### Symmetry requirements for identical particles

#### Problem:

The Hamiltonian for a system of two non-interacting particles is H = H1 + H2.   Particle 1 is in the normalized eigenstate |ψ1> of H1 and particle 2 is in the normalized eigenstate |ψ2> of H2.  Construct the two-particle eigenstates of H.  Does it depend on the nature of particles 1 and 2?

Solution:

• Concepts:
Identical particles
• Reasoning:
For distinguishable particles, we do not have to worry about symmetrization requirements.
For identical bosons, the state must be symmetric under the exchange of the two particles.
For identical fermions, the state must be anti-symmetric under the exchange of the two particles.
• Details of the calculation:
distinguishable particles:         |ψ> = |ψ1(1)>⊗|ψ2(2)> = |ψ12>         (notation)
identical bosons:                     |ψ> = 2(|ψ12> + |ψ21>)
identical fermions:                  |ψ> = 2(|ψ12> - |ψ21>)
The state of a single particle includes all its degrees of freedom.  It is determined by the measurement of a complete set of commuting observables for the single particle.  For the electron in the hydrogen atom, for example, ψi = ψn,l,m,ms.
We can also write the two-particle state in terms of wave functions.
distinguishable particles:         ψ(r1,r2) = ψ1(r12(r2)
identical bosons:                     ψ(r1,r2) = 21(r12(r2) + ψ2(r11(r2))
identical fermions:                  ψ(r1,r2) = 21(r12(r2) - ψ2(r11(r2))

#### Problem:

Suppose you have two particles, one in single particle states state Φ1 and Φ2, respectively.
Assuming that Φ1 and Φ2 are orthonormal, construct a two-particle wave function assuming that the particles are
(a)  distinguishable,
(b)  bosons, and
(c)  fermions.

Solution:

• Concepts:
Indistinguishable particles
• Reasoning:
For distinguishable particles, we do not have to worry about symmetrization requirements.
The two-particle wave function of indistinguishable bosons must be symmetric under the exchange of the two particles, and the two-particle wave function of indistinguishable fermions must be anti-symmetric under the exchange of the two particles.
• Details of the calculation:
(a)  Since the particles are distinguishable, I assume we know which particle is in state Φ1 and which particle is in state Φ2,
We then have  ψ(1,2) =  Φ1(1)Φ2(2)  or ψ(1,2) = Φ1(2)Φ2(1) depending on the information we have about the particles.
(b)  ψ(1,2) = 21(1)Φ2(2) + Φ1(2)Φ2(1))
(c)  ψ(1,2) = 21(1)Φ2(2) - Φ1(2)Φ2(1))
The subscript i is a placeholder for the quantum numbers that specify the state Φi.
The number in parenthesis denotes the particle.

#### Problem:

Consider the center of mass system of two interacting identical fermions with spin ½.
(a)  What is the consequence of the Pauli exclusion principle on the two-particle wave function?
(b)  Let S1 and S2 be the spin operators of the two individual neutrons.
Show that the operators  P± = ½ ± ¼ ± S1∙S22  are the projection operators of the triplet states and the singlet states of the spin wave functions.
(c)  Using the Pauli exclusion principle and the symmetry properties of the spin and relative orbital angular momentum L, find the allowed values of L for any bound triplet state of the two-particle system.
(d)  Again using the Pauli exclusion principle and the symmetry properties of the space coordinate, show that the particles in a triplet state can never scatter through an angle of 90 degrees in their center of mass system.

Solution:

