5 BCS Theory

5.1 Electron-Electron Attraction¹⁶

Despite the Coulomb repulsion between electrons, interactions between electrons and the lattice can provide a mechanism for a weak attractive force between electrons. At low temperatures the metal lattice will have little thermal agitation, but each lattice point will act like a harmonic oscillator with a maximum mode of oscillation given by the Debye frequency $(\omega _{D})$. As a conduction electron passes through the lattice, there will be a Coulomb attraction between the negatively charged electron and the positively charged cations at each lattice point. Nearby cations will move toward the electron and, after the electron leaves, the cations will relax back to their equilibrium location in a time approximately equal to period of the natural oscillation of the lattice, $\frac{2\pi }{\omega _{D}}$. Before their relaxation, the cations will be closer together, but the electron will have left the region, creating an increased charge density that will attract nearby electrons. By the time the cations return to their equilibrium positions, the first electron will have traveled a distance of $\nu _{e}\frac{2\pi }{\omega _{D}}$, where $\nu _{e}$ is the velocity of conduction electrons in the lattice. Since $\nu _{e}\sim 10^{6}$$\text {m/s}$ and $\frac{2\pi }{\omega _{D}}\sim 10^{-13}$$\text s$, the resulting distance is $\sim 10^{-7}$$\text m$ or about 1000Å, which is on the order of Pippard's coherence length. The net force on electrons separated by $\sim$1000Å is attractive: the Coulomb repulsive force is completely screened at this distance since the spacing between lattice points is only a few Å.

5.2 Cooper Pairing¹⁷

We will now show that this attractive force can cause electrons to form a bound pair that has an energy below the Fermi surface. We will model the attractive force between the electrons as the exchange of a packet of momentum (a phonon) from one electron to another. First, a thought experiment: take a non-interacting Fermi gas of electrons and cool it down to $T=0$. Because of the Pauli exclusion principle, no two fermions can occupy the same quantum state and so instead of having all the electrons with k=0, where k is the wavevector and $\frac{\hbar ^{2}k^{2}}{2m}$ is the energy of the electron, the wavevectors of the electrons will stack up to a maximum value of $k_{F}$ at the Fermi surface with an energy given by the Fermi energy $(E_{F})$ (see figure 9).

Figure 9: Behavior of bosons and fermions at T=0.

Now add two electrons to the system with wavevectors $\mathbf{k}_{1}$ and $\mathbf{k}_{2}$, each with energy just above the Fermi surface. The first electron then gives off a phonon q which represents the loss of momentum from an interaction with the lattice: $\mathbf{k}_{1}=\mathbf{k}_{1}^{'}+\mathbf{q}$, where $\mathbf{k}_{1}^{'}$ is the new wavevector of the electron. The second electron absorbs this phonon and gains momentum because of its attraction to the electron through the lattice, $\mathbf{k}_{2}+\mathbf{q}=\mathbf{k}_{2}^{'}$. Thus we have conservation of momentum:

$$\mathbf{k}_{1}+\mathbf{k}_{2}=\mathbf{k}_{1}^{'}+\mathbf{k}_{2}^{'}=\mathbf{K}.$$ (26)

The phonon $\mathbf{q}$ has a maximum energy on the order of the $\hbar \omega _{D}$ where $\omega _{D}$, the Debye frequency, is the maximum natural frequency of oscillation of the lattice. The maximum momentum of the phonon $\mathbf{q}$, defined as $\Delta k$, is determined by its maximum energy. Because all states below $k_{\mathbf{F}}$ are filled, they are unavailable by the Pauli exclusion principle, so

$$k_{F}\leq k'_{1},k'_{2}\leq k_{F}+\Delta k.$$ (27)

Figure 10: K-space: the circular rings are possible scattering states that obey equation (27). The shaded region also obeys conservation of momentum (equation 26).
Source: Ibach and Lüth (1996), page 230.

Figure 10 gives a graphical representation of the scattering of $k_{1}$ and $k_{2}$ into states $k_{1}^{'}$ and $k_{2}^{'}$ in momentum space. The circular rings represent the range in momentum where $k_{1}$ and $k_{2}$ can be scattered according to equation (27) and the shaded region is the area that also satisfies equation (26). As K get smaller, the shaded region gets larger and at $\mathbf{K}=0$ the two circles coincide and the shaded region is maximized. Thus this scattering process is strongest when $\mathbf{K}=0$, i.e. when

$$\mathbf{k}\equiv \mathbf{k}_{1}=-\mathbf{k}_{2}.$$ (28)

BCS theory makes the approximation that all electron-electron interactions occur when $\mathbf{K}=0$, which turns out to be a valid approximation for most superconductors. To summarize, in the BCS approximation, the attractive electron-electron interaction only occurs when the two electrons have equal and opposite momentum with a magnitude no more than $\Delta k$ above $k_{F}$.

