Density Operator in Statistical Mechanics

Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton

(Date: October 14, 2016)

It may be useful to rewrite ensemble theory (micro canonical, canonical, and grand canonical

ensemble) in a language that is more natural to a quantum-mechanical treatment, namely the

language of the operators and the wavefunctions. As far as the statistics are concerned, this

rewriting of the theory may not seem to introduce any new physical ideas as such; nonetheless, it

provides us with a tool that is highly suited for studying typical quantum systems. And once we

set out to study these systems in detail, we encounter a stream of new, and altogether different,

physical concepts

1. Fundamental

We introduce the density operator . In thermal equilibrium, the expectation value of

a macroscopic observable B ,

]ˆ)(ˆ[ BtTrB ,

has to be time independent. This is only possible if )(ˆ t is time independent:

]ˆ,ˆ[ˆ Ht

i ℏ , (Liouville theorem)

This equation is similar to the Heisenberg’s equation of motion for the Hermitian

operator, except for the sign. Since 0ˆ t

(in thermal equilibrium), 0]ˆ,ˆ[ H . This

equation implies that is a conserved quantity and is a function of the Hamiltonian H .

)ˆ(ˆˆ H .

_______________________________________________________________________

2. Density operator for the canonical ensemble

The density operator for the canonical ensemble is given by

)]ˆ[exp(

)ˆexp()ˆexp(

)(

1ˆ

HTr

HH

ZCC

,

where TkB

1 . H is the Hamiltonian of the system. )(CZ is the partition function and is

defined by

)]ˆ[exp()( HTrZC

We use the basis of energy eigenstate n , where

nnn EH ˆ

Then we get

n

nn

E

n

nn

H

ne

eH

ˆ)ˆexp(

where we use the closure relation. Then the density operator is

n

nnn

n

nn

E

C

C PeZ

n

)(

1ˆ

where nP is a probability and is given by

nE

C

n eZ

P

)(

1

The partition function is

n

E

n

H

n

n

H

C

ne

e

eTrZ

ˆ

ˆ][)(

The average value of operator B

)ˆ()(

1)ˆˆ(

ˆ

ˆ

H

HCCeBTr

eTrBTrB

The average energy is

)(ln)ˆˆ(

CCCZHTrE

which can be rewritten as

n

E

n

Cn

n

H

n

CC

neEZ

eHZ

E

)(

1ˆ)(

1 ˆ.

The energy fluctuation:

22

222

]ˆˆ[]ˆˆ[ HTrHTr

EEE

Representation of the density operator under the basis of r

]2

)'(exp[

)2(

1

)(

1

)'()2

exp()''exp()exp('

)2(

1

)(

1

)''exp(')2

exp()exp('

)2(

1

)(

1

'''exp[)exp(')2(

1

)(

1

'''exp[')(

1'ˆ

2

3

2

3

2

3

3

m

id

Z

m

iidd

Z

i

m

idd

Z

Hi

ddZ

HddZ

C

C

C

C

C

prrpp

ppp

rprppp

rpppp

rppp

rppprppp

rpppprpprr

ℏℏ

ℏℏℏ

ℏℏℏ

ℏℏ

where we use the transformation function

)exp()2(

12/3

rppr ℏℏ

i

rprppr )exp()2(

12/3

*

ℏℏ

i

2ˆ2

1ˆ pm

H (Hamiltonian for the free particle).

)'(' pppp

Here we use the spherical co-ordinate (p co-ordinate) for the integral

]2

'exp[

2

)'

sin()2

exp(

'

4

]cos)'(exp[)2

exp(2

]2

)'(exp[

2

2/3

0

2

00

22

2

ℏ

ℏ

ℏ

ℏ

ℏ

rr

rr

rr

rr

prrpp

mm

pm

ppdp

pi

dpm

pp

m

idI

or

]2

'exp[

2)(

1

]2

'exp[

2

)2(

1

)(

1'ˆ

2

2/3

2

2

2/3

3

ℏℏ

ℏℏ

rr

rrrr

mm

Z

mm

Z

Note that

3

2/3

3

2

0

2

3

2

3

2

2

)2(

]2

exp[4

)2(

]2

exp[

)2(

)2

exp(

)][exp()(

T

V

mV

m

pdpp

V

md

V

m

HTrZ

ℏ

ℏ

ℏ

pp

p

p

where T is the thermal length,

mT

22 ℏ

Then we have

]2

'exp[

1'ˆ

2ℏ

rr

rr

m

V

The diagonal element:

V

1ˆ rr

are just the probabilities for finding a particle at r . The iff-diagonal elements have no

vlassical analogy.

