Self Assesment Quiz

We can define creation and annihilation operators $b_{\vec{k}}^{+}$ and $b_{\vec{k}}$ using the electron creation and annihilation operators $C_{\vec{k}\uparrow}^{+}$ and $C_{-\vec{k}\downarrow}$ as follows

$\displaystyle b_{\vec{k}}^{+}$	$\displaystyle =$	$\displaystyle C_{\vec{k}\uparrow}^{+}C_{-\vec{k}\downarrow}^{+}$
$\displaystyle b_{\vec{k}}$	$\displaystyle =$	$\displaystyle C_{-\vec{k}\downarrow}C_{\vec{k}\uparrow}$

We note the following fundamental commutation relations between creation and annihilation operators for Bosons and Fermions,

$\displaystyle [C_{\vec{k}_{1}}^{+},C_{\vec{k}_{2}}^{+}]$	$\displaystyle =$	$\displaystyle C_{\vec{k}_{1}}^{+}C_{\vec{k}_{2}}^{+}+C_{\vec{k}_{2}}^{+}C_{\vec{k}_{1}}^{+}$
	$\displaystyle =$	0

That is, Fermion creation operators anticommute. Similarly, the Fermion annihilation operators anticommute,

$\displaystyle [C_{\vec{k}_{1}},C_{\vec{k}_{2}}]$	$\displaystyle =$	$\displaystyle C_{\vec{k}_{1}}C_{\vec{k}_{2}}+C_{\vec{k}_{2}}C_{\vec{k}_{1}}$
	$\displaystyle =$	0

The third commutation relation is between the creation and the annihilation operators ,

$\displaystyle [C_{\vec{k}_{1}},C_{\vec{k}_{2}}^{+}]$	$\displaystyle =$	$\displaystyle C_{\vec{k}_{1}}C_{\vec{k}_{2}}^{+}+C_{\vec{k}_{2}}^{+}C_{\vec{k}_{1}}$
	$\displaystyle =$	$\displaystyle \delta_{\vec{k}_{1}\vec{k}_{2}}$

Using the above anticommutation relations, we have for the Cooper pair operators the following three relations,
                                         $\displaystyle [b_{\vec{k}}^{+},b_{\vec{k}'}^{+}]_{-}=0$
                                         $\displaystyle [b_{\vec{k}},b_{\vec{k}'}]_{-}=0$
                                         $\displaystyle [b_{\vec{k}},b_{\vec{k}'}]_{-}=\delta_{\vec{k}\vec{k}'}(1-n_{-\vec{k}\downarrow}-n_{\vec{k}\uparrow})$
Thus, the Cooper pair creation and annihilation operators are not quite Bosons either. They are also not genuine Fermions!

2. For the Cooper pair wavefunction show that
$\displaystyle g_{\sigma}(\vec{k})=-g_{-\sigma}(-\vec{k})$

We write the Cooper pair wavefunction as

$\displaystyle \vert\psi$	$\displaystyle >$	$\displaystyle =\frac{1}{\sqrt{2}}\sum_{k\sigma}g_{\sigma}(\vec{k})C_{\vec{k}\sigma}^{+}C_{-\vec{k}-\sigma}^{+}\vert FS>$
	$\displaystyle =$	$\displaystyle \frac{1}{\sqrt{2}}\sum_{\vec{k}'\sigma'}g_{-\sigma'}(-\vec{k'})C_{-\vec{k'}-\sigma'}^{+}C_{\vec{k}'\sigma'}^{+}\vert FS>$

Relabeling, we get
                                                 $\displaystyle \vert\psi>=\frac{1}{\sqrt{2}}\sum_{\vec{k}\sigma}g_{-\sigma}(-\vec{k})C_{-\vec{k}-\sigma}^{+}C_{\vec{k}\sigma}^{+}\vert FS>$
Using the anticommutation rule, $[C_{\vec{k}\sigma}^{+},C_{\vec{k}'\sigma'}^{+}]_{+}=0$ , we have
                                                $\displaystyle \vert\psi>=-\frac{1}{\sqrt{2}}\sum_{\vec{k}\sigma}g_{-\sigma}(-\vec{k})C_{\vec{k}\sigma}^{+}C_{-\vec{k}-\sigma}^{+}\vert FS>$
Thus, we must have
                                             $\displaystyle g_{\sigma}(\vec{k})=-g_{-\sigma}(-\vec{k})$

3. Can the Cooper pair binding energy be expanded in a power series in the strength of the interaction

? If not, then what does it imply ?