Now we will be put this information into the Schrödinger equation. The Schrödinger equation for two particles is

$$\frac{-\hbar ^{2}}{2m}(\nabla _{1}^{2}+\nabla _{2}^{2})\psi (\mat... ...bf{r}_{2})+V(\mathbf{r}_{1},\mathbf{r}_{2})\psi (\mathbf{r}_{1},\mathbf{r}_{2}) = E\psi (\mathbf{r}_{1},\mathbf{r}_{2})$$ (29)

$$= (\varepsilon +2E_{F}^{0})\psi (\mathbf{r}_{1},\mathbf{r}_{2}),$$

where $E_{F}^{0}$ is the Fermi energy at $T=0$, $\varepsilon$ is the energy of the two electrons above the Fermi energy, and E is the total energy of the two electrons. Since V is small, we will assume that in first-order approximation, the eigenstates are the same as when $V=0$. Assuming the wavefunction is separable, $\psi (\mathbf{r}_{1},\mathbf{r}_{2})=\psi _{1}(\mathbf{r}_{1})\psi _{2}(\mathbf{r}_{2})$, the eigenstates are plane waves:

$$(\frac{1}{\sqrt{L^{3}}}e^{i\mathbf{k}_{1}\cdot \mathbf{r}_{1}})(\... ...r}_{2}})=(\frac{1}{L^{3}}e^{i\mathbf{k}\cdot (\mathbf{r}_{1}-\mathbf{r}_{2})}),$$ (30)

where $\frac{1}{\sqrt{L^{3}}}$ is from normalizing over the volume. Since the form of $\psi _{1}$ and $\psi _{2}$ in the above equation is symmetric and fermions must have anti-symmetric wavefunctions, the spin matrices for the two electrons must be anti-symmetric, so the two electrons have opposite spins. Because of the simplification in equation (28), this wavefunction can be reduced from two variables $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$ to one variable, $\mathbf{r}\equiv \mathbf{r}_{1}-\mathbf{r}_{2}$:

$$\frac{1}{L^{3}}e^{i\mathbf{k}\cdot \mathbf{r}}$$ (31)

The reduction from two variables to one variable is equivalent to switching to the center of mass in a central force problem in classical mechanics. The most general solution to the Schrödinger equation is a superposition of these eigenstates,

$$\psi (\mathbf{r})=\frac{1}{L^{3}}\sum _{k}g(k)e^{i\mathbf{k}\cdot \mathbf{r}},$$ (32)

where $g(k)$ is function that gives the probability amplitude of finding $\psi (\mathbf{r})$ in a momentum eigenstate $\frac{1}{L^{3}}e^{i\mathbf{k}\cdot \mathbf{r}}$. The function $g(k)$ must be zero for k-values whose momentum is outside the boundary range in equation (27):

$$\begin{array}{ccc} g(k)=0 & for & \left\{ \begin{array}{c} k<k_{F}\\ \\ k<k_{F}+\Delta k\\ \end{array} \right.\\ \end{array}$$ (33)

Substituting the general solution in equation (32) back into the Schrödinger in equation (29) yields,

$$\frac{-\hbar ^{2}}{2m}(\nabla _{1}^{2}+\nabla _{2}^{2})\frac{1}{L... ...E_{F}^{0})\frac{1}{L^{3}}\sum _{\mathbf{k}}g(k)e^{i\mathbf{k}\cdot \mathbf{r}}.$$ (34)

The eigenstates are normal to each other, so

$$\begin{array}{cc} \int e^{i\mathbf{k}\cdot \mathbf{r}}*e^{-i\... ... \mathbf{k}'\\ \end{array} \\ \end{array} \right.\\ \end{array}$$ (35)