3. Density operator for the Grand canonical ensemble

The density operator for the grand canonical ensemble is given by

)ˆˆ(

),(

1ˆ NH

G

G eZ

where the partition function is given by

][),()ˆˆ( NH

G eTrZ

Here, we use the energy-particle representation. Since 0]ˆ,ˆ[ NH , we have simultaneous

eigenkets such that

)()()(ˆ NENENEH iii ,

)()(ˆ NENNEN ii

Then we get the partition function as

0

N

0 ][

)(N

0 ][

)(

0 ][

))((

0 ][

)ˆˆ(

)ˆˆ(

)()(

][),(

N

CN

N Ni

NE

N Ni

NEN

N Ni

NNE

N Ni

i

NH

i

NH

G

Z

e

ee

e

NEeNE

eTrZ

i

i

i

or

0

)(),(N

CN

N

G ZZ

with

e

The average value of operator B

]ˆ[),(

1)ˆˆ( )ˆˆ( NH

G

GGeBTr

ZBTrB

In the energy-particle representation, we get

0 ][

)(N

0 ][],[

,

)(N

0 ][],[

)ˆˆ(

0 ][

)ˆˆ(

)(ˆ)( ),(

1

)(ˆ)( ),(

1

)()()(ˆ)( ),(

1

)(ˆ)( ),(

1

N Ni

NE

ii

G

N NjNi

ji

NE

ji

G

N NjNi

i

NH

jji

G

N Ni

i

NH

i

GG

i

i

eNEBNEZ

eNEBNEZ

NEeNENEBNEZ

NEeBNEZ

B

If CN

B is the average value of B in the canonical ensemble,

CNCN

Ni

NE

ii BZeNEBNE i

][

)()(ˆ)( .

then we have

0

N )( ),(

1

NCNNC

GG

BZZ

B

The average number:

GG

NHNH

G NZeNTreNTrZ ),(]ˆ[]ˆ[),( )ˆˆ()ˆˆ(

or

GGBGZTkN

),(ln

where G is the grand potential and is defined by

),(ln GBG ZTk

The number fluctuation is evaluated as follows.

GG

NHNH

G NZeNTreNTrZ 22)ˆˆ(22)ˆˆ(22

2

2

ˆ),(]ˆ[]ˆ[),(

leading to

),(),(

1)(ˆ

2

222

G

G

BG

ZZ

TkN

The number fluctuation:

GB

GB

G

GGG

B

G

G

G

G

B

GGG

NTk

ZTk

Z

ZZZ

Tk

Z

Z

Z

ZTk

NNN

),(ln)(

})],([

]),(

[),(

),(

{)(

}]),(

[)],([

1),(

),(

1{)(

ˆˆ

2

2

2

2

2

2

2

22

22

22

or

GBG NTkN

2

4. Entropy and density operator for the canonical ensemble

We show that the entropy in the canonical ensemble, is expressed in terms of the

density operator C as

]ˆlnˆ[ CCBTrkS .

((Proof))

We note that

FH

C eeTrZ ][)(ˆ

where F is the Helmholtz free energy and is given by

)(ln1

)(ln

CCB ZZTkF

Thus we can write the density operator as

HF

C eeˆ

ˆ

Then we have

)ˆ(ˆln HFC

SSTT

EFT

HFkk BCB )(1

)(1

)(ˆln

or

]ˆlnˆ[ˆln CCBCB TrkkS

5. Entropy and density operator for the canonical ensemble

We show that the entropy in the grand canonical ensemble, is expressed in terms of

the density operator G as

]ˆlnˆ[ GGBTrkS

((Proof))

We note that

JNH

G eeTrZ ][)( )ˆˆ(

where G is the grand potential and is given by

)(ln1

)(ln

GGBG ZZTk

Thus we can write the density operator as

)ˆˆ(ˆ NH

G ee G

Then we have

)ˆˆ(ˆln NHGG

SSTT

HFT

NHT

NHkk

G

GBGB

)(1

)(1

)(1

ˆˆˆln

or

]ˆlnˆ[ˆln GGBGB TrkkS

So the entropy is independent of the representation of . The entropy is a genuine

mathematical invariant of a quantum system.

