A low temperature expansion for matrix quantum mechanics
Received: October
A low temperature expansion for matrix quantum
YingHsuan Lin 0 1 3 4
ShuHeng Shao 0 1 2 4
Yifan Wang 0 1 2 4
Xi Yin 0 1 3 4
0 77 Massachusetts Ave. , Cambridge, MA 02139 , U.S.A
1 17 Oxford St. , Cambridge, MA 02138 , U.S.A
2 Center for Theoretical Physics, Massachusetts Institute of Technology
3 Jefferson Physical Laboratory, Harvard University
4 Open Access , c The Authors
We analyze solutions to looptruncated SchwingerDyson equations in massless N = 2 and N = 4 WessZumino matrix quantum mechanics at finite temperature, where conventional perturbation theory breaks down due to IR divergences. We find a rather intricate low temperature expansion that involves fractional power scaling in the temperature, based on a consistent soft collinear approximation. We conjecture that at least in the N = 4 matrix quantum mechanics, such scaling behavior holds to all perturbative orders in the 1/N expansion. We discuss some preliminary results in analyzing the gauged supersymmetric quantum mechanics using SchwingerDyson equations, and comment on the connection to metastable microstates of black holes in the holographic dual of BFSS matrix quantum mechanics.
matrix; quantum; Supersymmetric gauge theory; Gaugegravity correspondence; M(atrix) The

mechanics
Contents
1 Introduction 2
The action
2.2 SchwingerDyson equations
N = 2 SUSY Ward identities
3 The low temperature limit
A softcollinear approximation
Continuum limit
Free energy
Higher loop corrections
Action, SchwingerDyson equations, and free energy
5 Towards the low temperature expansion of BFSS
Higher loops
N = 2 gaugefixed BFSS
The N = 2 ghost determinant
5.3 The free energy
N = 4 gaugefixing
5.5 Phases of BFSS
6 The high temperature limit SchwingerDyson equations The free energy
6.3 Size of the wave function
7 Further discussions 7.1 More supersymmetric gauges mensions
A Hawking decay rate of D0branes
A.1 Greybody factor
A.2 Partial wave summation
A.3 Decay rate
B.1 Superspace
B.2 Vector multiplet
Connections
Supersymmetry transformations
Field strengths
B.3 Matter multiplet
C 1D N = 4 SUSY
C.1 Superspace and convention
C.2 Chiral multiplet
C.3 Vector multiplet
N = 4 SUSY Ward identities
D Computation of xn
Computation of the free energy
BFSS G.1 Subleading corrections to the selfenergies G.2 Free energy in the high temperature limit
G High temperature analysis of oneloop truncated N
= 2 gaugefixed
Introduction
limits where the Monte Carlo computation becomes costly.
supersymmetries.
N units of D0brane charge [25]
C1 = f 1
f
c0gsN ls7
1 + A
c0 = 603, A = 1 r7
0
The Hawking temperature of the black hole is
TH =
(gY2MN )1/2
(gY2MN )1/3
low temperature limit.
infinity (exponentially) in the infinity N limit.
probe D0brane in the background (1.1)
SD0 = TD0
dt f 1A1/2p1 + f A2r2
dt f 1 1 + A
namically stable. The latter is expected when r0
the decay rate is exponentially suppressed.
N 1/9lP . This is also the regime where
using SchwingerDyson equations.
This is the approach of [2628], where the authors
T /(gY2MN )1/3
results in the N
the cubic superpotential W
We refer to such a theory as an
of the selfenergies of all other modes.
hr = sign(r)
sign(n)Cn +
sign(n)Cn n
sign(r)Cr + O(1),
where Cn = P
7/5 + O(2).
meanfield approximation [26]. The result is
F = const 2 3
N 2Nf 6/5 + O(9/5).
free energy interpolates between T 9/5 and T 6/5.
and Model II, the N
is left for the future.
= 4 WessZumino
matrix theory is analyzed in appendix G.
The action
real superfields1
interacting through a cubic superpotential
a = a + ia + if a2,
W = 6c
After integrating out the fermionic coordiantes, the action is3
S =
1The i in front of f gives the right sign for the kinetic term of f .
{Q, } = i f + i,
[Q, f ] = i .
SchwingerDyson equations
nonperturbatively.
