Many Body Theory Problem Set 1

Rudolph J. Magyar

October 23, 2000

In the following problems, I use a unit system so that h̄ = 1 and e = 1. Problem One

a. The equation of motion is

\frac{da(0)}{dt} = −i[a, H]

Substitute

a = \frac{1}{\sqrt{2}} [\frac{1}{\sqrt{mω}} p̂ − i\sqrt{mω}x̂]

a^{†} = \frac{1}{\sqrt{2}} [\frac{1}{\sqrt{mω}} p̂ + i\sqrt{mω}x̂]

into

H = ω(a^{†}a + \frac{1}{2})

to get

H = \frac{1}{2m}p̂^{2} + \frac{1}{2}mω^{2}x̂^{2}

using [x̂_{j}, p̂_{k}] = iδ_{jk}. This shows that that a and a^{†} can be considered as an operator factorization of H. Note that these so called creation and annihilation operators satisfy the following commutation relationship,

[a, a^{†}] = 1

Write out the commutator explicitly and commute the a and a^{†} in the first term.

\frac{da(0)}{dt} = −iω(aa^{†}a + \frac{1}{2}a − a^{†}aa − \frac{1}{2}a)

= −iω(a^{†}aa − a^{†}aa + a) = −iωa

Upon integration

a(t) = a(0)e^{−iωt}

b. Given

a^{†}(t) = a^{†}(0)e^{iωt}

And from part a

a^{†}(0) = \frac{1}{\sqrt{2}} [\frac{1}{\sqrt{mω}} p̂(0) + i\sqrt{mω}x̂(0)]

so

a(t) = \frac{1}{\sqrt{2}} [\frac{1}{\sqrt{mω}} p̂(0) − i\sqrt{mω}x̂(0)] e^{−iωt}

= [\frac{1}{\sqrt{2mω}} p̂(t) − i\sqrt{\frac{1}{2}mω}x̂(t)]

a^{†}(t) = \frac{1}{\sqrt{2}} [\frac{1}{\sqrt{mω}} p̂(0) + i\sqrt{mω}x̂(0)] e^{iωt}

= [\frac{1}{\sqrt{2mω}} p̂(t) + i\sqrt{\frac{1}{2}mω}x̂(t)]

Add these

a(t) + a^{†}(t) = \sqrt{\frac{2}{mω}} p̂(t) = \frac{1}{\sqrt{2mω}} p̂(0) [e^{−iωt} + e^{iωt}] + i\sqrt{\frac{mω}{2}} x̂(0) [e^{iωt} − e^{−iωt}]

Solve for p̂

p̂(t) = p̂(0) cos ωt − mωx̂(0) sin ωt

To get x̂, subtract

a^{†}(t) − a(t) = \sqrt{2mω}x̂(t) = \frac{1}{\sqrt{2mω}} p̂(0) [e^{iωt} − e^{−iωt}] + i\sqrt{\frac{mω}{2}} x̂(0) [e^{iωt} + e^{−iωt}]

Solve for x̂

x̂(t) = x̂(0) cos ωt + \frac{1}{mω} p̂(0) sin ωt

c. We know that ⟨q⟩ and ⟨p⟩ obey the Erenfest theorem and we can write:

\frac{⟨q⟩}{dt} = \frac{⟨p⟩}{m}
\frac{⟨p⟩}{dt} = −⟨V′⟩.

And they are equivalent to classical equations when ⟨V′⟩ = ⟨V′_{cl}⟩. If dispersion (Δq)^{2} = ⟨q^{2}⟩ − ⟨q⟩^{2} of wave packet is small then we can write:

V′(q) = V′_{cl}(⟨q⟩) + (q − ⟨q⟩)V′′_{cl} + (q − ⟨q⟩)^{2}V′′′_{cl} + ...

We see that V′ = V′_{cl} if all term higher that V′′ are equal to zero. Harmonic oscillator potential has quadratic form and hence quantum and classical equation of motion look similar.