In the actual evaluation of thermodynamic average, the similarity of

)ˆexp(1

ˆ HZC

C ,

to the expression for the time evolution operator

)ˆexp(ˆ tHi

Tℏ

.

has often been used. Thus, by putting

ℏ

it .

One can treat the inverse temperature as an imaginary time variable.

6. Density operator for the microcanonical ensemble

Since the density operator for the microcanonical ensemble is characterized by a

stationary distribution, we have

0]ˆ,ˆ[ HMC

leading to the simultaneous eigenkets of H and MC . Here we use the energy eigenstates

to get the expression of MC .

nnn EEEH ˆ

with

mnmn EE ,

For an isolated system, the principle of a priori probability states that all possible states of

the system have the same probability. Therefore we can write

m

mmmMC EEP

with

0

1

)(

1

EPm

otherwise

for δEEEE m

where the constant Pm can be obtained from the normalization condition

1]ˆ[ MCTr

We define

EEEE

m

EEEE

m

mm

mm

EETr

EVNEW

EVNEE

1] [

),,,(

),,()(

1] [)(

1]ˆ[

EEEE

m

mmMC

m

EETrE

Tr

Then we have

EEEE

m

mmMC

m

EEE

)(

1ˆ

The expectation value of the operator B in the microcanonical ensemble is

]ˆ [)(

1

EEEE

m

mm

m

EEBTrE

B

The average energy:

E

EE

EEHTrE

E

EEEE

m

m

EEEE

m

mm

m

m

)(

1

]ˆ [)(

1

The entropy S is given by

)(ln EkS B

________________________________________________________________________

7. Probability distribution function for the Microcanonical ensemble (classical

mechanics

We define the State density function in the phase space as

),...,,,,...,,(),()( 321321 NN

N pppqqqfpqff

)( Nf

0

1

otherwise

tofor δEEE

Using the property of the Dirac delta function, )( Nf can be rewritten as

EpqHEf N

N )),(()(

where ),( pqH N is the N-particle Hamiltonian. The total number of states for E to

EE

)(),,(),,,()(NN

fdEVENEVENWE

),,( VEN is the density of states (units of erg-1). We define the probability distribution

function

EVEN

EpqHE

EVENW

f N

NN

MC

),,(

)),((

),,,(

)()(

where

1)( NNd

The average value of the physical quantity

)()(N

MC

NN

MCBdB

The average energy:

EpqHdEN

MCN

N

MC )(),(

NNdNN

MC

N

MC )(

The entropy ),,( VENS is defined by

),,,(ln),,( EVENWkVENS B

8. Exmaple-1

R. Kubo, Statistical Mechanics: An Advanced Course with Problems and

Solutions (North-Holland, 1965).

Chapter 2, Problem 2-29 (p.138)

The Hamiltonian of an electron in magnetic field B is given by

Bσ ˆBH

where σ is the Pauli’s spin operator and B stands for the Bohr magneton. Evaluate (i) the

density matrix in the diagonalized representation of z , (ii) the density matrix in the

diagonalized representation of x and (iii) the averages of z in these representations, the z-axis

being taken along the field direction.

((Solution))

The magnetic moment of spin 1/2 is given by

σS

μ ˆˆ2

ˆBB

ℏ

.

The Zeeman energy:

BH ZB ˆˆˆ Bμ

The density operator is given by

)ˆexp()(

1)ˆexp(

)(

1ˆ

ZB

CC

C BZ

HZ

U is the unitary operator

zUx ˆ , zUx ˆ

where

11

11

2

1U ,

11

11

2

1U .

The partition function )(CZ is

)cosh(2)exp()exp()]ˆ[exp()( BBBBTrZ BBBZBC

(a) The density matrix under the basis of { z- ,z }.

B

B

B

zB

B

e

e

B

0

0

)cosh(2

1ˆ

(b) The density matrix under the basis of { x- ,x }.

1)tanh(

)tanh(1

2

1

)cosh(2)sinh(2

)sinh(2)cosh(2

)cosh(4

1

11

11

2

1

0

0

11

11

2

1

)cosh(2

1

ˆˆˆˆ

B

B

BB

BB

B

e

e

B

UU

B

B

BB

BB

B

B

B

B

zx

B

B

(c)

The average of z under the basis of { z- ,z }

)tanh(][ BTr Bzzz

since

B

B

B

B

B

B

zz

B

B

B

B

e

e

B

e

e

B

0

0

)cosh(2

1

10

01

0

0

)cosh(2

1

(c)

The average of z under the basis of { x- ,x }.

