The Breakdown of the Coherent State Path Integral
Justin H. Wilson
Presentation Based on:
Advisor: Victor Galitski
PRL 106, 110401 (2011)
University of Maryland, College Park, MD
Preprint: arXiv:1012.1328
Quantum Vacuum Meeting
17 May 2012
Introduction
Path integrals have a jaded mathematical history.
Semiclassics have produced incorrect results (fixed with the
discovery of a phase anomaly).
We will find that we get quantitatively wrong results with an
exact calculation.
We try to fix this a posteriori.
Nothing incorrect appears when the Hamiltonian is a
linear sum of generators of the Lie algebra
Glauber coherent states are an
over-complete set of states.
h4 = {1, a, a† , a† a}
|z = e
a |z = z |z
[a, a† ] = 1
|z|2 /2 za†
e
|0
Z
d2 z
|z⇥ z| = 1
Partition function/Path integral
( Z
)
Z
H
2
⇤
Z = tr e
= D z exp
[z ż + ⇥z|H|z⇤]
0
R. J. Glauber, Phys. Rev. 131, 2766 (1963)
The single-site Bose-Hubbard model can
be written in terms of a Hamiltonian or a
path integral.
Hamiltonian:
U
µn̂ + n̂(n̂ 1)
2
†
† †
a a
a a aa
H=
Partition Function:
Z = tr e
H ?
=
Z
2
D ze
R
0
4
|z|
d⇥ [z ⇤ ż µ|z|2 + U
]
2
=: Z
0
The evaluation of the path integral...
Z
⇥
Z :=
z=
0
p
Z =
0
Z =
Z
2
D ze
ne
Z
R
0
d
2
i
DnD e
4
⇤
2
U
|z|
z
ż
µ|z|
+
[
]
2
D z = Dn D
R
0
2
n
d⇥ [in ˙ µn+ U
]
2
Integrate by parts
DnD e
i(n( )⇥( ) n(0)⇥(0))
R
0
2
n
d⇤ [ iṅ⇥ µn+ U
]
2
Method:
A. Alekseev et al., J. Geom. Phys. 5, 391 (1988)
D.C. Cabra et al., J. Phys. A 30, 2699 (1997)
0
Z =
0
Z
Z =
DnD e
Z
X
k
0
Z =
n( )
n(0) = 0
⇥( )
⇥(0) = 2⇤k
DnD e
Z
X
k
i(n( )⇥( ) n(0)⇥(0))
i2⇥kn(0)+i
Z
D⇥(⇤ )e
Dn (ṅ) e
R
0
R
i2 kn(0)
2
n
d⇤ [ iṅ⇥ µn+ U
]
2
R
ṅ
0
i
R
0
2
d⇤ [ µn+ U
n
]
2
d⇥ f (⇥ ) (⇥ )
0
R
0
= (f )
2
d⇥ [ µn+ U
n
]
2
0
Z =
0
Z
Z =
DnD e
Z
X
k
0
Z =
n( )
n(0) = 0
⇥( )
⇥(0) = 2⇤k
DnD e
Z
X
k
i(n( )⇥( ) n(0)⇥(0))
i2⇥kn(0)+i
Z
D⇥(⇤ )e
Dn (ṅ) e
R
0
R
i2 kn(0)
2
z
n
d⇤ [ iṅ⇥ µn+ U
]
2
R
ṅ
0
i
R
0
2
d⇤ [ µn+ U
n
]
2
d⇥ f (⇥ ) (⇥ )
0
R
0
= (f )
2
d⇥ [ µn+ U
n
]
2
0
Z =
0
Z
Z =
DnD e
Z
X
k
0
Z =
n( )
n(0) = 0
⇥( )
⇥(0) = 2⇤k
DnD e
Z
X
k
i(n( )⇥( ) n(0)⇥(0))
i2⇥kn(0)+i
Z
D⇥(⇤ )e
Dn (ṅ) e
R
0
2
z
n
d⇤ [ iṅ⇥ µn+ U
]
2
k=1
R
ṅ
0
i
R
R
i2 kn(0)
0
2
d⇤ [ µn+ U
n
]
2
d⇥ f (⇥ ) (⇥ )
0
R
0
= (f )
2
d⇥ [ µn+ U
n
]
2
0
Z =
0
Z
Z =
DnD e
Z
X
k
0
Z =
n( )
n(0) = 0
⇥( )
⇥(0) = 2⇤k
DnD e
Z
X
k
i(n( )⇥( ) n(0)⇥(0))
i2⇥kn(0)+i
Z
D⇥(⇤ )e
Dn (ṅ) e
R
0
2
z
n
d⇤ [ iṅ⇥ µn+ U
]
2
k=2
R
ṅ
0
i
R
R
i2 kn(0)
0
2
d⇤ [ µn+ U
n
]
2
d⇥ f (⇥ ) (⇥ )
0
R
0
= (f )
2
d⇥ [ µn+ U
n
]
2
0
Z =
0
Z
Z =
DnD e
Z
X
k
0
Z =
n( )
n(0) = 0
⇥( )
⇥(0) = 2⇤k
DnD e
Z
X
k
