Physics 715: Bogoliubov transformation and the Fermi vacuum

Simple model Hamiltonians for quantum particles of all types have free-particle parts such as

$H_0=\sum_k E_k a_k^\dagger a_k, \qquad H_0=\sum_{k,s} E_k b_{k,s}^\dagger b_{k,s}$

(for bosons we would have)

$[b_{k,s}^\dagger,b_{k',s'}]=\delta_{s,s'}\delta_{k,k'}$

which simply counts up the number of excitations in the system and multiplies by the energy of the excitation, and interaction parts that contain more creation/annihilation operators. We will study simple interacting systems by first establishing what their ground states look like, and then studying low-lying excitations by various linearization techniques. We do so systematically, we first study the ground states.

Let’s denote the (fermion) absolute vacuum by $|0\rangle$ , and a quantum state containing one electron of momentum $\mathbf{k}$ and spin $s$ by $|\mathbf{k},s\rangle$ and a two electron state with momenta/spin $(\mathbf{k}, s)$ and $(\mathbf{q}, s')$ by $|\mathbf{k}, s; \, \mathbf{q}, s'\rangle$ .
Note that a proper fermion state must obey the exclusion principle

$|\mathbf{k}, s; \, \mathbf{q}, s'\rangle=-|\mathbf{q}, s'; \, \mathbf{k}, s\rangle, \qquad \langle x_1, x_2|\mathbf{k}, s; \, \mathbf{q}, s'\rangle={1\over \sqrt{2}}\Big(\psi_{\mathbf{k},s}(x_1) \psi_{\mathbf{q},s'}(x_2)-\psi_{\mathbf{k},s}(x_2) \psi_{\mathbf{q},s'}(x_1)\Big)$

Let’s define a “first in, first out” notation

$b^\dagger_{\mathbf{k},s}|0\rangle=|\mathbf{k},s\rangle, \qquad b^\dagger_{\mathbf{q},s'} \, b^\dagger_{\mathbf{k},s}|0\rangle=|\mathbf{q}, s';\mathbf{k},s\rangle, \qquad b_{\mathbf{k},s}|\mathbf{q}, s'\rangle=\delta^3(\mathbf{k}-\mathbf{q}) \, \delta_{s,s'} \, |0\rangle$

The required Pauli exclusion principle is satisfied if the annihilation/creation operators anti-commute

$\{b^\dagger_{\mathbf{q}, s'}, b^\dagger_{\mathbf{k}, s}\}=0, \qquad \{b_{\mathbf{q}, s'}, b_{\mathbf{k}, s}\}=0, \qquad \{b^\dagger_{\mathbf{q}, s'}, b_{\mathbf{k}, s}\}=\delta^3(\mathbf{q}-\mathbf{k}) \, \delta_{s,s'}$

Note that

$b_{\mathbf{l}, s''} \, b^\dagger_{\mathbf{q},s'} \, b^\dagger_{\mathbf{k},s}|0\rangle=b_{\mathbf{l}, s''} \,|\mathbf{q}, s';\mathbf{k},s\rangle=\delta^3(\mathbf{q}-\mathbf{l}) \, \delta_{s'',s'} \, |\mathbf{k},s\rangle-\delta^3(\mathbf{k}-\mathbf{l}) \, \delta_{s'',s} \, |\mathbf{q},s'\rangle$

Let a system of $N$ independent non-relativistic ( $E_{\mathbf{k}}=E_{-\mathbf{k}}={k^2\over 2m}$ ) fermions have quantum Hamiltonian

$H=\sum_{\mathbf{k}} \, E_{\mathbf{k}} \, \Big(\creup{k}\annup{k}+\cred{k}\annd{k}\Big)=\sum_{\mathbf{k},s} \, E_{\mathbf{k}} \, b^\dagger_{\mathbf{k},s}b_{\mathbf{k},s}$