Problem Two

Ŝ_{x} = \frac{1}{2} (c_{↑}^{†}c_{↓} + c_{↓}^{†}c_{↑})

Ŝ_{y} = −\frac{1}{2}i (c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↑})

Ŝ_{z} = \frac{1}{2} (c_{↑}^{†}c_{↑} − c_{↓}^{†}c_{↓})

Check commutators

[Ŝ_{x}, Ŝ_{y}] = −\frac{i}{4} [c_{↑}^{†}c_{↓}c_{↑}^{†}c_{↓} + c_{↓}^{†}c_{↑}c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↑}c_{↓}^{†}c_{↑} − c_{↑}^{†}c_{↓}c_{↓}^{†}c_{↑}]

+\frac{i}{4} [c_{↑}^{†}c_{↓}c_{↑}^{†}c_{↓} + c_{↑}^{†}c_{↓}c_{↓}^{†}c_{↑} − c_{↓}^{†}c_{↑}c_{↓}^{†}c_{↑} − c_{↓}^{†}c_{↑}c_{↑}^{†}c_{↓}]

Use

{c_{σ}, c_{σ′}^{†}} = δ_{σσ′}

Normal order. Because c_{σ}c_{σ} = 0 and c_{σ}^{†}c_{σ}^{†} = 0, several terms vanish.

[Ŝ_{x}, Ŝ_{y}] = −\frac{i}{4} [−c_{↓}^{†}c_{↑}^{†}c_{↑}c_{↓} + c_{↓}^{†}c_{↓} + c_{↑}^{†}c_{↓}^{†}c_{↓}c_{↑} − c_{↑}^{†}c_{↑} + c_{↑}^{†}c_{↓}^{†}c_{↓}c_{↑} + c_{↓}^{†}c_{↓} − c_{↓}^{†}c_{↑}^{†}c_{↑}c_{↓} − c_{↑}^{†}c_{↑}]

= \frac{i}{2} [c_{↑}^{†}c_{↑} − c_{↓}^{†}c_{↓}] = iŜ_{z}

[Ŝ_{z}, Ŝ_{y}] = −\frac{i}{4} [c_{↑}^{†}c_{↑}c_{↑}^{†}c_{↓} + c_{↓}^{†}c_{↓}c_{↓}^{†}c_{↑} − c_{↓}^{†}c_{↓}c_{↑}^{†}c_{↓} − c_{↑}^{†}c_{↑}c_{↓}^{†}c_{↑}]

+\frac{i}{4} [c_{↑}^{†}c_{↓}c_{↑}^{†}c_{↑} + c_{↓}^{†}c_{↑}c_{↓}^{†}c_{↓} − c_{↑}^{†}c_{↓}c_{↓}^{†}c_{↓} − c_{↓}^{†}c_{↑}c_{↑}^{†}c_{↑}]

= −\frac{i}{4} [−c_{↑}^{†}c_{↑}^{†}c_{↑}c_{↓} + c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↓}^{†}c_{↓}c_{↑} + c_{↓}^{†}c_{↑} + c_{↓}^{†}c_{↑}^{†}c_{↓}c_{↓} + c_{↑}^{†}c_{↓}^{†}c_{↑}c_{↑}]

−\frac{i}{4} [+c_{↑}^{†}c_{↑}^{†}c_{↓}c_{↑} + c_{↓}^{†}c_{↓}^{†}c_{↑}c_{↓} − c_{↑}^{†}c_{↓}^{†}c_{↓}c_{↓} + c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↑}^{†}c_{↑}c_{↑} + c_{↓}^{†}c_{↑}]

= −\frac{1}{2}i (c_{↑}^{†}c_{↓} + c_{↓}^{†}c_{↑}) = −iŜ_{x}

[Ŝ_{z}, Ŝ_{x}] = \frac{1}{4} [c_{↑}^{†}c_{↑}c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↓}c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↓}c_{↓}^{†}c_{↑} + c_{↑}^{†}c_{↑}c_{↓}^{†}c_{↑}]

−\frac{1}{4} [c_{↑}^{†}c_{↓}c_{↑}^{†}c_{↑} − c_{↑}^{†}c_{↓}c_{↓}^{†}c_{↓} − c_{↓}^{†}c_{↑}c_{↓}^{†}c_{↓} + c_{↓}^{†}c_{↑}c_{↑}^{†}c_{↑}]

= \frac{1}{4} [−c_{↑}^{†}c_{↑}^{†}c_{↑}c_{↓} + c_{↑}^{†}c_{↓} + c_{↓}^{†}c_{↑}^{†}c_{↓}c_{↓} + c_{↓}^{†}c_{↓}^{†}c_{↓}c_{↑} − c_{↓}^{†}c_{↑} − c_{↑}^{†}c_{↓}^{†}c_{↑}c_{↑}]

+\frac{1}{4} [+c_{↑}^{†}c_{↑}^{†}c_{↓}c_{↑} − c_{↑}^{†}c_{↓}^{†}c_{↓}c_{↓} + c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↓}^{†}c_{↓}c_{↓} + c_{↓}^{†}c_{↑}^{†}c_{↑}c_{↑} − c_{↓}^{†}c_{↑}]

