Condensed Matter VI « Physics Made Easy

Condensed Matter VI

Diamagnetism arises from closed atomic shells of electrons. When a B-field is applied, these electrons set up a screening current that opposes the applied field.

Start by thinking of electrons in a circular orbit of radius ρ in the xy plane.

Recall

$\nabla \times \bold{E}=\frac{-\partial \bold{B}}{\partial t}$

$\int |\nabla \times \bold{E}| dA=-A\frac{-\partial \bold{B}}{\partial t}$

$\oint \bold{E}.d\bold{l}=-A\frac{-\partial \bold{B}}{\partial t}$

Flux $\phi=BA$

$\frac{\partial \phi}{\partial t}=A\frac{-\partial \bold{B}}{\partial t}$

$\frac{\partial \phi}{\partial t}= \oint \bold{E}.d\bold{l}$

$\frac{\partial \phi}{\partial t}= -2\pi \rho E=V$

Force on an electron $\frac{dp}{dt}=-eE$

$\frac{dv}{dt}=\frac{-eE}{m}$

$\frac{dv}{dt}=\frac{-eV}{2\pi \rho m}$

$\frac{dv}{dt}=\frac{-eA}{2\pi \rho m}\frac{\partial \bold{B}}{\partial t}$

$\Delta v=\frac{-eA}{2\pi \rho m}B$

Current due to Z electrons in an atom $I=\frac{-Ze}{\tau}=\frac{-Ze\Delta v}{l}$

So $I=\frac{-Ze^2AB}{(2\pi \rho)^2m}$

Magnetisation per atom $M=IA$

$A=\pi \rho^2$ where $\rho^2=x^2+y^2=\frac{2}{3}r^2$ for a spherically symmetrical orbit (note that strictly we should be using the average values <ρ> and <r>).

So $M=\frac{-Ze^2A^2B}{4\pi^2\rho^2 m}=\frac{-Ze^2r^2B}{6m}$

Total magnetisation for N atoms per unit volume $M=\chi H=\frac{\chi B}{\mu_0}=\frac{-Ze^2r^2BN}{6m}$

-> diamagnetic susceptibility $\chi=\frac{-Ze^2\mu_0r^2N}{6m}$

Again note that strictly not all atoms will have the same r² so really we need to consider $\sum_{n=1}^Nr_n^2$

Gyromagnetic ratio, μ/l

$\mu=IA$

$\mu=\frac{-e}{\tau}\pi r^2$ classical

$\mu=\frac{-ev}{2\pi r}\pi r^2$

$\mu=\frac{-e}{2m}mvr$

$\mu=\frac{-e}{2m}l$

Similarly, in quantum mechanics $\bold{\mu}=\frac{-e}{2m}\bold{l}$

Magnetic levitation can occur when magnetic force balances gravity

U_m=-M.B

$\frac{-\partial U_m}{\partial x}=V\rho g$

$M\frac{-\partial B}{\partial x}=V\rho g$

$\frac{V\chi B}{\mu_0}\frac{-\partial B}{\partial x}=V\rho g$

$\frac{\chi B}{\mu_0}\frac{-\partial B}{\partial x}=\rho g$

Using this principle, scientists have managed to levitate Brazilian tree frogs, which isn’t as impressive as it sounds when you realise how small they are. Unfortunately, the need for a magnetic field gradient makes it hard to levitate larger items, so there won’t be any floating elephants just yet.

Paramagnetism: unfilled atomic shells have an overall magnetic moment $\bold{\mu}=-g_J\mu_B\bold{J}$ which aligns along the direction of an applied field.

States with different M_J are shifted in energy by $\Delta E=-<\bold{\mu}.\bold{B}>=g_J\mu_BM_JB$

For now, we will consider only spin-1/2 systems.

Take N spin-1/2 atoms per unit volume- J=1/2, g_J=2

Boltzmann factors

$\frac{N_2}{N}=\frac{e^{\frac{-\mu_B B}{kT}}}{ e^{\frac{+\mu_B B}{kT}}+ e^{\frac{-\mu_B B}{kT}}}$

$\frac{N_1}{N}=\frac{e^{\frac{\mu_B B}{kT}}}{ e^{\frac{+\mu_B B}{kT}}+ e^{\frac{-\mu_B B}{kT}}}$

Total magnetisation

$M= \mu_B B(N_1-N_2)$

$M=\mu_B B N\frac{e^{\frac{\mu_B B}{kT}}-{e^{\frac{-\mu_B B}{kT}}}{e^{\frac{\mu_B B}{kT}}+{e^{\frac{-\mu_B B}{kT}}}</p> <p>$ latex \frac{\chi B}{\mu_0}=\mu_B B N \tanh\frac{\mu_B B}{kT}$

In the high-temperature, low-field limit, $\tanh\frac{\mu_B B}{kT} \rightarrow \frac{\mu_B B}{kT}$

-> $\chi=\frac{\mu_0\mu_B^2N}{kT}=\frac{C}{T}$ Curie’s Law

Collective magnetisation: so far we have considered the atoms individually and not worried about interactions between them. The next types of magnetism all arise because of interactions of atoms and so fall under the heading of collective magnetism.