The result of $L^{3}$ is because when $\mathbf{k}=\mathbf{k}^{'}$ , the integrand becomes one and the integral is just $\int d\mathbf{r}$, a volume integral. By multiplying equation (34) by $e^{-i\mathbf{k}'\cdot \mathbf{r}}$ and taking the integral with respect to $\mathbf{r}$, the first and last summation terms go to zero except when $\mathbf{k}^{'}=\mathbf{k}$. Additionally $(\nabla _{1}^{2}+\nabla _{2}^{2})e^{i\mathbf{k}\cdot \mathbf{r}}=2k^{2}e^{i\mathbf{k}\cdot \mathbf{r}}$ since the $\nabla ^{2}$ operator is just a second derivative, so equation (34) becomes

$$\frac{\hbar ^{2}k^{2}}{m}g(k')+\frac{1}{L^{3}}\int V(\mathbf{r})\... ...hbf{k}-\mathbf{k}')\cdot \mathbf{r}}d\mathbf{r}=(\varepsilon +2E_{F}^{0})g(k'),$$ (36)

which can be written as

$$\frac{\hbar ^{2}k^{2}}{m}g(k')+\frac{1}{L^{3}}\sum _{\mathbf{k}}V_{k'k}*g(k)=(\varepsilon +2E_{F}^{0})g(k')$$ (37)

$$\begin{array}{cc} \text {where} & V_{k'k}=\int V(\mathbf{r})e^{i(\mathbf{k}-\mathbf{k}')\cdot \mathbf{r}}\\ \end{array} dr$$ (38)

$V_{k'k}$ is the potential energy from the interaction between the two electrons. As we saw in equation (33), this interaction is confined to a narrow range of momentum states, $k_{F}\leq k\leq k_{F}+\Delta k$, so $V_{k'k}$ is zero for momentum states outside this range. Because the momentum range is narrow, BCS theory makes the approximation that $V_{k'k}$ is constant:

$$V_{k'k}=\left\{ \begin{array}{ccccc} -V_{0} & \text {for} & k_{F}... ...k_{F}+\Delta k\\ & & & & \\ 0 & \text {otherwise,} & & & \\ \end{array} \right.$$ (39)

where $V_{0}$ is positive¹⁸.

A quick summary: assuming that the interaction between two electrons is positive but limited to when the electrons have equal and opposite momentums, substituting the most general solution to the Schrödinger equation when $V=0$ back into the full Schrödinger equation, yields equation (37). Assuming that the attractive interaction is limited to a narrow energy range and is constant in that narrow energy range yields equation (39).

Rearranging equation (37) yields,

$$\left(\frac{-\hbar ^{2}k^{2}}{m}+\varepsilon +2E_{F}^{0}\right)g(k)=-\frac{V_{0}}{L^{3}}\sum _{\mathbf{k}}g(k).$$ (40)

Remembering that $\sum _{\mathbf{k}}g(k)$ is a constant, the above equation can be rewritten as

$$1=\frac{V_{0}}{L^{3}}\sum _{\mathbf{k}}\frac{1}{\hbar ^{2}k^{2}/m-\varepsilon -2E_{F}^{0}}$$ (41)

Assuming that the above summation is roughly constant between adjacent terms in the sum, we can approximate the summation as an integral. Remembering that $\mathbf{k}$ is a three dimensional vector, and that the distance between each point in k-space is $\frac{2\pi }{L}$, the volume of each point in k-space is $(\frac{2\pi }{L})^{3}$, so

$$\sum _{\mathbf{k}}\rightarrow \frac{L^{3}}{(2\pi )^{3}}\int d\mathbf{k}.$$ (42)

Equation (41) then becomes,

$$1=V_{0}\frac{1}{(2\pi )^{3}}\int _{k_{F}}^{k_{F}+\Delta k}\frac{d\mathbf{k}}{\hbar ^{2}k^{2}/m-\varepsilon -2E_{F}^{0}}$$ (43)

Because the integral is spherically symmetric, the volume integral $d\mathbf{k}$ can be replaced by a one-dimensional integral $dk$ by multiplying by the density of states, $Z(k)$:¹⁹

$$1=V_{0}\int _{k_{F}}^{k_{F}+\Delta k}Z(k)\frac{dk}{\hbar ^{2}k^{2}/m-\varepsilon -2E_{F}^{0}}$$ (44)

Expressing the integral in terms of energy instead of momentum yields²⁰

$$1=V_{0}\int _{E_{F}^{0}}^{E_{F}^{0}+\hbar \omega _{D}}\frac{Z(E)\: dE}{2E-\varepsilon -2E_{F}^{0}}$$ (45)