The ground state $|\Phi_0\rangle$ (illustrated below) is the state with

$H|\Phi_0\rangle =\sum_{|\mathbf{k}|<k_f} \, 2E_{\mathbf{k}}|\Phi_0\rangle$

$\[ \sum_{\mathbf{k}} \, \Big(\creup{k}\annup{k}+\cred{k}\annd{k}\Big)|\Phi_0\rangle =N \, |\Phi_0\rangle$

Let’s prove that the modified Hamiltonian

$\tilde{H}=H-\sum_{|\mathbf{k}|\le k_f,s} E_{\mathbf{k}}$

has zero vacuum expectation value

$\langle\Phi_0|\tilde{H}|\Phi_0\rangle=0$

For the Fermi gas illustrated,

$E=\langle\Phi_0|H|\Phi_0\rangle=2\cdot 2\Big(E_0+E_1+E_2+E_3+E_4\Big), \qquad E_i=E_{\mathbf{k}_i}$

and since $E_f=E_4$ , $\langle \Phi_0|\tilde{H}|\Phi_0\rangle=0$ .
Properties of the Fermi-Dirac vacuum/GS
$\bullet$ What is the total spin of the Fermi ground state $|\Phi_0\rangle$

$\langle \Phi_0|S_z|\Phi_0\rangle={\hbar\over 2}\cdot\Big(2(1-1)+2(1-1)+2(1-1)+2(1-1)+2(1-1)\Big)=0$

$\bullet$ What is the total momentum of the Fermi ground state $|\Phi_0\rangle$

$\langle \Phi_0|\mathbf{k}|\Phi_0\rangle=2\cdot\Big((\mathbf{k}_0-\mathbf{k}_0)+(\mathbf{k}_1-\mathbf{k}_1)+(\mathbf{k}_2-\mathbf{k}_2)+(\mathbf{k}_3-\mathbf{k}_3)+(\mathbf{k}_4-\mathbf{k}_4)\Big)=0$

so the filled Fermi-sea ground state has no net spin or momentum,
$\bullet$ and using $\tilde{H}$ to measure its energy it has no net energy.
This is a rather good model of a filled energy band in a crystalline solid. It certainly explains the electrical conductivity (none!) of a filled band.

Bogoliubov transformations: a quantum canonical transformation
We invent new operators

$\tilde{b}_{\mathbf{k}, \ua}=\Big{\{}\begin{array}{cc} \annup{k} & k>k_f\\ \cred{-k} & k\le k_f\end{array}, \qquad \tilde{b}_{\mathbf{k}, \da}=\Big{\{}\begin{array}{cc} \annd{k} & k>k_f\\ \creup{-k} & k\le k_f\end{array}$

$\tilde{b}^\dagger_{\mathbf{k}, \ua}=\Big{\{}\begin{array}{cc} \creup{k} & k>k_f\\ \annd{-k} & k\le k_f\end{array}, \qquad \tilde{b}^\dagger_{\mathbf{k}, \da}=\Big{\{}\begin{array}{cc} \cred{k} & k>k_f\\ \annup{-k} & k\le k_f\end{array}$

$\{\tilde{b}^\dagger_{\mathbf{k}, s}, \tilde{b}_{\mathbf{q}, s'}\}=\delta^3(\mathbf{k}-\mathbf{q}) \, \delta_{s,s'}$

for $|\mathbf{k}|$ both above and below the Fermi momentum.
To see this, rewrite the operators in terms of an angle variable

$\tilde{b}_{\mathbf{k}, s}=\cos\theta_k \, b_{\mathbf{k}, s}+\sin\theta_k \, b^\dagger_{-\mathbf{k}, \bar{s}}, \qquad \theta_k=0, \quad k>k_f, \quad \theta_k=\pi/2, \quad k\le k_f\] \[\tilde{b}^\dagger_{\mathbf{k}, s}=\cos\theta_k \, b^\dagger_{\mathbf{k}, s}+\sin\theta_k \, b_{-\mathbf{k}, \bar{s}}, \qquad \theta_k=0, \quad k>k_f, \quad \theta_k=\pi/2, \quad k\le k_f$

in which $s=\uparrow$ or $\downarrow$ , and $\bar{s}=\uparrow$ when $s=\downarrow$ , $\bar{s}=\downarrow$ when $s=\uparrow$ . This gives us the commutation relations we seek, and we can easily compute

