Question:In Chapter 4. we learned that the uncertainty principle is a powerful tool. Here we use it to estimate the size of a Cooper pair from its binding energy. Due to their phonon-borne attraction, each electron in a pair (if not the pair's center of mass) has changing momentum and kinetic energy. Simple differentiation will relate uncertainty in kinetic energy to uncertainty in momentum, and a rough numerical measure of the uncertainty in the kinetic energy is the Cooper-pair binding energy. Obtain a rough estimate of the physical extent of the electron's (unknown!) wave function. In addition to the binding energy, you will need to know the Fermi energy. (As noted in Section 10.9, each electron in the pair has an energy of about EF.) Use 10^-3 eV and 9.4 eV, respectively, values appropriate for indium.

The kinetic energy in terms of momentum can be expressed as, $KE=\frac{p^2}{2m}$.

We know on basis of Heisenberg uncertainity relation that product of uncertainties of the momentum and the position is expressed as below.

$\mathit{\Delta }\mathit{x}\mathbf{·}\mathit{\Delta }\mathit{p}\mathbf{⩾}\frac{\mathbf{h}}{\mathbf{2}}$

Where, h represents reduced Planck's constant,$\mathit{\Delta }\mathit{p}$ represents uncertainty in momentum and $\mathit{\Delta }\mathit{x}$represents uncertainty in position.

The kinetic energy in terms of momentum can be expressed as, $KE=\frac{p^2}{2m}$.

$p=\sqrt{2\left(KE\right)\left(m\right)}$

Thus, the momentum can be expressed in terms of kinetic energy as, .

Take the differentiation on both sides of the above equation.

$$\begin{gathered} \Delta KE=\frac{2p}{2m}\Delta p \\ =\frac{p}{m}\Delta p \end{gathered}$$

Thus, the change in moment can be expressed,$\Delta p=\frac{m\Delta KE}{p}$.

Uncertainty principle states that tile product of uncertainties of momentum and position is order of Planck's constant.

It can be expressed as, $\Delta x⩾\frac{h}{\Delta p}$.

Therefore, the size of the Cooper pair (or uncertainty in position) can be expressed as, $\Delta x·\Delta p⩾h$.

Substitute the equation $\Delta p=\frac{m\Delta KE}{p}$ in the above equation.

$$\begin{gathered} \Delta x⩾\frac{h}{\left(\frac{m\Delta KE}{p}\right)} \\ ⩾\frac{ph}{m\Delta KE} \\ ⩾\frac{h\left[\sqrt{2\left(KE\right)\left(m\right)}\right]}{2m\Delta KE} \end{gathered}$$

So the size of the Cooper pair is given by:

$$\begin{gathered} \Delta x=\frac{1.055\times {10}^{-34} \text{Js}\sqrt{2(9.2 \text{eV})\left(1.6\times {10}^{-19} \text{JleV}\right)\left(9.11\times {10}^{-31} \text{kg}\right)}}{\left(9.1\times {10}^{-31} \text{kg}\right)\left({10}^{-3} \text{eV}\right)\left(1.6\times {10}^{-19} \text{JleV}\right)} \\ =1.0\times {10}^{-6} \text{m}\left(\frac{\text{l}\mu \text{m}}{{10}^{-6} \text{m}}\right) \\ =1 \mu \text{m} \end{gathered}$$

The size of the cooper pair is $1 \mu \text{m}$.