Because this integral is over the narrow range of energy near $E_{F}^{0}$ where $V_{0}$ is non-zero, we can make the approximation that the density of states, $Z(E)$, is a constant, so

$$1=V_{0}Z(E_{F}^{0})\int _{E_{F}^{0}}^{E_{F}^{0}+\hbar \omega _{D}... ...}{2}V_{0}Z(E_{F}^{0})\ln (\frac{\varepsilon -2\hbar \omega _{D}}{\varepsilon })$$ (46)

$$\varepsilon =\frac{2\hbar \omega _{D}}{1-e^{-2/(V_{0}Z(E_{F}^{0}))}}$$

For a weak interaction, $V_{0}Z(E_{F}^{0})\ll 1$,

$$\varepsilon \approx -2\hbar \omega _{D}e^{-2/V_{0}Z(E_{F}^{0})}$$ (47)

The negative sign means that the energy of the two extra electrons is lower than the Fermi surface, and inside the filled Fermi sea. This apparent violation of the Pauli exclusion principle comes from the fact that the two electrons become correlated by their attractive force and act as a single pseudo-particle, called a Cooper pair, which is a boson. Bosons do not obey the Pauli exclusion principle, so the Cooper pair is able to achieve an energy inside the filled Fermi sea.

Notice the form of equation (47): it is not an analytic function at $V=0$ so it cannot be expanded in terms of powers of V. Therefore, equation (47) cannot be obtained by perturbation theory, which helps to explain why it took five years from the discovery of an attractive force between the electrons to achieve this result.

5.2.0.1 Summary

At $T=0$, the Fermi sea is filled. Adding two electrons with opposite momenta and energy just above the Fermi sea with an attractive force between them causes the two electrons to form a Cooper pair and to have a collective energy below the Fermi surface. This means that, at sufficiently low temperatures, the Fermi sea is unstable since it is energetically favorable for some electrons to form Cooper pairs.

5.3 The BCS Ground State²¹

With an attractive force between electrons, the Fermi sea becomes unstable and will form Cooper pairs to minimize energy. Determining exactly how many Cooper pairs are formed is a difficult task because it involves comparing the relative energy loss from the formation of an additional Cooper pair (see equation 47) with the energy gain from creating holes in the Fermi sea. Using first-order perturbation theory, we will find out how many Cooper pairs are formed to minimize energy and calculate the ground state energy of a superconductor .

At $T=0$, the only kinetic energy is from the Cooper pairs, so the total kinetic energy is simply the sum over all states of the kinetic energy (energy above the Fermi surface) for each wavevector, weighted by probability of having an Cooper pair in that state ($w_{k}$) (with a factor of two because each Cooper pair contains two electrons):

$$E_{kin}=2\sum _{\mathbf{k}}=w_{k}\xi _{k}, \qquad \text {where}\; \quad \xi _{k}=\frac{\hbar ^{2}k^{2}}{m}-E_{F}^{0}.$$ (48)

But this assumes that the energy of formation of each Cooper energy pair is independent of the number of the Cooper pairs. We can approximate the energy shift from interactions between Cooper pairs and the Fermi sea with first-order perturbation theory. We will treat the scattering of Copper pairs from $(\mathbf{k}\uparrow ,-\mathbf{k}\downarrow )$ to $(\mathbf{k}'\uparrow ,-\mathbf{k}'\downarrow )$ as the annihilation of a Cooper pair in the state $(\mathbf{k}\uparrow ,-\mathbf{k}\downarrow )$ and the creation of a Cooper pair in the state $(\mathbf{k}'\uparrow ,-\mathbf{k}'\downarrow )$. Let us define some notation: for each wavevector, a Cooper pair can either exist or not exist. We will denote $\vert\rangle _{\mathbf{k}}$ for an unoccupied state and $\vert 1\rangle _{\mathbf{k}}$ for an occupied state. The most general state for a pair with wavevector $\mathbf{k}$ is a superposition of the two states.

$$\vert\psi \rangle _{\mathbf{k}}=\mu _{\mathbf{k}}\vert\rangle _{\mathbf{k}}+\nu _{\mathbf{k}}\vert 1\rangle _{\mathbf{k}},$$ (49)

where $\mu$ and $\nu$ are probability amplitudes²², so $\nu _{\mathbf{k}}^{2}=w_{\mathbf{k}}$ and $\mu _{\mathbf{k}}^{2}+\nu _{\mathbf{k}}^{2}=1$. If we assume that the Cooper pairs do not interact with each other, then the eigenfunctions of the system are unchanged, and we can approximate the ground state wavefunction as

$$\vert\Phi _{BCS}\rangle \simeq \prod _{\mathbf{k}}(\mu _{\mathbf{k}}\vert\rangle _{\mathbf{k}}+\nu _{\mathbf{k}}\vert 1\rangle _{\mathbf{k}}).$$ (50)