We will make a
equations, and then discuss the validity of such truncations.
f a =
nZ
rZ+ 12
nZ
S = Tr
1 X 2ir a
1 X fanfn
X abcfankbnck +
X abc ars,r,s .
b c
Let us denote the exact propagators by
hanbmi nabn,m,
h,r,si igrabr,s,
a b
= 1 +
1 X knk,
2 X kgrk.
2 X k nk +
2 X grgnr,
tional perturbation theory.
n ( 2n )2 + n
written in terms of the selfenergies, are
hr =
2 X
2 X
h( 2k )2 + ki [1 + nk]
2 X
10
the fermion propagator gr, respectively.
to the SchwingerDyson equations (2.12).
continuous, we write the exact twopoint functions (2.8) as
If SUSY is not spontaneously broken, then
In momentum space, (2.14) and (2.15) read
hf (p)f (p)i =
p + h(p)
11
hr = s
1
The low temperature limit
A softcollinear approximation
numbers n, r are assumed to be O(1))
modes as hard and zero modes as soft.
12
hr =
+ Cr,
According to the scaling ansatz (3.1),
X k nk + X grgnr ,
k6=0
k6=0,n
2 X kgrk,
1 X kk
k6=0
k6=0
0 1 + n 4/5,
2n Bn 3/5,
hr = sign(r)
13
sn(1 + xn),
2n sn(1 + xn)1,
sn sign(n)
directly from the SchwingerDyson equations, and find
xn =
+ O(6/5) 3/5,
in accordance with (2.16).
n6=0 s2n
in (3.2), and we find
1 X
n6=0 42n2 s2n(1 + xn)2 + O(3/5) =
For sn and hr, let us write
sn = sign(n)
hr = sign(r)
+ s(n1) + s(n2) + O(8/5),
+ h(r1) + h(r2) + O(8/5),
g
, Ck
X sign(`)sign(k `)
`6=0,k
14
h(r1), and h(r2) can be found in appendix E. The results are
s(n1) =
h(r1) =
Cn + O(1),
+ sign(n)
22n2 + O(8/5),
k6=0,2n
82 Cn2 42 k6=0,n
X ()ksign n 2
82 Cr2 42 k6=0
+ sign(r)
k6=0
k6=0,n
X sign(k)sign(r k)xk + O(8/5).
from which we obtain
X sn2
X hr2 = g
n6=0
16
3 304 + O(16/5),
52 57 + O(2).
Continuum limit
X sign(k)sign(n k)xk
15
of the holonomy matrix along the thermal circle is involved.
Free energy
truncated SchwingerDyson equations:
S0 = Tr
F = F0 + hS2 S0i0 2 hS32i0.
h i0
DDDf eS0 .
action, i.e.,
1 X ln n 2 n
1 X ln n +
1 X ln gr2,
for the nonzero modes can be written as
(p) =   1 +
=   1 +
k=1 2k p + O(2)
at x = 0, taking p 0 gives
hS2 S0i0 =
1 X
1 X mk m+k +
1 X grgsr+s.
1 X( n 1)
gr 1 ,
hS2 S0i0 = 1 22
n
n6=0
n6=0
n
X n
X n
(0)3 + O(9/5),
(0)3 + O(9/5).
F = const 4 3
N 2Nf 6/5 + O(9/5),
where we have restored the overall N 2Nf factor.
We have also
convention of [26]).
Higher loop corrections
the fermion propagator gr, respectively.
term6
1 +
3 1 +
scaling solution in the low temperature limit.
18
The YangMills action in the WZ gauge is
SYM = 2
2 D0 2 i[Xi, ]
2 D0XiD0Xi 4 [Xi, Xj][Xi, Xj] ,
component fields to be
= Xi + 21 i,
= A0 i 2
= + 2 2
[A0, ] 2i i[Xi, ] +
2 [d, ] 2 [, ]
Matter multiplet
transformations are
= + i + if 2.
= if + i,
The super gauge transformation is
Written in terms of the component fields, we have
The action we will consider is
SM =
4 aa
and write them in terms of the selfenergies as
sn = ( 2n )2 + nG
vector :
matter :
ghost :
ar =
2n =
un =
The oneloop truncated SchwingerDyson equations are
3 X
r ( 2r + rM )( 2(nr) + nMr)
r ( 2r + rG)( 2(nr) + nGr)
rV = 3 2r X
+ 2 1 + (2n)2nM ,
8 1
2r + rM ,
2 X
n (1 + (2n)2nM)((2(rn))3 + rVn)
4 X
nG = 1 2n 2 X
7 X
r ((2r)3 + rV )(2(nr) + nMr)
4 X
n (2n + 2nVn)2
F0 = 2 log
X log l2 + X log ar 2 l
l r
7 X log l2 + 7 X log gr
7 X log l2 + X log sl 2 X log tr + X log ul,
l6=0
l2 1 + X
7 X
l6=0
tr 1
X (ul 1) ,
i
2 X 2r 2s
aras +
7 X 2
1 X l2m2 + 14 X l2m2
4 X
14 X
7 X
2 X
7 X g
1 X tr +
l2trts ultras +
1 X
sltras
and quartic couplings.