i(n( )⇥( ) n(0)⇥(0))
i2⇥kn(0)+i
Z
D⇥(⇤ )e
Dn (ṅ) e
R
0
R
i2 kn(0)
2
n
d⇤ [ iṅ⇥ µn+ U
]
2
R
ṅ
0
i
R
0
2
d⇤ [ µn+ U
n
]
2
d⇥ f (⇥ ) (⇥ )
0
R
0
= (f )
2
d⇥ [ µn+ U
n
]
2
Z
X
0
Z =
k
R
i2 kn(0)
Dn (ṅ) e
0
2
d⇥ [ µn+ U
n
]
2
n(t) is a constant in time
0
Z =
n(t) = x
XZ
k
1
dx e
2
µx+ U
x
]
2
[
i2⇥kx
0
X
0
e
2 ikx
=
0
Z =
n
1
dx (x
n)e
[
0
0
Z =
X
n
e
µn
(x
n
k
Z
X
X
U
2
n2
2
µx+ U
x
]
2
n)
They are not the same
expression.
U
0
Z =
Z=
1
X
e
1
X
e
µn
U
2
n2
1
1+e
e
U/2
n=0
µn
U
2
n(n 1)
1+e
n=0
=)
µ
Z = Z0
µ
+ e2
µ
e
+ ···
U
+ ···
This is a new issue.
⇥
Z :=
Z
2
D ze
R
⇤
z
ż
[
0
4
|z|
µ|z|2 + U
]
2
:= tr e
H0
Semiclassics suggest3,4:
U
H = HW = µn̂ + n̂(n̂ + 1) (Up to a constant)
2
Exact calculation suggests:
U 2
0
H = µn̂ + n̂
2
Original Hamiltonian:
0
H=
U
µn̂ + n̂(n̂
2
1)
3 Kochetov, J. Phys. A. 31, 4473
(1998)
For spin-path integral:
4 M. Stone et al., J. Math. Phys. (N.Y.) 41, 8025 (2000)
How to “fix” the path integral
Z=
Z
2
D ze
R
0
d
2
2
⇤
2
U
|z|
(|z|
1)]
z
ż
µ|z|
+
[
2
Prescription:
Replace operator n in
2
|z|
Hamiltonian by
Path integral used previously:
⇥
Z :=
Z
2
D e
R
0
d
4
⇤
2
U
|z|
z
ż
µ|z|
+
[
]
2
How to “fix” the path integral
Z=
Z
2
D ze
R
0
d
2
2
⇤
2
U
|z|
(|z|
1)]
z
ż
µ|z|
+
[
2
Prescription:
Replace operator n in
2
|z|
Hamiltonian by
Equality when doing our exact calculation.
Path integral used previously:
⇥
Z :=
Z
2
D e
R
0
d
4
⇤
2
U
|z|
z
ż
µ|z|
+
[
]
2
How to “fix” the path integral
Z=
Z
2
D ze
R
0
d
2
2
⇤
2
U
|z|
(|z|
1)]
z
ż
µ|z|
+
[
2
Prescription:
Replace operator n in
2
|z|
Hamiltonian by
Equality when doing our exact calculation.
Drawbacks:
1. No a priori reason to suspect this.
2. Semiclassics give the wrong result.
Path integral used previously:
⇥
Z :=
Z
2
D e
R
0
d
4
⇤
2
U
|z|
z
ż
µ|z|
+
[
]
2
The calculation works when the
Hamiltonian is a linear sum of
generators.
Same problem comes up in the spin coherent state path
integral
e.g.
H = Sz2
for spin-1
Everything works out for Hamiltonians that are linear in
generators of the algebra
e.g.
Hspin = ASx + BSy + CSz + D
†
†
HHO = E + F a + F a + Ga a
Therefore, it always works for spin-1/2.
Conclusion
Path integral breaks down by the exact calculation shown.
This extends to other models (spin path integrals, general
Bose Hubbard, etc.).
It works when the Hamiltonian is a linear sum of Lie
algebra generators.
Agreement can be achieved by naively replacing the
2
|z|
n
operator by
in the path integral.
In the time discretized path integral (before any continuity
assumption), everything works out OK.