• Concepts:
Indistinguishable particles, identical fermions
• Reasoning:
A state of many identical particles is totally antisymmetric if they are fermions.
• Details of the calculation:
(a)  The total state vector must be anti-symmetric under the exchange of the two particles.  If the two particles are in the singlet state, then the space wave function is symmetric, if they are in the triplet state, then the space wave function is anti-symmetric.
(b)  A set of basis vectors for the spin space are {|S,MS> = |1,1>, |1,0>, |1,-1>, |0,0>}.
S1S2 = ½(S2 - S12 - S22).
S2|S,MS> = S(S + 1)ħ2|S,MS>.
S12|S,MS> = ¾ħ2|S,MS>.
S22|S,MS> = ¾ħ2|S,MS>.
P±|S,MS> = (½ ± ¼ ± (½(S(S + 1)/ħ2 - ¾ - ¾))|S,MS>.
For S = 0, the singlet state:
P+|0,0> = (½ + ¼ - ¾)|0,0> = 0.
P-|0,0> = (½ - ¼ + ¾)|0,0> = |0,0>.
For S = 1, the triplet state:
P+|1,MS> = (½ + ¼ + 1 - ¾)|1,MS > = |1,MS >.
P-|1,MS> = (½ - ¼ - 1  + ¾)|1,MS > = 0.
Any state vector for the system can be written as a linear combination of the basis vectors.  Therefore P+ is the projector onto the triplet state and P- is the projector onto the singlet state.
(c)  The relative motion of two interacting particles in their CM frame is treated by treating the motion of a fictitious particle of reduced mass m in an external potential U(r).  Here r points from particle 2 to particle 1, r = r1-r2.
Triplet state: The space wave function is anti-symmetic,
ψ(r12) = -ψ(r21) or ψ(r) = -ψ(-r).
We therefore need l = odd.  (Property of the spherical harmonics.)
(d)  For a scattering problem we have Φk(r,θ) = eikz + fk(θ)(eikr)/r, σk(θ) = |fk(θ)|2.
For identical fermions the state vector must be anti-symmetric.  For the triplet state the spin function is symmetric, so the space function must be anti-symmetric under exchange.  We have
Φk(r,θ) = eikz - e-ikz + [fk(θ) - fk(π-θ)](eikr)/r.
We therefore have σk(θ) = |fk(θ) - fk(π-θ)|2.  σk(θ) = 0 if θ = π/2.

#### Problem:

Consider a coupled pair of one-dimensional, distinguishable, simple harmonic oscillators with equal masses, equal individual potentials U(x1) = ½Cx12, U(x2) = ½Cx22, C > 0, and a coupling potential Uc(x1,x2) = ½k(x2-x1)2, k > 0.
(a)  Separate the Hamiltonian in center of mass and relative variables R = ½(x1+x2) and r = (x2-x1).
(b)  Show that the total eigenfunction can be written as a product of two functions and determine the energy eigenvalues of the coupled system of distinguishable particles.
(c)  Now assume that the particles are indistinguishable so that they are either Bosons or Fermions.  Using the symmetry properties of the product functions determined in part (b), determine which of the energy levels found in part (b) are associated with each type of particle.
(d)  Show that for Fermions with their spins aligned the probability of finding the two particles at at the same position is zero.
(Note: In this problem, you may use your knowledge of the wave functions and eigenvalues of the simple harmonic oscillator without derivation or proof.)

Solution:

• Concepts:
Coupled Harmonic oscillators, indistinguishable particles
• Reasoning:
We have two interacting identical particles in a harmonic oscillator potential.
• Details of the calculation:
(a)  T = ½m[(dx1/dt)2 +  (dx2/dt)2]
= ¼m[(dx1/dt + dx2/dt)2 + (dx1/dt - dx2/dt)2]
= m(dR/dt)2 + ¼m(dr/dt)2 = mV2 + ¼mv2.
U = ½Cx12 + ½Cx22 + ½k(x2 - x1)2
= ¼C[(x1 + x2)2 + ¼C(x1 - x2)2] + ½k(x2 - x1)2
CR2 + ¼Cr2 + ½kr2.
L = T - U

P = ∂L/∂V = 2mV, p = ∂L/∂v = ½mv.
H = P2/(2M) + p2/(2μ) + ½AR2 + ½Br2.
Here M = 2m, μ = m/2, A = 2C and B = C/2 + k.
To make the transition to Quantum Mechanics we let the canonical variables R and P and the canonical variables r and p become operators.
Hψ(R,r) = [(-ħ2/(2M))(∂2/∂R2) + (-ħ2/(2μ))(∂2/∂r2) + ½AR2 + ½Br2]ψ(R,r) = Eψ(R,r)
is the eigenvalue equation for H.
Separation of variables is possible.
Let ψ(R,r) = χ(R)Φ(r).
Then
[(-ħ2/(2M))(∂2/∂R2)+ ½AR2]χ(R) = ERχ(R),  ER = (nR + ½)ħωR, ωR2 = A/M.
[(-ħ2/(2μ))(∂2/∂r2) + ½Br2]Φ(r) = ErΦ(r), Er = (nr + ½)ħωr, ωr2 = B/μ.
The solutions are the solutions of the 1D harmonic oscillator.
E = ER + Er, nR = 0, 1, 2, ...,  nr = 0, 1, 2, ...  .
(c)  For identical bosons the total wave function must be symmetric under the exchange of the two particles.
Under exchange R --> R, r --> -r.
Assume the spin function is symmetric, as it must be for spin 0 bosons.  Φnr(r) is symmetric if nr = even.  The allowed energy levels are E = ER + Er, nR = 0, 1, 2, ...,  nr = even.