We can express our state vectors in matrix form

$$\vert\rangle _{\mathbf{k}}=\left(\begin{array}{c} 0\\ 1\\ \end{ar... ...e _{\mathbf{k}}=\left(\begin{array}{c} 1\\ 0\\ \end{array} \right)_{\mathbf{k}}$$ (51)

Using the Pauli matrices

$$\sigma _{\mathbf{k}}^{(1)}=\left(\begin{array}{cc} 0 & 1\\ 1 & 0\... ...{\mathbf{k}}^{(2)}=\left(\begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array} \right),$$ (52)

creation and annihilation operators can be defined:

Notice that the matrices have the same form as spin operators although they represent different actions. Working out the matrix multiplication one can verify the following relationships:

$$\sigma _{\mathbf{k}}^{+}\vert 1\rangle =0,\quad \; \; \sigma _{\m... ...rt 1\rangle =\vert\rangle ,\quad \; \; \sigma _{\mathbf{k}}^{-}\vert\rangle =0.$$ (55)

The creation operator acting on a filled state destroys the state, the creation operator action on an empty state fills the state, the annihilation operator acting on a filled state empties the state, and the annihilation operator acting on an empty state destroys the state. In operator form, the Hamiltonian for the scattering of Cooper pairs from ( $\mathbf{k}\uparrow -,\mathbf{k}\downarrow$) to ( $\mathbf{k}'\uparrow -,\mathbf{k}'\downarrow$) is the destruction of the $\mathbf{k}$-state and the creation of the $\mathbf{k}'$-state, summed over all states, and multiplied by the associated energy reduction from the scattering, $-\frac{V_{0}}{L^{3}}$:

$$\mathcal{H}=-\frac{V_{0}}{L^{3}}\sum _{\mathbf{kk}'}\sigma _{\mathbf{k}}^{-}\sigma _{\mathbf{k}'}^{+}$$ (56)

In first-order perturbation theory, $\Delta E=\langle \psi \vert\mathcal{H}\vert\psi \rangle$. To find the change in energy from the effect of forming numerous Cooper pairs, we must calculate $\langle \Phi _{BCS}\vert\mathcal{H}\vert\Phi _{BCS}\rangle$.

$$\langle \Phi _{BCS}\vert\mathcal{H}\vert\Phi _{BCS}\rangle =-\fra... ...mathbf{q}}\rangle +\nu _{\mathbf{q}}\vert 1_{\mathbf{q}}\rangle \right)\right],$$ (57)

which can be simplified (see appendix A) to

$$\langle \Phi _{BCS}\vert\mathcal{H}\vert\Phi _{BCS}\rangle =-\fra... ...bf{kk}'}\mu _{\mathbf{k}}\nu _{\mathbf{k}}\mu _{\mathbf{k}'}\nu _{\mathbf{k}'}.$$ (58)

Adding this energy shift to the kinetic energy of the unperturbed state in equation (48), we get:

$$W_{BCS}=2\sum _{\mathbf{k}}\nu _{\mathbf{k}}^{2}\xi _{\mathbf{k}}... ...bf{kk}'}\mu _{\mathbf{k}}\nu _{\mathbf{k}}\mu _{\mathbf{k}'}\nu _{\mathbf{k}'},$$ (59)

where $W_{BCS}$ is the energy above the Fermi sea caused by Cooper pairing according to BCS theory. Because $\mu _{\mathbf{k}}^{2}+\nu _{\mathbf{k}}^{2}=1$, we can express the two coefficients in terms of a single phase $\theta _{\mathbf{k}}$:

The angle $\theta _{\mathbf{k}}$ represents how many Cooper pairs states are filled. The number of states filled is determined by what minimizes the energy of the system, which occurs when $\frac{\partial W_{BCS}}{\partial \theta _{\mathbf{k}}}=0$. Substituting equation (60) into equation (59) gives us