= \frac{1}{2} (c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↑}) = i(−\frac{i}{2}) (c_{↑}^{†}c_{↓} − c_{↓}^{†}c_{↑}) = iŜ_{y}

Problem Three

We start with

H = ∑_{i,j} a_{i}^{†}a_{j}⟨i|\frac{p̂^{2}}{2m}|j⟩ + \frac{1}{2} ∑_{i,j,k,n} a_{i}^{†}a_{j}^{†}a_{k}a_{n}⟨i, j|\frac{A}{4π} \frac{1}{|r − r′|}e^{−λ|r−r′|}|k, n⟩

Insert the completeness relationship once in the first term, four times in the next term. Rewrite the result equation in terms of wave-functions.

H = − ∑_{i,j} ∫ d^{3}r \frac{1}{2m}a_{i}^{†}a_{j}ϕ_{i}^{∗}(r)∇^{2}ϕ_{j}(r)

+\frac{A}{8π} ∑_{i,j,k,n} a_{i}^{†}a_{j}^{†}a_{k}a_{n} ∫ d^{3}r d^{3}r′ \frac{1}{|r − r′|}e^{−λ|r−r′|}ϕ_{i}^{∗}(r)ϕ_{j}^{∗}(r′)ϕ_{k}(r′)ϕ_{n}(r)

Use the definitions of

ψ(r) = ∑_{j} ϕ_{j}(r)a_{j}

ψ^{†}(r) = ∑_{i} ϕ_{i}^{∗}(r)a_{i}^{†}

to do the sums over the i,j, etc. We get

H = − ∫ d^{3}r \frac{1}{2m}ψ^{†}(r)∇^{2}ψ(r) + \frac{A}{8π} ∫ d^{3}r d^{3}r′ ψ^{†}(r)ψ^{†}(r′)\frac{1}{|r − r′|}e^{−λ|r−r′|}ψ(r′)ψ(r)

which is what we wanted.

b. Fourier transform everything. The following will be useful:

ψ(r) = \frac{1}{(2π)^{3}} ∫ d^{3}k c_{k} e^{ik·r}

V (r − r′) = \frac{1}{(2π)^{3}} ∫ d^{3}q \frac{A}{q^{2} + λ^{2}} e^{iq·(r−r′)}

δ(k_{2} − k_{1}) = \frac{1}{(2π)^{3}} ∫ d^{3}r e^{ir·(k_{2}−k_{1})}

Plug these into the result from part a.

H = \frac{1}{(2π)^{6}} ∫ d^{3}r d^{3}k_{1} d^{3}k_{2} \frac{1}{2m}k_{2}^{2}c_{k_{1}}^{†}c_{k_{2}}e^{i(k_{2}−k_{1})·r}

+\frac{A}{16π^{3}} \frac{1}{(2π)^{6}} ∫ d^{3}r d^{3}r′ d^{3}k_{1} d^{3}k_{2} d^{3}k d^{3}k′ d^{3}q

\frac{1}{q^{2} + λ^{2}} c_{k_{1}}^{†}c_{k_{2}}^{†}c_{k′}c_{k}e^{i(k_{1}+q−k)·r}e^{i(k_{2}−q−k′)·r′}

Pick out delta functions and do integrations over them.

H = \frac{1}{(2π)^{3}} ∫ d^{3}k \frac{1}{2m}k^{2}c_{k}^{†}c_{k}

+\frac{A}{16π^{3}} ∫ d^{3}k d^{3}k′ d^{3}q \frac{1}{q^{2} + λ^{2}} c_{k−q}^{†}c_{k′+q}^{†}c_{k′}c_{k}

Problem Four Show

e^{−a(n_{↑}− \frac{1}{2} )(n_{↓}− \frac{1}{2} )} = \frac{1}{2}e^{− \frac{a}{4}} ∑_{σ=±} e^{aσ(n_{↑}−n_{↓})}

To prove, I’ll manipulate each side into suggestive forms and then show that both sides are the same. First, consider the left hand side of the equality.

l.h.s. = e^{−a(n_{↑}− \frac{1}{2} )(n_{↓}− \frac{1}{2} )} = e^{− \frac{a}{4} + \frac{1}{2} a(n_{↑}+n_{↓})−an_{↑}n_{↓}}

r.h.s. = \frac{1}{2}e^{− \frac{a}{4}} ∑_{σ=±} e^{aσ(n_{↑}−n_{↓})}

= e^{− \frac{a}{4}} cosh [a|n_{↑} − n_{↓}|]

I put the absolute value sign to remind us that cosh(x) = cosh(−x). For fermions, n_{σ} = 0 or 1. If n_{↑} = 1 and n_{↓} = 1, then |n_{↑} − n_{↓}| = 0 If n_{↑} = 0 and n_{↓} = 0, then |n_{↑} − n_{↓}| = 0 If n_{↑} = 1 and n_{↓} = 0, then |n_{↑} − n_{↓}| = 1 If n_{↑} = 0 and n_{↓} = 1, then |n_{↑} − n_{↓}| = 1 In the first two cases, the hyperbolic cosine term, cosh(a|n_{↑} − n_{↓}|), gives one.

