Download Many Particle Systems

Many Particle Systems
• can write down the Schrodinger Equation for a many particle system
H ( x1 , x2  xn )  i
 ( x1 , x2  xn )
t
• with xi being the coordinate of particle i (r if 3D)
• the Hamiltonian has kinetic and potential energy
1 2
1 2
H   (

)  V ( x1 , x2  xn )
2m1 x 2
2m2 x 2
2
• if only two particles and V just depends on separation then can treat as
“one” particle and use reduced mass (ala classical mech. or H atom)
• in QM, H does not depend on the labeling. And so if any i  j and j
i, you get the same observables or state this as (for 2 particles)
H(1,2)=H(2,1)
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Exchange Operator
• for now keep using 2 particle as example. Use 1,2 for both space coordinates and
quantum states (like spin)
P12u (1,2)  u ( 2,1)
• can define the exchange operator
P12 ( P12u (1,2))  P12u ( 2,1)  u (1,2)
•
 eigenvalue s :   1
eigenvalues of H do not depend on 1,2, implies that P12 is a constant of motion
H , P1 2   0 
P1 2
0
t
• can then define symmetric and antisymmetric states. If start out in an eigenstate,
then stays in it at future times N = normalization. so for 2 and 3 particle systems
S 
1
( (1,2)   ( 2,1))
N
A 
1
( (1,2)  ( 2,1))
N
1
( (1,2,3)   ( 2,1,3)   ( 2,3,1)   (3,2,1)   (3,1,2)   (1,3,2))
N
1

( (1,2,3)   ( 2,1,3)   ( 2,3,1)   (3,2,1)   (3,1,2)   (1,3,2))
N
S 
A
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Schrod. Eq. For 2 Particles
• have kinetic energy term for both electrons (1+2)
12 T  2m 22 T  VT T  ET T


 
VT  V ( r1 )  V ( r2 )  V12 ( r1  r2 )

2
2m
2
• let V12 be 0 for now (can still make sym/antisym in any case)
• easy to show: can then separate variables and the wavefunction is:
 


 T ( r1 , r2 )   ( r1 ) ( r2 )
ET  E1  E2
• where these are (identical) single particle wavefunctions (i.e.
Hydrogen, infinite box)
• define format. 1 (2) is particle 1’s (2’s) position and ,b are the
quantum numbers for that eigenfunction
  (1)
  (2)
 b (1)
P460 - many particles
 b (2)
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Identical Particles
• Particles are represented by wave packets. If packet A has mass = .511
MeV, spin=1/2, charge= -1, then it is an electron
A
• any wave packet with this feature is indistinguishable
T0
t1>t0
t2>t1
• can’t tell the “blue” from the “magenta” packet after they overlap
• an ensemble (1 or more) of spin ½,3/2...particles (Fermions) have
antisymmetric wavefunctions while spin 0,1,2...particles (Bosons) are
described by symmetric. Note also holds for 1 particle (remember
rotating spin ½ by 360 degrees. gave phase of -1)
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Identical Particles
• Create wave function for 2 particles
 T    (1) b (2)  T   b (1)  (2)  E  E
• the 2 ways of making the wavefunction are degenerate--have the same
energy--and can use any linear combination of the wavefunctions
• Want to have a wavefunction whose probability (that is all measured
quantities) is the same if 1 and 2 are “flipped”
*
|  T |2   * (1) b
( 2)  (1) b ( 2)
*
1  2  * ( 2) b
(1)  ( 2) b (1)
• These are NOT the same. Instead use linear combinations (as
degenerate). Have a symmetric and an antisymmetric combination
S 
A 
1
2
1
2
[  (1) b ( 2)    ( 2) b (1)]
[  (1) b ( 2)    ( 2) b (1)]
 S (1  2)   S
|  |2 unchanged
 A (1  2)   A
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2 Identical Particles in a Box
• Create wave function for 2 particles in a box
  cos nax or sin nax
quantum states :   n  1, b  n  2
 S  A[cos ax sin 2ax  cos ax sin 2ax ]
1
 A  A[cos ax sin
1
2
2x2
a
2
 cos
x2
a
1
sin
2x1
a
]
• the antisymmetric term = 0 if either both particles are in the same
quantum state (Pauli exclusion) OR if x1 = x2
• suppression of ANTI when 2 particles are close to each other.
Enhancement of SYM when two particles are close to each other
• this gives different values for the average separation <|x2 –x1 |> and so
different values for the added term in the energy that depend on
particle separation (like e-e repulsion)….or different energy levels for
the ANTI and SYM spatial wave functions (the degeneracy is broken)
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Particles in a Box and Spin
• Have spatial wave function for 2 particles in a box which are either
symmetric or antsymmetric
• there is also the spin. assume s= ½
 s ( S z  1)  
 s ( S z  0) 
1
(    )
2
 s ( S z  1)  
 A ( S z  0) 
1
(    )
2
S=1
S=0
• as Fermions need totally antisymmetric:
spatial Asymmetric + spin Symmetric (S=1)
spatial symmetric + spin Antisymmetric (S=0)
• if Boson, need totally symmetric and so symmetric*symmetric or
antisymmetric*antisymmetric
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Multiparticle eigenstates
• Need an antisymmetric wave function (1,2,3 are positions; i,j,k…are
quantum states). Can make using determinant:
 (1,2,3....n) 