1D N = 4 SUSY
43
Superspace and convention
= = (iI, ~).
Q = + ,
D = .
The only nontrivial anticommutators are
= , = ,
In particular,
= =
= = = .
= 21 , =
= 2 ,
Chiral multiplet
We can solve this constraint by first introducing
a(y, ) = a(y) +
in the action positive.
The SUSY transformations are
[Q, a] = i2a, {Q, a} =
{Q , } = i2 a, [Q , f a] =
a
L =
Vector multiplet
the following component field expansion
The gauge transformation is
e2V
e2i e2V e2i
The fieldstrength superfield is defined by
The kinetic action for the vector multiplet is then given by
L = Tr
fa(p)f b(p)i =
a
h (p)b(p)i = i p + h(p)
Going to the momentum space, this implies
that is
Similarly, we can consider
0 = h{Q, a( )fb( 0)}i =
2 hf a( )fb( 0)i
In the momentum space it is
Using = , we have
In summary, we have obtained the exact relation
hfb(p)f a(p)i = ip
b
h(p)a(p)i.
h(p) =
1 R is set to zero in the zero mode truncation. We will write
The fieldstrength superfield is
similarly for D. Using
dse2sRAe2(1s)R Z 1
usual WessZumino gauge.
= e2R +
ds e2sR 2 Ae2(1s)R
= e2R + 2
ds e2sR Ae2(1s)R
du e2uRAe2(su)RAe2(1s)R,
47
we then expand
dse2sRAe2(1s)R Z 1
We can simplify the above equation by using the identity
The zero mode terms in the Lagrangian can then be written as
L =
3 gY2M
dse2sRAe2(1s)R Z 1
) + ] A+AA+A,
k6=0,n
nk
1+xnk +X
k sksnk 1+xk
r hrhnr 2n0
1 + O(6/5)i .
r 20h1+O(6/5)i
We have used
Computation of xn
(SL)(SL) = 1 (
Using (3.5), (3.7) and (3.8), the equation for xn is
0 2n + sn(1 + xn)1 + An
#1
(1 + xn)2 =
An =
Bn =
The SchwingerDyson equation for xn can thus be written as
2n (1 + xn)2 1
= An (1 + xn)2Bn + O(11/5),
48
k6=0,n ksksnk
1 + xnk
1 + xk
(1 + xn)2
(1 + xk)(1 + xnk)
1 + xnk + X
k6=0,n sksnk 1 + xk
r hrhnr
= (1 + xn)2 1 + O(3/5).
xn)2Bn 1. Write
sk = sign(k)
+ s(k1), hr = sign(r)
where s
(k1), h(r1) are of order 2/5.
We have
hr =
1 X
n6=0 ksk(1 + xk)hrk
s(k1) =
h(r1) =
`6=0,k
The equation for xn can be written as
(1 + xn)2 1
2
k6=0,n
k6=0,n
k6=0,n
0 X sign(k)sign(n k)
1 + xnk
1 + xk
1 + xk
1
(1 + xn)2
(1 + xk)(1 + xnk)
r 20+O(3/5).
X sign(nk)s(k1)
X sign(nr)h(r1)+
X sign(n k)s(k1)
X sign(n r)h(r1)
k6=0,n
4 `6=0   k6=0,n,`
sign(n k)sign(k `)
unless the r.h.s. vanishes to leading order.
xn =
0 2 an + O(6/5),
k6=0,n
0 = 2n
X sign(k)sign(n k)
(ank an) +
`6=0   k6=0,n,`
= X 1
`6=0  
an now obeys the equation
sign(n k)sign(k `)
X sign(n r)sign(r `)
k6=0,n
sign(n k)sign(k `)
X sign(n r)sign(r `)
4 sign(n)HN (n 1) +
an = 3
+ 2 X sign(k)ak
k6=0,n n k
HN (m)
k=1 k
an =
xn =
22n2 (1+xn)2 1 =
`6=0   k6=0,n,`
Here we used
that is,
A solution is given by
50
1 X
0 hr 20hr2 +
k6=0,n ksksnk
k6=0 ksk(1 + xk)hrk
1 X
k6=0,n sksnk
0 s2n (An + Bn) + O(8/5),
r hrhnr
1 X
k6=0,n ksksnk
g =
(xk + xnk) + O(8/5).
sn and hr to the relevant order.
sn =
hr =
An =
Bn =
An =
Bn =
+ Bn(1 + xn) + O(8/5)
We can then write
sn = sign(n)g1 +
hr = sign(r)g1 +
X (1)ksign n 2
X n k
X sign(k)sign(n k)
k6=0,n
(xk + xnk) + O(g8).