In the latter case, we can use the relation cosh ±a = e^{\frac{1}{2} a} and rewrite the r.h.s. Equate the r.h.s. to the l.h.s and factor out common multiples of e^{− \frac{a}{4}} . Take the log of both sides. We’ll be left with

−an_{↑}n_{↓} + \frac{1}{2}a(n_{↑} + n_{↓}) = \frac{1}{2}a|n_{↑} − n_{↓}|

Finally, I show that for all pairing of n_{σ} = 0, 1 this equality will be satisfied. If n_{↑} = 1 and n_{↓} = 1, then −a + a = 0 If n_{↑} = 0 and n_{↓} = 0, then 0 + 0 = 0 If n_{↑} = 1 and n_{↓} = 0, then 0 + \frac{1}{2}a = \frac{1}{2}a If n_{↑} = 0 and n_{↓} = 1, then 0 + \frac{1}{2}a = \frac{1}{2}a so our relation is true for fermions. All of this could have been rewritten in matrix notation.

Problem Five

H_{0} = ∑_{α} ϵ_{α}a_{α}^{†}a_{α}

a. I use γ instead of what’s on the homework assignment so that I can keep the labels straight. β has a negative sign so that the partition function is bounded.

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = ∑_{{n_{i}}}⟨{n_{i}}| [e^{−βH_{0}}a_{γ}^{†}a_{γ}] |{n_{i}}⟩

where |{n_{i}}⟩ stands for a many particle state with a set, {n_{i}}, of occupation numbers. a_{γ}^{†}a_{γ} acting on |{n_{i}}⟩ returns 0 or 1 depending on whether there is a fermion in state γ - Alas, the notation is a bit degenerate - We can likewise get the eigenvalues for the operators in the H-potential. I will use e∑stuff = ∏ estuff, and write

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = ∑_{{n_{i}}} ∏_{α}⟨{n_{i}}| [e^{−βϵ_{α}a_{α}^{†}a_{α}}a_{γ}^{†}a_{γ}] |{n_{i}}⟩

Now, I let the density operators act on the various γ states.

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = ∑_{{n_{i}}} n_{γ} ∏_{α} [e^{−βϵ_{α}n_{α}}]

Take the term which was generated by the state γ out of the product.

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = ∑_{{n_{i}}} n_{γ}e^{−βϵ_{γ}n_{γ}} ∏_{α≠γ} [e^{−βϵ_{α}n_{α}}]

Do the sum over occupation numbers {n_{i}} with n_{i} = 1 or 0.

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = e^{−βϵ_{γ}} ∏_{α≠γ} [1 + e^{−βϵ_{α}}]

b. Solve

Trace [e^{−βH_{0}}a_{γ}a_{γ}^{†}] = −Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] + Trace [e^{−βH_{0}}]

using {a_{α}, a_{α}^{†}} = 1. We know the first term on the left from part a. The second term doesn’t take too much effort to get.

Trace [e^{−βH_{0}}] = ∑_{{n_{i}}} ∏_{α}⟨{n_{i}}| (e^{−βϵ_{α}n_{α}}) |{n_{i}}⟩ = ∏_{α} (1 + e^{−βϵ_{α}})

So

Trace [e^{−βH_{0}}a_{γ}a_{γ}^{†}] = −e^{−βϵ_{γ}} ∏_{α≠γ} [1 + e^{−βϵ_{α}}] + ∏_{α} (1 + e^{−βϵ_{α}})

= [−\frac{e^{−βϵ_{γ}}}{1 + e^{−βϵ_{α}}} + \frac{1 + e^{−βϵ_{α}}}{1 + e^{−βϵ_{α}}} ] ∏_{α} (1 + e^{−βϵ_{α}})

Which can be rewritten

Trace [e^{−βH_{0}}a_{γ}a_{γ}^{†}] = \frac{1}{1 + e^{−βϵ_{γ}}} ∏_{α} (1 + e^{−βϵ_{α}})

c. Noting that

e^{−βH_{0}}a_{γ}^{†}a_{γ} = −\frac{1}{β} \frac{∂}{∂ϵ_{γ}} e^{−βH_{0}}

We can find

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = −\frac{1}{β} \frac{∂}{∂ϵ_{γ}} Trace [e^{−βH_{0}}]

¿From earlier,

Trace [e^{−βH_{0}}] = ∏_{α} [1 + e^{−βϵ_{α}}]

whence it follows

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = e^{−βϵ_{γ}} ∏_{α≠γ} [1 + e^{−βϵ_{α}}]

which is what I got in part a.

d.