(1) $\begin{eqnarray*}\{\tilde{b}^\dagger_{\mathbf{k}, s}, \tilde{b}_{\mathbf{k}', s'}\}&=&\{\cos\theta_k \, b^\dagger_{\mathbf{k}, s}-\sin\theta_k \, b_{-\mathbf{k}, \bar{s}}, \, \cos\theta_{k'} \, b_{\mathbf{k}', s'}-\sin\theta_{k'} \, b^\dagger_{-\mathbf{k}', \bar{s}'}\}\nonumber\\ &=&\cos\theta_k \, \cos\theta_{k'} \, \{b^\dagger_{\mathbf{k}, s}, \, b_{\mathbf{k}', s'}\}+\sin\theta_k \, \sin\theta_{k'} \, \{b_{-\mathbf{k}, \bar{s}}, \, b^\dagger_{-\mathbf{k}', \bar{s}'}\}\nonumber\\ &=&\Big(\cos\theta_k \, \cos\theta_{k'}+\sin\theta_k \, \sin\theta_{k'}\Big) \, \delta^3(\mathbf{k}-\mathbf{k}') \, \delta_{s,s'}\nonumber\\ &=&\delta^3(\mathbf{k}-\mathbf{k}') \, \delta_{s,s'}\end{eqnarray*}$

since $\delta_{s,s'}=\delta_{\bar{s}, \bar{s}'}$ and the anticommutator of two $b$ ‘s or two $b^\dagger$ ‘s are zero.

Describe the states

$\[ \tilde{b}^\dagger_{\mathbf{k}_1, \ua} \, |\Phi_0\rangle, \qquad \tilde{b}^\dagger_{\mathbf{k}_2, \da} \, |\Phi_0\rangle, \qquad \tilde{b}^\dagger_{\mathbf{k}_5, \ua} \, |\Phi_0\rangle$

for the system represented in the figure above, illustrated in the $|\Phi_0\rangle$ state. For example we can see that

$\tilde{b}^\dagger_{\mathbf{k}_1, \ua} \, |\Phi_0\rangle=b_{-\mathbf{k}_1, \da} \, |\Phi_0\rangle$

is a state with a filled vacuum but with a vacancy or hole in the $-\mathbf{k}_1$ position. $\tilde{H} \, \tilde{b}^\dagger_{\mathbf{k}_1, \uparrow} \, |\Phi_0\rangle=E_1\tilde{b}^\dagger_{\mathbf{k}_1, \uparrow} \, |\Phi_0\rangle$ , and this state has spin $0-(-{\hbar\over 2})={\hbar\over 2}$ and momentum $0-(-\mathbf{k}_1)=\mathbf{k}_1$ .

$\tilde{b}^\dagger_{\mathbf{k}_2, \da} \, |\Phi_0\rangle=b_{-\mathbf{k}_2, \ua} \, |\Phi_0\rangle$

is a state with a filled vacuum but with a vacancy or hole in the $-\mathbf{k}_2$ position. $\tilde{H} \, \tilde{b}^\dagger_{\mathbf{k}_2, \downarrow} \, |\Phi_0\rangle=E_2\tilde{b}^\dagger_{\mathbf{k}_2, \downarrow} \, |\Phi_0\rangle$ , and this state has spin $0-({\hbar\over 2})=-{\hbar\over 2}$ and momentum $0-(-\mathbf{k}_2)=\mathbf{k}_2$ .
The $\tilde{b}^\dagger_{k,s}$ operators create holes in the otherwise filled Fermi sea if $k<k_f$ , these holes have positive energy relative to $\tilde{H}$ .