Therefore,

$$\frac{\partial W_{BCS}}{\partial \theta _{\mathbf{k}}}=-2\xi _{\m... ...^{3}}\sum _{\mathbf{k}'}\cos 2\theta _{\mathbf{k}}\sin 2\theta _{\mathbf{k}'}=0$$

$$\xi _{\mathbf{k}}\tan 2\theta _{\mathbf{k}}=-\frac{1}{2}\frac{V_{0}}{L^{3}}\sum _{\mathbf{k}'}\sin 2\theta _{\mathbf{k}'}.$$ (62)

For convenience, we will define the the following quantity $\Delta$:

$$\Delta \equiv \frac{1}{2}\frac{V_{0}}{L^{3}}\sum _{\mathbf{k}'}\sin 2\theta _{\mathbf{k}'}$$ (63)

Substitution with equation (62) yields,

$$\tan 2\theta _{\mathbf{k}} = -\frac{\Delta }{\xi _{\mathbf{k}}}$$ (64)

Using the trigonometric identity $\cos ^{2}2\theta +\sin ^{2}2\theta =1$:

$$2\mu _{\mathbf{k}}\nu _{\mathbf{k}} = \sin 2\theta _{\mathbf{k}} = \frac{\Delta }{\sqrt{\Delta ^{2}+\xi _{\mathbf{k}}^{2}}}$$ (65)

$$\nu _{\mathbf{k}}^{2}-\mu _{\mathbf{k}}^{2} = \cos 2\theta _{\mathbf{k}} = -\frac{\xi _{\mathbf{k}}}{\sqrt{\Delta ^{2}+\xi _{\mathbf{k}}^{2}}}$$ (66)

Since the probability that a Cooper pair is occupied ( $w_{\mathbf{k}}$) equals $\nu _{\mathbf{k}}^{2}$, we can solve the above equations for $\nu _{\mathbf{k}}$ with the identity $\cos ^{2}\theta =\frac{1}{2}(1+\cos 2\theta )$,

$$w_{\mathbf{k}}=\cos ^{2}\theta _{\mathbf{k}}=\frac{1}{2}\left(1-\frac{\xi _{\mathbf{k}}}{\sqrt{\xi _{\mathbf{k}}^{2}+\Delta ^{2}}}\right).$$ (67)

Substituting the above equation into equation (61) yields the ground BCS state ( $W_{BCS}^{0}$) because it minimizes the energy. The ground state can be simplified to (see appendix B),

$$W_{BCS}^{0}=-2\sum _{\mathbf{k}}\nu _{\mathbf{k}}^{4}\sqrt{\left(\Delta ^{2}+\xi _{\mathbf{k}}^{2}\right)}.$$ (68)

We have calculated the ground energy state of a superconductor. To explain the meaning of $\Delta$ and find the energy gap of a superconductor, we will need to consider the first excited superconducting state, $W_{BCS}^{1}$. Take a state $\mathbf{k}'$ with a Cooper pair ( $\nu _{\mathbf{k}'}=1$); the first excited superconducting state, $W_{BCS}^{1}$, is achieved by breaking that one Cooper pair, i.e. changing $\nu _{\mathbf{k}'}$ from 1 to 0. The new state is identical to the above equation except that there is one less Cooper pair:

$$W_{BCS}^{1}=-2\sum _{\mathbf{k}\neq \mathbf{k}'}\nu _{\mathbf{k}}^{4}\sqrt{\left(\Delta ^{2}+\xi _{\mathbf{k}}^{2}\right)}.$$ (69)

The energy gap between the ground state and the first excited state is (remember for $W_{BCS}^{0}$, $\nu _{\mathbf{k}'}=1$),

$$\Delta E=W_{BCS}^{1}-W_{BCS}^{0}=2\sqrt{\Delta ^{2}+\xi _{\mathbf{k}'}^{2}}.$$ (70)

Remembering that $\xi _{\mathbf{k}}$ is the energy of one electron above the Fermi Surface, we know that $\xi _{\mathbf{k}}$ can be arbitrarily small, thus:

$$\textrm{Energy Gap}=\Delta E_{min}=2\Delta$$ (71)

The minimum energy difference between excited states in the BCS state, $2\Delta$, is the minimum energy required to break one Cooper pair. Since the formation of Cooper pairs lowers the energy of the system, the energy of the BCS ground state $(W_{BCS}^{0})$, should be lower than the normal ground state $(W_{n}^{0})$, which does not have any Cooper pairs. Remembering that the energy of the normal ground state is just the sum of energies of the electrons in the filled states below the Fermi surface,

$$W_{n}^{0}=\sum _{k<k_{f}}\xi _{k},$$ (72)

it can be shown that (see Appendix C)

$$W_{BCS}^{0}-W_{n}^{0}=-\frac{1}{2}Z(E_{F}^{0})\Delta ^{2}L^{3}.$$ (73)