Z = Trace [e^{−βH_{0}}] = ∏_{α} [1 + e^{−βϵ_{α}}]

and

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = e^{−βϵ_{γ}} ∏_{α≠γ} [1 + e^{−βϵ_{α}}]

= \frac{e^{−βϵ_{γ}}}{1 + e^{−βϵ_{γ}}} ∏_{α} [1 + e^{−βϵ_{α}}]

so

⟨n_{γ}⟩ = Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] /Trace [e^{−βH_{0}}]

= \frac{1}{1 + e^{βϵ_{γ}}}

e. For bosons,

[a, a^{†}] = 1

First, we get the partition function. Notice that all for each n_{i}, all positive integer values are allowed. The sum is a geometric series and can be done explicitly.

Trace [e^{−βH_{0}}] = Z = ∑_{{n_{i}}} ∏_{α} e^{−βϵ_{α}n_{α}}

= ∏_{α} \frac{1}{1 − e^{−βϵ_{α}}}

Then, we use a nice little trick.

Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] = −\frac{1}{β} \frac{∂}{∂ϵ_{γ}} Trace [e^{−βH_{0}}]

= \frac{e^{−βϵ_{α}}}{1 − e^{−βϵ_{α}}} ∏_{α} \frac{1}{1 − e^{−βϵ_{α}}}

⟨n_{γ}⟩ = Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] /Trace [e^{−βH_{0}}]

= \frac{−1}{1 − e^{βϵ_{γ}}}

When one of the energies is zero or β → 0, the bose function diverges, and we get a condensate.

Finally, I need to find Trace [e^{−βH_{0}}a_{γ}a_{γ}^{†}]. Use the commutator to shuffle terms.

Trace [e^{−βH_{0}}a_{γ}a_{γ}^{†}] = Trace [e^{−βH_{0}}a_{γ}^{†}a_{γ}] + Trace [e^{−βH_{0}}] = ⟨n_{γ}⟩Z + Z

f. We work with bosons.

[a_{γ}, H_{0}] = ∑_{α} ϵ_{α}[a_{γ}, a_{α}^{†}a_{α}]

= ∑_{α≠γ} ϵ_{α}[a_{γ}a_{α}^{†}a_{α} − a_{α}^{†}a_{α}a_{γ}] + ϵ_{γ}[a_{γ}, a_{γ}a_{γ}^{†}]

= ϵ_{γ}a_{γ}

[a_{γ}^{†}, H_{0}] = ∑_{α} ϵ_{α}[a_{γ}^{†}, a_{α}^{†}a_{α}]

= ∑_{α≠γ} ϵ_{α}[a_{γ}^{†}a_{α}^{†}a_{α} − a_{α}^{†}a_{α}a_{γ}^{†}] + ϵ_{γ}[a_{γ}^{†}, a_{γ}^{†}a_{γ}]

= −ϵ_{γ}a_{γ}^{†}

So [H_{0}, a_{γ}] = −ϵ_{γ}a_{γ} and [H_{0}, a_{γ}^{†}] = ϵ_{γ}a_{γ}^{†}.

( \begin{matrix}a_{γ}^{†}(τ) \\ a_{γ}(τ) \end{matrix} ) = e^{τH_{0}} ( \begin{matrix}a_{γ}^{†} \\ a_{γ} \end{matrix} ) e^{−τH_{0}}

= ( \begin{matrix}a_{γ}^{†} e^{ϵ_{γ}τ} \\ a_{γ} e^{−ϵ_{γ}τ} \end{matrix} ) e^{τH_{0}−τH_{0}}

So commuting the a and a^{†}s past the H, we find

( \begin{matrix}a_{γ}^{†}(τ) \\ a_{γ}(τ) \end{matrix} ) = ( \begin{matrix}a_{γ}^{†} e^{ϵ_{γ}τ} \\ a_{γ} e^{−ϵ_{γ}τ} \end{matrix} )