$\tilde{b}^\dagger_{\mathbf{k}_5, \ua} \, |\Phi_0\rangle=b^\dagger_{\mathbf{k}_5, \ua} \, |\Phi_0\rangle$

is a state with a filled vacuum plus an additional quanta above the Fermi level. $\tilde{H} \, \tilde{b}^\dagger_{\mathbf{k}_5, \uparrow} \, |\Phi_0\rangle=E_5\tilde{b}^\dagger_{\mathbf{k}_5, \upaarrow} \, |\Phi_0\rangle$ , and this state has spin $0+({\hbar\over 2})={\hbar\over 2}$ and momentum $0+(\mathbf{k}_5)=\mathbf{k}_5$ .

Let’s work on writing $\tilde{H}$ in terms of the new annihilation/creation operators, lets start with

(2) $\begin{eqnarray*}H=\sum_{k,s}E_k \, b^\dagger_{k,s} b_{k,s}&=&\sum_{k\le k_f,s}E_k \, b^\dagger_{k,s} b_{k,s}+\sum_{k> k_f,s}E_k \, b^\dagger_{k,s} b_{k,s}\nonumber\\ &=&\sum_{k\le k_f,s}E_k \, \tilde{b}_{-k,\bar{s}} \tilde{b}^\dagger_{-k,\bar{s}}+\sum_{k> k_f,s}E_k \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\nonumber\\ &=&\sum_{k\le k_f,s}E_k \, \tilde{b}_{k,s} \tilde{b}^\dagger_{k,s}+\sum_{k> k_f,s}E_k \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\nonumber\\ &=&\sum_{k\le k_f,s}E_k \, \Big(1-\tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\Big)+\sum_{k> k_f,s}E_k \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\end{eqnarray*}$

now consider that any particle moved out of the set $k\le k_f$ must be moved into the set $k>k_f$ ,

(3) $\begin{eqnarray*}\tilde{H}&=&\sum_{k\le k_f,s}E_k \, \Big(1-\tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\Big)+\sum_{k> k_f,s}E_k \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}-\sum_{k\le k_f,s}E_k\nonumber\\ &=&\sum_{k\le k_f,s}(-E_k) \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}+\sum_{k> k_f,s}E_k \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\nonumber\\ &=&\sum_{k\le k_f,s}(-E_k) \, n_{hole, \mathbf{k}, s}+\sum_{k> k_f,s}E_k \, n_{particle, \mathbf{k}, s}\end{eqnarray*}$

and so the number of holes equals the number of particles excited above the Fermi level, so

$0=\sum_{k\le k_f}n_{hole, \mathbf{k}, s}-\sum_{k> k_f}n_{particle, \mathbf{k}, s}$

and we can add zero onto $\tilde{H}$ to get

(4) $\begin{eqnarray*}\tilde{H}&=&\sum_{k\le k_f,s}(E_f-E_k) \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}+\sum_{k> k_f,s}E_k \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\nonumber\\ &=&\sum_{k\le k_f,s}(-E_k) \, n_{hole, \mathbf{k}, s}+\sum_{k> k_f,s}(E_k-E_f) \, n_{particle, \mathbf{k}, s}\nonumber\\ &=&\sum_{k,s}|E_k-E_f| \, \tilde{b}^\dagger_{k,s} \tilde{b}_{k,s}\end{eqnarray*}$

in effect the “zero point” of the energy has been moved to the Fermi sea surface and the Fermi vacuum has zero energy. In statistical physics and condensed matter physics we study phenomena that preserve $N$ , so any process creating a hole in the Fermi sea creates a filled state or quanta above it. In other words $dN_p=dN_h$ where the “p,h” refer to particle excitation and hole.

This means that in any equilibrium process $\mu_p=-\mu_h$ , a property that you will many times in particle and condensed matter physics.

If we use only the new annihilation/creation operators and $\tilde{H}$ , the Fermi ground state is effectively a “vacuum”, which has two interesting types of excitations, “particles” and “holes”. Note that creating a hole is an excitation with a positive energy.

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Physics 715: Bogoliubov transformation and the Fermi vacuum

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