For identical fermions the total wave function must be antisymmetric under the exchange of the two particles.
Assume the spin function is antisymmetric.  Φnr(r) is symmetric if nr = even.  The allowed energy levels are E = ER + Er, nR = 0, 1, 2, ...,  nr = even.
Assume the spin function is symmetric.  Φnr(r) is antisymmetric if nr = odd.  The allowed energy levels are E = ER + Er, nR = 0, 1, 2, ...,  nr = odd.
(d)  If the spins are aligned, the spin function is symmetric and Φnr(r) is antisymmetric.  Φnr(0) is 0 if nr = odd.  r = 0 --> x1 = x2.  The probability of finding the two particles at at the same position is zero.

#### Problem:

Two identical spin-½ fermions are placed in the one-dimensional harmonic potential
U(x) = ½mω2x2,
where m is the mass of the fermion and ω its angular frequency.
(a)  Find the energies of the ground and first excited states of this two-fermion system.  Express the eigenstates corresponding to these two energy levels in terms of harmonic oscillator wave functions and spin states.
(b)  Calculate the square of the separation of the two fermions,
<(x1 - x2)2> = <(x12 + x22 - 2x1x2)>
for the lowest energy state of the two-fermion system.
(c)  Repeat the calculations for the first excited states.

Hint:
x1 =  (a0/√2)(a1 + a1),   x2 =  (a0/√2)(a2 + a2),   a0 = (ħ/(mω))½.

Solution:

• Concepts:
Indistinguishable particles
• Reasoning:
We have two non-interacting identical particles in a harmonic oscillator potential.
• Details of the calculation:
(a)  Denote the space-eigenstates of H by {|n>}, n = 0, 1, 2, ... .
Let X0 denote the singlet spin state of two spin ½ particles and X1 a linear combination of the triplet states.
X0 = 2 (|+-> - |-+>),  X1 = a|++> + b 2 (|+-> + |-+>) + c|-->, a, b, c arbitrary constants.

The ground state of two spin ½ fermions is |00>⨂X0.  The space function is symmetric so the spin function must be anti-symmetric.  The energy of the ground state is ħω.
The energy eigenvalue of the first excited state is degenerate.  The energy is 2ħω.
The eigenstate is a linear combination of 2(|01> + |10>)⨂X0 and  2(|01> - |10>)⨂X1.

(b)  The position operator does not operate on the spin.
<x12> = (a02/2)<00|a1a1 + a1a1 + a1a1 + a1a1|00> = (a02/2)<00| a1a1 |00> = (a02/2).
Similarly <x22> = (a02/2).
<x1x2> = (a02/2)<00|a1a2 + a1a2 + a1a2 + a1a2|00> = 0.
So <(x1 - x2)2> = a02.

(c)  For 2(|01> - |10>)⨂X1 (triplet state)
<x12> = (a02/4)((<01| - <10|)a1a1 + a1a1 + a1a1 + a1a1(|01> - |10>))
=  (a02/4)(<01|a1a1|01> + <10|a1a1|10>  + <10|a1a1|10>) = (a02/4)(1 + 2 + 1) = a02.
Similarly <x22> = a02.
<x1x2> = (a02/4)((<01| - <10|)a1a2 + a1a2 + a1a2 + a1a2(|01> - |10>))
= -(a02/4)(<01|a1a2|10> + <10|a1a|01>) = -(a02/4)(1 + 1) = -(a02/2).
So <(x1-x2)2> = a02 + a02 -2(-(a02/2)) = 3a02.

For 2(|01-> + |10>)⨂X0 (singlet state) we obtain the same result as for the ground state proceeding in the same way.