)tanh(

]ˆˆˆˆˆˆ[

]ˆˆ[

B

UUUUTr

Tr

B

zz

zzz

Note that

)tanh(1

1)tanh(

2

1

01

10

1)tanh(

)tanh(1

2

1ˆˆˆˆˆˆ

B

B

B

BUUUU

B

B

B

B

zz

and

01

10

11

11

11

11

2

1

11

11

2

1

10

01

11

11

2

1ˆˆˆ UU z

9. Example-1

The entropy is given by

]ˆlnˆ[ CCBTrkS

In the above example, the density operator is given by

B

B

B

CB

B

e

e

B

0

0

)cosh(2

1ˆ

Using the Mathematica the entropy is obtained as

)tanh()]cosh(2ln[

]2

)tanh(1ln[

1)

1

1ln(

1

12

2

22

BBB

B

e

e

eeS

BBB

B

B

B

BB B

B

BB

((Mathematica))

Clear "Global` " ;

11

2 Cosh

Exp 0

0 Exp;

M1 Tr 1 . MatrixLog 1 FullSimplify

Log1

1 22

Log1

21 Tanh

12

M2 Log 2 Cosh Tanh ;

h1 Plot M1 . 1 x, x, 0, 6 ,

PlotStyle Red, Thick ;

h2 Plot M2 . 1 x, x, 0, 6 ,

PlotStyle Blue, Thick ;

h3

Graphics

Text Style "y S kB", Black, Italic, 12 ,

0.5, 0.65 ,

Text Style "x kBT BB", Black, Italic, 12 ,

5.5, 0.05 ;

Show h1, h3

Fig. Entropy vs temperature at the fixed B.

y S kB

x kBT BB

1 2 3 4 5 6

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Show h2, h3

y S kB

x kBT BB

1 2 3 4 5 6

0.1

0.2

0.3

0.4

0.5

0.6

0.7

10. Example-2

R. Kubo, Statistical Mechanics: An Advanced Course with Problems and

Solutions (North-Holland, 1965).

y S kB

x BB

0.5 1.0 1.5 2.0 2.5 3.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Show h2, h3

y S kB

x BB

0.5 1.0 1.5 2.0 2.5 3.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Chapter 2, Problem 2-30 (p.138)

Evaluate the matrix elements of the density matrix

H

He

eTr

ˆ

ˆ][

1ˆ

where

222 ˆ2

1ˆ

2

1ˆ qmpm

H

of a one-dimensional harmonic oscillator in the q representation. Discuss, in particular, the

limiting case of

1

ℏℏ

TkB

Hint: The eigenfunction )(qn for the eigenvalue

ℏ)2

1( nEn

is given by

2/

4/12

!2

)()(

en

Hmq

n

nn

ℏ with q

m

ℏ

The Hermite polynomials )(nH are defined by

uiu

n

n

n eiudue

ed

deH

22

2

22

)2()1()(

Use the last expression in the integral form.

((Solution))

222 ˆ2

1ˆ

2

1ˆ qmpm

H (one-dimensional simple harmonics)

nnnH )2

1(ˆ ℏ

The density matrix in the basis of q

n

nn

nn

nn

ynnxn

ynnxnnn

ynnHnnxyHx

])2

1(exp[

''])2

1(exp[

'')ˆexp()ˆexp(

',

',

',

ℏ

ℏ

The wave function is given by

)()2

exp(!2

124/1

xm

Hxmm

nnx n

n ℏℏℏ

with

]2exp[)2()exp(

)(2

2

ixuuiudux

xHn

n

Then we get

n

nnn

n

ym

Hxm

Hnn

yxmm

ynnxnyHx

)()(])2

1(exp[

!2

1

]2

)(exp[

])2

1(exp[)ˆexp(

222/1

ℏℏℏ

ℏℏ

ℏ

where

)(2exp[)2()2(

])(

exp[1

)()(

22

22

yvxum

ivuiviudvdu

yxmy

mHx

mH

nn

nn

ℏ

ℏℏℏ

Thus we have

]2)(2exp[

)]22

)(exp[

1

)2(!