i
(1) j ( 2) k (3).....
a llterms
  n!
n 3
terms / antisymmet rical
1
6
 a (1)
 b (1)
 c (1)
 a ( 2)
 b ( 2)
 c ( 2)
 a (3)
 b (3)
 c (3)
• while it is properly antisymmetric for any 1j. For atoms,
practically only need to worry about valence effects. Solids, different
terms lead to energy bands
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Density of States
• any system determine density of states D=dN/dE
• can do for a gas of uninteracting but overlapping identical particles
• density of states the same for Bosons or Fermions but how they are
filled (the probability) and so average energy, etc will be different
(quantum statistics – do in 461)
• for Fermions (i.e. electrons), Pauli exclusion holds and so particles fill
up lower states
• at T=0 fill up states up to Fermi Energy EF. Fermi energy depends on
density as gives total number of particles available for filling up states
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2014:Skip rest of Density of States
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Density of States II
• # of available states (“nodes”) for any wavelength
• wavelength  momentum  energy
• “standing wave” counting often holds:often called “gas” but can be
solid/liquid. Solve Scrd. Eq. In 1D
2L
 
d
n
0
L
  a  0
dx
n
2
2
 ( x  0)   ( x  L)  0
n 2
L
2 L
 kn 

 n  k 
L
n

2
• go to 3D. ni>0 and look at 1/8 of sphere
kx 
n x
L
ky 
n y
L
kz 
n z
L
1 4n 3
# nodes 
 Degeneracy
i.e. 2s+1
8 3
1 4
( 2 L)3

 Deg 
take derivative
8 3
3
4V
 N ( )d 
 ( Deg )  d
4

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Density of States III
• convert to momentum
p
h


hn
2L
D( p )dp 
# nodes 
1 4
8 3
2L 3 2
(
2
s

1
)(
2
h ) p dp

• convert to energy depends on kinematics
E  pc
• non-relativistic
p2
E
2m
relativistic
dE  cdp
D ( E )dE 

( 2 s  1)( 2 hLp )3

2
( 2 s  1)( 2hcL )3 E 2 dE
8V 2
E dE
3 3
c h
( for  )
p
dp
m

2L 3
m
D ( E ) dE 
( 2 s  1)(
) 2mE
dE
2
h
2mE

2L 3

( 2 s  1)(
)
2m 3 / 2 E 1/ 2 dE
2
h

dE 
P460 - many particles
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Density of States IV
• can find Fermi energy (assume T=0). Find number of states (n)
Nonrelativistic n  D ( E )dE    2 L  2m E dE
EF


0
0
 2L 
n 

 h 
E
3/ 2
F
3
EF


 h 
3/ 2
1/ 2
3
3/ 2
2m 3 / 2 E F
3h 3
h2  3

 EF 



3/ 2
8m  
8 2 m
2

2/3
• compare to lowest energy of particle in a box of size a with 2a
h
(good way of remembering EF)
E 
2
8ma 2
• and this in some sense gives the condition on overlapping (particle
indistinguishability). two identical particles in the same quantum state
have

separation 
P460 - many particles
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