No. The result cannot be obtatained from the perturbation theory.

4. What is the essential difference between the bound states of two electrons above the Fermi sea and another in the absence of a Fermi sea ?

In the absence of a Fermi sea the density of states varies as the square root of energy, and a minimum strength of interaction is needed before the two electrons can form a bound state.

5. In order to obtain an effective electron-electron interaction from the electron-phonon interaction we have carried out a canonical transformation
$\displaystyle \tilde{H}=e^{-S}He^{S}$
where we require $S^{+}=-S.$ Why?

$e^{S}$ represents a unitary transformation. This implies that
$\displaystyle (e^{S})^{+}=(e^{S})^{-1}$
So that,
$\displaystyle S^{+}=-S$
One can check that our transformation satisfies the above constraint.

6. Show that for a coherent state
$\displaystyle <\alpha\vert(a^{+})^{p}a^{q}\vert\alpha>=(\alpha*)^{p}\alpha^{q}$

It can easily be checked by noting that
$\displaystyle a\vert\alpha>=\alpha\vert\alpha>$

7. Is the BCS wavefunction an eignefunction of the superconducting Hamiltonian ?

No. It is a variational wavefunction which is used to obtain an estimate of the ground state energy using the superconducting Hamiltonian.

8. Does the BCS wavefunction $\vert\psi_{BCS}>$ contain a fixed number of particles? If not then what does it mean to have $<\psi_{BCS}\vert\psi_{BCS}>=1$ ?

It does not contain a fixed number of particles.

9. At

, what are the values of $u_{\vec{k}}^{0}$ and $v_{\vec{k}}^{0}$ for $\vec{k}<\vec{k}_{F}$ and $\vec{k}>\vec{k}_{F}$ ?

The values are $u_{\vec{k}}^{0}$ and $v_{\vec{k}}^{0}$ for $\vec{k}<\vec{k}_{F}$ and $\vec{k}>\vec{k}_{F}$
$\begin{displaymath} u_{\vec{k}}^{0}=\begin{cases} \begin{array}{c} 0\\ 1 \... ...}<\vec{k}_{F}\\ \vec{k}>\vec{k}_{F} \end{array}\end{cases} \end{displaymath}$
and
$\begin{displaymath} v_{\vec{k}}^{0}=\begin{cases} \begin{array}{c} 1\\ 0 \... ...}<\vec{k}_{F}\\ \vec{k}>\vec{k}_{F} \end{array}\end{cases} \end{displaymath}$

10. We have shown that the phase and the number of particle operator are dynamically conjugate variables like position and momentum. Do they satisfy some uncertainty relation similar to that of position and momentum ?

Yes. They satisfy the uncertainty relation given by
$\displaystyle \Delta\phi\Delta N\geq2\pi$

11. In the variational determination of the energy of the superconducting state how do we include the constraint that the number of particles is fixed.

We use the method of Lagrange's multipliers to ensure that the number of particles is fixed in the calculation.

12. What is the minimum energy needed to create a single particle excitation in the BCS ground state ?

13. For $\Delta=0,$ interpret the Bogoliubov-Valatin operators.

For $\Delta=0$ , the Bogoliubov-Valatin operators become
$\begin{displaymath} \gamma_{\vec{k}\uparrow}^{+}=\begin{cases} \begin{array}{c... ...}>\vec{k}_{F}\\ \vec{k}<\vec{k}_{F} \end{array}\end{cases} \end{displaymath}$
and
$\begin{displaymath} \gamma_{\vec{k}\uparrow}=\begin{cases} \begin{array}{c} C... ...}>\vec{k}_{F}\\ \vec{k}<\vec{k}_{F} \end{array}\end{cases} \end{displaymath}$
when $\Delta\ne0,$ the Bogoliubov-Valatin operators contain a mixture of electron-like, $C_{\vec{k}\uparrow}^{+}$ , and hole-like, $-C_{-\vec{k}\downarrow}$ , operators.

Self Assessment Quiz