`6=0,k
C k2
k6=0,n
2g6Cn2
k6=0,n
2sign(n) 7
X (1)ksign n 2
k6=0,2n
k6=0,n
= 1
(ghr)2 = 1 hr
= 1
k6=0
g4sn X sign(k)sign(n k)
(xk + xnk)
k6=0,n
The equations for sn and hr are equivalent to
(gsn)4 = 1 sn
2g4snCn
+ g4AnBns2n + g2sn(An + Bn) + O(g9)
k6=0,2n
+ sign(r)
3 X sign(k)sign(r k)xk + O(g9),
k6=0,n
It follows that
X(gsn)2
n6=0
X(ghr)2 = X(1)`
3 X(1)`sign(`)C ` +
2
`6=0
`6=0
X(1)`C2`
52
sign(r) X sign(k)sign(r k)
k6=0 kskhrk
sign(r)Cr +
s(n2) and h(r2) can then be solved to be
X ()ksign n 2
3 X sign(k)sign(nk)xk +
k6=0
sign(r) X sign(k)sign(r k)
g6
22 22 sign(r) X sign(r k)Ck + sign(k)Crk
k
k6=0,n
k6=0
n6=0, k2
n6=0
X(1)`sign(`) X
k6=0,n
13 g6 + O(g9)
= 1 2 6
where we have used
sign(n k2 )sign(n)
`6=0
sign( 2` k)Ck + sign(k)C 2` k
+sn(1+xn) +Xln
+sn(1+xn)1
X n
2Xln
(0)3 + O(9/5),
X(1)`sign(`) X
`6=0
k6=0, 2`
n6=0
X()nCn2/2 = 3
sign(k)C 2` k
= 3
Computation of the free energy
energy for the trial action S0,
+Xln
= const + g
k6=0, 2`
k6=0
n6=0
n
by the terms in 21 hS32i0.
1
To compute 2 hS32i0, it will be convenient to recall our solutions for the propagators,
(n0) + (n1) + (n2) + O(1),
(n1) 2n
g gs(n2) +
2n + sn(1 + xn)1
(n0) + (n1) + (n2) + O(3),
2 Cn + gs(n2) (1 xn + x2n) + g
n
2 Cn xn ,
= g sign(r) 1+
2 Cr +gh(r2) +g
 
gr(0) +gr(1) +gr(2) +O(2),
gr(0) g sign(r), g(1)
r sign(r)
2 Cr, gr(2) g sign(r) gh(r2) +
  42 C2r g
n =
n
n
Using the above,
1 X
`6=0 n6=0,`
`6=0
1 X
`6=0 n6=0,`
`6=0 r
n6=0
= g
n
rg
X n
(0)3 +O(9/5)
54
X n n 0 r>0
`6=0 r
1 X X grg`r`
2 X ()k +
(1 + 0) 42 n=1 n2 +
X (n0) (n1) 0 r>0
X (n0) (n2) 0 r>0
X gr(2)gr(0) 0 r>0
`6=0
n6=0,`
X gr(0)g`(0)r(`1) + O(9/5)
1 X
`6=0 n6=0,`
2 X X gr(1)g(0) (0)
`r `
`6=0 r
3
where we have used
Subleading corrections to the selfenergies
equations (see appendix B.4), with a, b, c, d defined in (6.2),
4 X
4 X
2 (0V )2
= 3
b2
2 X
m6=0
3 X
2 (0V )2 (0M )2
= 3
b2
0 0
12 a2(d + 1) 1/2 a(d + 1)2
12 X
13 X
m6=0 2m2 = 3 ,
The solution to these algebraic equations are,
(1)M = 2.52, (1)V = 0.32, (1)M = 0.683/2, (1) = 5.53.
0 0 0
Now for nonzero modes,
4 X
4 X
+ r 2r 2(n r)
2(1) 8(1)
2(n r) 2 + 2 + 8
(0V )2
3 X
b2
+ r 2r 2(n r)
Vn (0V )2 (0M )2
0 0
5 X
7 X
(0V )2 (0M )2
3
m6=0
1 X
We have used
2(a 6b)
r r(n r)
= 0,
r r3(n r)
3 2n2 a2 + 6
4(a 6b)
Therefore, from (6.3) and (G.3),
= 5.24183 +
= 4.86502 +
1 b2
Free energy in the high temperature limit
F (0) = f () f
7 X log l2
+ 7log sinh
kinetic energy),
l6=0
l6=0
7 X M
2 l6=0
l
l6=0
l
7 X lM + X
l6=0
7 X
2 l6=0
7 X
l 2 X
r 2r r 2 1 + 0M
l 7 X
Note that to this order
7(7a + 36b)
3/2(a2 + 6)2 a(a2 + 6)
Up to an additive constant,
105b4 2338b2 2220 3/2
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