Magnetic dipole-dipole interactions: neighbouring atoms exert a force on each other which tries to align dipoles m.

Interaction energy $U=-\bold{m}.\bold{b}$

Take $U\simeq \frac{-\mu_0\bold{m}_1.\bold{m}_2}{4\pi r^3}$

For the purposes of making a quick calculation, we’ll use typical values $\bold{m}_1=\bold{m}_2=\mu_B$ and r~0.3nm, giving us U~10^-24 J

This corresponds to an ordering temperature of tenths or hundredths of Kelvin (i.e. very small). At room temperature we wouldn’t expect the spins to be aligned because of magnetic dipole interactions- and yet we do see magnetism at room temperature, so we need to look for other mechanisms.

Exchange interaction: this is the reason why spins align at temperatures as high as 1000K.

Consider two adjacent atoms and two electrons with a total wavefunction $\psi(r_1, r_2, s_1, s_2)$

Electrons are fermions, so \psi(r_1, r_2, s_1, s_2)= -\psi(r_2, r_1, s_2, s_1)$

Electrons with the same spin ‘repel’ each other and form symmetric and antisymmetric wavefunctions.

What we end up with are

Singlet $\psi_s(1,2)=[\phi_a(r_1) \phi_b(r_2)+ \phi_a(r_2) \phi_b(r_1)]\chi_s(\uparrow\downarrow)$ S=0

Triplet $\psi_t(1,2)=[\phi_a(r_1) \phi_b(r_2)-\phi_a(r_2) \phi_b(r_1)]\chi_t(\uparrow\uparrow)$ S=1, M_S=0,±1

The triplet energy lies below that of the singlet. The energy difference between the singlet and triplet is $U=-2J\bold{S}_1\bold{S}_2$ where J is the exchange integral.

More will be added on the exchange integral at a later date.

Ferromagnetism and the molecular field model: ferromagnetism occurs when the exchange integral J>0 (electron-electron coupling). Spins align to create a spontaneous magnetisation at a critical temperature- we can see this using the molecular field (also called the mean field) model.

Say that the atoms look like they are in an effective flux $B_{eff}=B_0+\mu_0\lambda M$

(first term on RHS is an external field, second is a molecular ‘field’ due to exchange interactions). Note that B_eff is not a real magnetic field that we can measure, it’s just a way of thinking about these interactions.

Assume that Curie’s Law holds for the total field B_eff.

$\mu_0M=\frac{CB_{eff}}{T}=\frac{C(B_0+\mu_0\lambda M)}{T}$

$\mu_0M(T-\lambda C)=CB_0$

$\mu_0M=\frac{CB_0}{T-C\lambda}$

$\chi=\frac{\mu_0M}{B_0}=\frac{C}{T-C\lambda}=\frac{C}{T-T_C}$

T_C is the critical temperature where we see a spontaneous magnetisation.

Domains: pure iron only magnetises in the presence of an external field. Why don’t all the spins line up spontaneously in zero field?

We need to consider the three types of energy in a magnetic crystal.

Magnetostatic energy = $\int B_0dM$ . This will be large if all spins are lined up in the same direction.
Anisotropy energy: spins like to line up along particular crystal directions. There is an energy cost for any spins not aligned with this easy axis.
Exchange energy = $-\sum J\bold{S}_1.\bold{S}_2$ . There is an energy cost whenever spins are not aligned.

All spins are aligned along easy axis but magnetostatic energy is large.
Magnetostatic energy is reduced.
Further reduction of magnetostatic energy.
All magnetostatic energy now contained within the crystal but now there are lots of atoms perpendicular to the easy axis.
Better still- now only a few atoms perpendicular to the easy axis.

So why doesn’t this just go on until the domains get infinitely small? To answer this, we need to think about domain walls.

Domain walls (Bloch walls) look like this:

Most of the spins in the wall are not aligned with the easy axis; also, adjacent spins are at an angle θ to each other, so their exchange energy is higher. This means that each domain wall costs energy. What we have to do is find an optimum where we are not ‘spending too much’ on any one type of energy.