1)](2exp[

)]22

)(exp[

1)ˆexp(

22

222/1

22

222/1

ℏ

ℏ

ℏ

ℏ

ℏℏ

ℏ

ℏ

ℏℏ

uveyvxum

ivudvdu

yxmm

euvn

yvxum

ivudvdu

yxmmyHx

n

nn

where

])1(

)(2(exp[

)1(]2)(2exp[

2

22

2/1222

ℏ

ℏℏ

ℏℏ

ℏ

ℏ

e

yxexyem

euveyvxum

ivudvdu

((Mathematica)) Proof of the above integral using the Mathematica

______________________________________________________________________________

Then we have

}])sinh(

2)coth()({

2exp[

)sinh(2

])1(

)2(

2

)(exp[

1

])1(

)(2(

2

)(exp[

1)ˆexp(

22

2/1

2

2222

2

22/1

2

2222

2

22/1

ℏℏ

ℏ

ℏℏ

ℏℏ

ℏ

ℏℏ

ℏ

ℏ

ℏ

ℏ

ℏ

ℏ

ℏℏ

ℏ

ℏ

xyyx

m

m

e

yxyexmyxm

e

em

e

yxexyemyxm

e

emyHx

Note that

)}2

coth()()2

tanh(){(2

1

)sinh(

2)coth()( 2222

ℏℏ

ℏℏ yxyx

xyyx {

Thus we have

)}]2

coth()()2

tanh(){(4

exp[

)sinh(2)ˆexp(

22

2/1

ℏℏ

ℏ

ℏℏ

yxyxm

myHx

When 1ℏ ,

ℏℏ )sinh( , 2

)2

tanh( ℏℏ

,

ℏ

ℏ 2)

2coth( .

So we get the approximate form as

)( )2

exp(

])(2

)(8

exp[

2)ˆexp(

22

2

2

22

2/1

2

yxxm

yxm

yxm

myHx

ℏ

ℏℏ

Here we use the definition of the Dirac delta function

)(])(2

1exp[

2lim

2

0yxyx

.

______________________________________________________________________________

11. Example-3

R. Kubo, Statistical Mechanics: An Advanced Course with Problems and

Solutions (North-Holland, 1965).

Chapter 2, Problem 2-31 (p.139)

Evaluate 2q and

2p from the density matrix in the q representation obtained in the

Example-2 for a one-dimensional harmonic oscillator.

((Solution))

)]ˆexp(ˆ[)(

1 22 HxTrZ

x

, )]ˆexp(ˆ[)(

1 22 HpTrZ

p

2/3

2/1

22

2/1

2

22

]2

tanh[2)sinh(2

)]2

tanh(exp[)sinh(2

)ˆexp(

)ˆexp(ˆ)]ˆexp(ˆ[

ℏ

ℏℏℏ

ℏ

ℏℏℏ

mm

dxxm

xm

xHxdxx

xHxxdxHxTr

where

)]2

tanh(exp[)sinh(2

)ˆexp( 2

2/1

ℏ

ℏℏℏx

mmxHx

The partition function:

2/1

2/1

2

2/1

)2

tanh()sinh(2

)]2

tanh(exp[)sinh(2

)ˆexp(

)ˆexp()]ˆ[exp()(

ℏ

ℏ

ℏℏ

ℏ

ℏℏℏ

m

m

dxxmm

xHxdx

xHxdxHTrZ

Then we have

)2

coth(2

]2

tanh[2

)2

tanh(

)]ˆexp(ˆ[)(

1

2/3

2/1

22

ℏℏ

ℏ

ℏ

ℏ

ℏ

m

mm

HxTrZ

x

)2

tanh(

ln2

1

)sinh(2ln

2

1ln

ℏ

ℏ

ℏℏ m

mZ

)2

coth(2

ln

ℏℏ

Z

Since

222

2

1

2

1xmp

m

)2

coth(2

)2

coth(2

222222

ℏℏℏℏ m

mmxmp

((Note))

xHxxi

xxdxdx

xHxxi

xxdxdx

xHxxpxdxdx

xHpxdxHpTr

)ˆexp(""

)"("

)ˆexp(""

""

)ˆexp(""ˆ"

)ˆexp(ˆ)]ˆexp(ˆ[

2

2

2

22

ℏ

ℏ

where

)}]2

coth()"()2

tanh()"{(4

exp[

)sinh(2)ˆexp("

22

2/1

ℏℏ

ℏ

ℏℏ

xxxxm

mxHx

Using the Mathematica we get

)2

coth(2

)]ˆexp(ˆ[)(

1 22

ℏℏm

HpTrZ

p

((Mathematica))

______________________________________________________________________________