Now we have the tools to explain the basic properties of superconductivity. To quickly summarize, because the metal lattice is deformed when an electron moves near it, there is an increased charge density which attracts nearby electrons and causes certain electrons to be attracted to each other. By conservation of momentum, pairs of electrons with equal and opposite momentum and energy slightly above the Fermi surface are likely to form bound pairs because of this attractive interaction. These electron pairs, called Cooper pairs, behave as bosons and have an energy below the the Fermi surface. Numerous Cooper pairs form until the energy is minimized, resulting in a system with a net energy lower than the normal ground state. The lowest energy state is the BCS ground state, $W_{BCS}^{0}$. The next state is $W_{BCS}^{1}$ with one less Cooper pair and an energy $2\Delta$ higher, then $W_{BCS}^{2}$ with two less Cooper pairs and an energy $4\Delta$ above $W_{BCS}^{0}$, etc. This process continues until the BCS excited state has an energy above $W_{n}^{0}$, at which point there will be a transition to the normal state. As the energy increases, the system will go from $W_{BCS}^{0}$ with numerous Cooper pairs to $W_{BCS}^{1}$ with one less Cooper pair, etc, and then suddenly to $W_{n}^{0}$ with zero Cooper pairs. Since the transition is governed by Bose-Einstein statistics and occurs at low temperature, the jump from numerous Cooper pairs to zero is very steep. The Cooper pairs, bosons, are in one macroscopic quantum state, so when an electric current enters one end of the state, it will simply exit the other side of the macroscopic quantum state with no energy loss because the electron is not interacting with millions of electrons and atoms, but with the macroscopic quantum state; Cooper pairs are the carriers of superconductivity. The steepness of the transition to superconductivity (see Onnes' original data in figure 3) is because of the steepness of the transition from numerous Cooper pairs to zero Cooper pairs. At temperatures above $T_{c}$, the system is excited above the BCS states that have Cooper pairs. Additionally, since Cooper pairs have opposite spin, an external magnetic field will try to align these spins, raising the energy of the Cooper pairs until at some field strength, it is never energetically favorable to form Cooper pairs.

5.4 Experimental Verification²³

Calculating the numerous experimental predictions of BCS theory is a complicated task and beyond the scope of this paper, but BCS theory has had remarkable success in predicting numerous characteristics of many superconductors. One example of the close fit between theory and experiment is shown in figure 11.

Figure 11: BCS theory and experiment: the measured and predicted temperature dependence of the energy gap $\Delta (T)$ relative to $\Delta (0)$ for Indium, Tin and Lead.
Source: Ibach and Lüth (1996), page 243.

Superconductors that have strong electron-phonon interactions, however, are not precisely treated by some of the assumptions and approximations presented here and in the original formation of BCS theory. There are measurable discrepancies between theory and experiment for these kinds of superconductors, but more mathematically rigorous methods of applying BCS theory have been able to account for most of these discrepancies.

Footnotes

... Attraction¹⁶

This explanation comes from Ibach and Lüth (1996, p.228-233), and Buckel (1991, p.43-47).

...sub:Cooper-Pairing¹⁷

The explanation and derivation presented here is based almost entirely on Ibach and Lüth (1996, p.228-233). A similar but somewhat less accessible derivation is given in Tinkham (1996, p.44-46).

... positive¹⁸

An attractive force has a negative potential energy, so $V_{k'k}$ is negative.

...:¹⁹

Note that $Z(k)$ absorbs the constant $\frac{1}{(2\pi )^{3}}$.

... yields²⁰

The density of states can be expressed as function of momentum $Z(k)$, or a function of energy $Z(E)$. Both functions are equivalent, but in terms of different inputs.

... State²¹

This derivation is based almost entirely on Ibach and Lüth (1996, p.233-242). A similar but somewhat less accessible treatise is given in Tinkham (1996, p.48-62).

... amplitudes²²

We will assume that $\mu _{k}$ and $\nu _{k}$ are real, but in a more rigorous derivation, this assumption is not necessary.

... Verification²³

Source: Ibach and Lüth (1996, p.242-246).