Thickness of domain walls: energy is minimised by changing the spin slowly in N steps by a small angle $\theta=\frac{\pi}{N}$ .

For each line of spins, exchange energy cost $U=-\sum 2J\bold{S}_1.\bold{S}_2$

$\Delta U=2NJS^2(1-\cos\theta)$

(here we’re considering the difference in exchange energy when the N spins are at angle θ instead of being aligned).

θ is small, so do a power series expansion.

$\Delta U=2NJS^2(1-1+\frac{\theta^2}{2})$

$\Delta U=\frac{JS^2\pi^2}{N}$ per line of atoms.

We have $\frac{1}{a^2}$ lines per unit area (a is just the atomic spacing).

$\Delta U=\frac{JS^2\pi^2}{Na^2}$

Anisotropy energy cost = $\frac{aKN}{2}$

Total energy cost $U_{tot}=\frac{JS^2\pi^2}{Na^2}+\frac{aKN}{2}$

$\frac{\partial U_{tot}}{\partial N}=\frac{-JS^2\pi^2}{N^2a^2}+\frac{aK}{2}$

So U_tot is a minimum when $N=(\frac{2JS^2\pi^2}{Ka^3})^{\frac{1}{2}}$

For iron, a domain wall will be around 300 atoms thick (or 300a metres thick).

Impurities and hard magnets: pure iron is a soft magnetic material, i.e. it does not retain its magnetisation in zero field. In order to pin the domain walls and keep the magnetisation, we need to introduce impurities, e.g. carbon in steel.

Magnetisation of a hard magnet

Reversible boundary displacement- domain boundaries move a little but they don’t hit impurities so they can easily go back to their original position.
Irreversible boundary displacement- the applied field forces the domain walls through impurities. When the field is removed, they can’t pass back through the impurities.
Magnetisation rotation- all spins line up with the field, i.e. saturation.

Because (2) is irreversible, the system has a hysteresis and an overall magnetisation is retained when the applied field is reduced to zero. A reverse field is then required to demagnetise the material.

Antiferromagnetism: if the exchange integral J<0, adjacent spins line up antiparallel to each other- this is antiferromagnetism. To model this, we use the molecular field model, but this time we have two sublattices, A and B, each with N/2 atoms.

Mean field at A due to B $B_A=-\mu_0\lambda M_B$

Mean field at B due to A $B_B=-\mu_0\lambda M_A$

Total magnetisation $M=M_A+M_B$

Applied field B₀

$B_{eff}^A=B_0-\mu_0\lambda M_B$

$B_{eff}^B=B_0-\mu_0\lambda M_A$

As before, assume Curie’s Law holds.

$\mu_0 M_A=\frac{CB_{eff}^A}{2T}=\frac{C}{2T}(B_0-\mu_0\lambda M_B)$

$\mu_0 M_B=\frac{C}{2T}(B_0-\mu_0\lambda M_A)$

$\mu_0(M_A+M_B)=\frac{C}{2T}\{ 2B_0-\mu_0\lambda(M_A+M_B)\}$

$\mu_0 M(T+\frac{c\lambda}{2})=CB_0$

$\mu_0 M=\frac{CB_0}{ T+\frac{c\lambda}{2}}$

$\mu_0 M=\frac{CB_0}{ T+T_N}$ where $T_N=\frac{C\lambda}{2} is the Néel temperature.</p> <p class=\"MsoNormal\">So$ latex \chi=\frac{\mu_0 M}{B_0}=\frac{C}{T+T_N}$

Antiferromagnets have an anisotropic susceptibility:

Vertical B-field has little effect

Horizontal B-field ->

Magnetic resonance experiments: the basic idea is to split up the atomic energy levels by applying a magnetic field and then observe the transitions between them.

$\Delta E=-\bold{\mu}.\bold{b}=g_J\mu_BM_JB$

We’ll use a J=3/2 level for this example.

Selection rule $\Delta M_J=\pm 1$ for ΔJ=0.

Expect to see three peaks corresponding to the three allowed transitions.

You may not expect there to be a splitting at zero field, but remember that this isn’t an isolated atom- it is in a crystal, where there are other perturbations to consider.

Apparatus- Nuclear Magnetic Resonance (NMR)

For ESR (electron spin resonance), we are looking at atomic rather than nuclear transitions so we need a frequency in the GHz region. To achieve this, replace the coil with a cavity resonator. We also need to cool the sample to a low temperature to see a signal.

In both cases, either sweep the frequency through resonance whilst keeping the B-field constant, or vice versa.

