D, r] = ia[− rp+tH, r] = ia − r(−i) + ti∂t r = a − r + tṙ
X
2
2
2
(where we used the CCR [p, r] = −i). So
1
1
1
[D, H] = [− rp+tH, H] = − (r[p, H] + [r, H]p) = −
2
2
2
3
λ
p2
+2i 2 − 2i
= iH.
r
2m
This is the statement that the hamiltonian has definite scale dimension (namely
one). We found a conserved current, but it doesn’t commute with the Hamiltonian. What gives? The thing that’s true is that the charge is time independent –
the total time derivative vanishes. But the total time derivative has two parts:
d
D = ∂t D − i[H, D] = H − H = 0.
dt
(d) Describe the behavior of the solutions to this equation as r → 0. [Hint: in this
limit you can ignore the RHS. Make a power-law ansatz: ψ(r) ∼ r∆ and find ∆.]
Plugging in ψ(r) ∼ r∆ gives
r
1
1
∆(∆ − 1) + 2mλ = 0
=⇒
∆± = ± 2mλ + .
2
4
(e) What happens if 2mλ < − 14 ? It looks like there is a continuum of negative-energy
solutions (boundstates). This is another example of a too-attractive potential.
When 2mλ passes through − 14 from above the two roots ∆± collide and move
off into the complex plane. This means that the eigenfunctions oscillate near the
origin, like eia log(r) for some constant a, for any E. This is innocuous for E > 0
where there is already a continuum of scattering states. (Recall that the behavior
at large r satisfies −∂r2 ψ = Eψ, so
( √
ei Er , E > 0
r→∞
√
ψ(r) ∼
e− |E|r , E < 0
) But for E < 0, a good Schrodinger equation will specify a discrete set of energies
at which we can integrate the wavefunction in from r = ∞ without encountering
a singularity. This one (at 2mλ < − 14 ) instead allows any negative E. This
suggests a problem.
(f) A hermitian operator has orthogonal eigenvectors. We will show next that to make
H hermitian when 2mλ < − 14 , we must impose a constraint on the wavefunctions:
(ψE? ∂r ψE − ψE ∂r ψE? ) |r=0 = 0
(2)
There are two useful perspectives on this condition: one is that the LHS is the
probability current passing through the point r = 0.
The other perspective is the following. Consider two eigenfunctions:
HψE = EψE , HψE 0 = E 0 ψE 0 .
Multiply the first equation by ψE? 0 and integrate; multiply the second by ψE? and
integrate; take the difference. Show that the result is a boundary term which
must vanish when E = E 0 .
4
The boundary term is (the Wronskian):
ψE? ∂r ψE − ψE ∂r ψE? .
This is (i times) the probability current through the point r. For r → 0 the probability has nowhere to go, (and more generally for static solutions the probability
should not be moving) so if this doesn’t vanish there is trouble. We conclude from
this analysis that when eigenstates of H go like ψ(r) ∼ r∆ with complex ∆, that
H must not be Hermitian: probability is leaking into the hole in the potential at
r = 0! It does not describe a closed system.
(g) Show that the condition (2) is empty for 2mλ > − 41 . Impose the condition
(2) on the eigenfunctions for 2mλ < − 41 . Show that the resulting spectrum of
boundstates has a discrete scale invariance.
[Cultural remark: For some silly reason, restricting the Hilbert space in this way
is called a self-adjoint extension.] 1
An example of a condition which implies (2) is simply to impose that
a = ψ(r)|r=
for some constant a and 1 some UV cutoff on the potential. Any other choice
of boundary condition at r = will have the same qualitative effect. This is a
restriction on the Hilbert space on which the Hamiltonian above is Hermitian. It
is extra short-distance information (near the origin) about the potential, which
comes with a scale: . The UV cutoff breaks scale invariance. The original scale
invariant theory was not well defined (as a closed quantum system).
The result of imposing this boundary condition is the following. We need one
more piece of information about how the energy enters into the wavefunctions.
We could find this by solving the equation exactly (e.g. plug it into Mathematica).
The solution is some Bessel K (K is the one that goes to zero at large argument).
More instructive is to use dimensional analysis to notice that E scales like√r−2 ;
therefore, up to some overall factor, the wavefunction must be a function of Er.
So we want to impose :
√ 12 +iδ
√ 12 −iδ
a = ψ(r)|r= = A+
Er
+ A−
Er
q
where A± are the two (real) integration constants and δ ≡ |2mλ − 14 |. A more
convenient description is in terms of amplitude and phase
√
1
a = A0 2 cos δ log E + ϕ
which is satisfied for En such that
δ
log E2 = −2πn + ϕ
2
1
This model has been studied extensively, beginning, I think, with K.M. Case, Phys Rev 80 (1950) 797.
More recent literature includes Hammer and Swingle, arXiv:quant-ph/0503074, Annals Phys. 321 (2006)
306-317. It also arises as the scalar wave equation for a field in anti de Sitter space.
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Or
En ∝ −2 e−
4πn
δ
.
The energies of the allowed boundstates form a geometric series. Again we see
dimensional transmutation: the energy scale that determines the boundstate spectrum comes from the UV cutoff we were forced to introduce.
(h) [Extra credit] Consider instead a particle moving in IRd with a central 1/r2 potential, r2 ≡ ~x · ~x,
Z
λ
1 ˙ ˙
m~x · ~x − 2 .
S[~x] = dt
2
r
Show that the same analysis applies (e.g. to the s-wave states) with minor modifications.
[A useful intermediate result is the following representation of (minus) the laplacian in IRd :
L̂2
1
1
p~2 = − d−1 ∂r rd−1 ∂r + 2 , L̂2 ≡ L̂ij L̂ij , Lij = −i (xi ∂j − xj ∂i ) ,
r
r
2
2
i i
where r ≡ x x . By ‘s-wave states’ I mean those annihilated by L̂2 .]
2. Gaussian integrals are your friend
The solutions to these problems are in the Appendix in Zee Part I (2d edition).
(a) Show that
Z
∞
dxe
− 12 ax2 +jx
−∞
r
=
2π j2
e 2a .
a
(b) Consider a collection of variables xi , i = 1..N and a hermitian matrix aij . Show
that
Z Y
N
1
(2π)N/2 1 J i a−1
j
i
e 2 ij J .
dxi e− 2 xi aij xj +J xi = √
det a
i=1
(Summation convention in effect, as always.)
Q
[Hint: change into variables to diagonalize a. det a =
ai , where ai are the
eigenvalues of a.]
(c) Consider a Gaussian field X, governed by the (quadratic) euclidean action
Z
1 2
2 2
S[x] = dt Ẋ + Ω X .
2
Show that
R
R
1
he− dsJ(s)X(s) iX = N e+ 4 dsdtJ(s)G(s,t)J(t)
where G is the (Feynman) Green’s function for X, satisfying:
−∂s2 + Ω2 G(s, t) = δ(s − t).
Here N is a normalization factor which is independent of J. Note the similarity
with the previous problem, under the replacement
a = −∂s2 + Ω2 , a−1 = G.
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