Adiabatic demagnetisation (used for cooling): we can cool a solid containing magnetic ions by using a magnetic field. To do this, we apply a field, and then remove it adiabatically (keeping the populations of spin states the same).

$\frac{kT_f}{u_BB_f}=\frac{kT_i}{u_BB_i}$ T_f is limited by small interactions which split energy levels at T=0.

Superconductivity: in a superconductor, resistivity goes to zero below the critical temperature T_C. But a superconductor is not just a perfect conductor, it is also a perfect diamagnet (χ=-1)- it expels magnetic flux (Meissner effect).

Superconductivity can be destroyed by a critical magnetic field B_C or critical current J_C.

Meissner effect and flux expulsion: when a superconductor is cooled below T_C (which is usually in the range of mK to ~160K), surface currents are set up which expel magnetic flux. For a zero resistance metal cooled below T_C, the flux becomes trapped rather than expelled.

Type I superconductors

Sharp transition to normal state at B_C.

Type II superconductors e.g. Nb

Superconducting state.
Vortex state: a few lines of flux get through the superconductor surrounded by helical screening currents.

Type II are of more practical use.

Penetration depth: from London theory, surface currents go as $j=j_0e^{-\lambda x}$ where $\lambda=\sqrt{\frac{m}{\mu_0ne^2}}$ (penetration depth). Magnetisation energy occurs over this penetration depth.

BCS theory, Cooper pairs and energy gap: specific heat, IR absorption and tunnelling all indicate that superconductivity is related to an energy gap (this idea will be discussed further a bit later).

Electrons experience an attraction caused by the interaction with the crystal lattice, leading to binding in pairs (Cooper pairs). Electrons of wavevector k₁ and k₂ can exchange virtual phonons. This interaction is strongest when k₁= -k₂, so electrons bind together in pairs with momenta k_F and –k_F.

Wavefunction $\phi=\phi_s(r_1,r_2)\frac{1}{\sqrt{2}}(\uparrow\downarrow-\downarrow\uparrow)$ singlet

The pair has charge 2e, mass 2m and binding energy Δ per electron.

Pairs are destroyed when there is enough energy to excite electrons across the gap.

We have a perfect conductor because no scattering can occur until there is sufficient energy to excite pairs across the gap.

Coherence length: $\xi=v_F\tau$ say $\frac{\bar{h}}{\tau}=2\Delta$ (roughly)

So $\xi \simeq \frac{\bar{h}v_F}{2\Delta}=1-1000 nm$

Flux quantisation

Current density $\bold{j}(\bold{r})=\frac{i\bar{h}e}{2m}(\psi*\nabla\psi-\psi\nabla\psi*)-\frac{2e^2}{m}\psi*\psi\bold{A}$

General wavefunction of a Cooper pair, mass 2m, charge 2e

$\psi(\bold{r})=|\psi(\bold{r})|e^{i\theta(\bold{r})}$

-> \bold{j}(\bold{r})=\frac{-e}{2m}|\psi(\bold{r})|^2(\bar{h}\nabla \theta+2e\bold{A}$

Far from the surface of a superconductor in its Meissner state, j=0, so

$\bar{h}\nabla \theta=-2e\bold{A}$

Integrate around closed curve C inside superconductor.

$\bar{h}\oint_C \nabla\theta.d\bold{l}=-2e\oint_C\bold{A}.d\bold{l}$

$\bar{h}\Delta\theta=-2e\oint(\nabla \times \bold{A}).d\bold{S}$

$\bar{h}\Delta\theta=-2e\oint \bold{B}).d\bold{S}$

$\bar{h}\Delta\theta=-2e \Phi$

ψ(r) is single-valued, so around a closed loop, $\Delta\theta=2n\pi$

-> \Phi=\pm\frac{2\pi n \bar{h}}{2e}=\pm\frac{nh}{2e}=\pm n\Phi_0$

where Φ₀=2.07×10^-5Tm² is the flux quantum. Flux around a closed loop is quantised in units of Φ₀.

Evidence for 2Δ band gap in superconductors

Infrared absorption. Analogous to results for semiconductor band gap; absorption only occurs when hν>2Δ
Tunelling between two superconductors with a thin barrier (~10^-9m)

Tunnel current shows features due to alignment of energy levels either side of the barrier- measures energy gaps.

Form of heat capacity indicates a 2Δ band gap- $C \propto e^{\frac{-\Delta}{kT}}$ below T_C.

Experiment to measure the flux quantum

Place sample in a magnetic field of ~10μT.

Cool sample through T_C.

Vibrate sample between search coils to find the trapped flux.

Physics Made Easy

Condensed Matter VI

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