Positive Hall coefficient in metals

[Moreno]

How Positive Hall coefficient in metals could be explained?

[Enthalpy]

It happens in materials where electrons have a negative mass, in metals just as in semiconductors.

Edited July 11, 2016 by Enthalpy

[Moreno]

It happens in materials where electrons have a negative mass, in metals just as in semiconductors.

 

I have doubt in it. A positive Hall coefficient could be often found in pure metals. For example in Bismuth. Does it mean electrons have negative effective mass in Bismuth? As well as in Zinc, Cadmium etc. I have large doubts in it.

[studiot]

The simple expectation for the Hall coefficient and conductivity/resistivity is based on the 'free electron theory' or Drude theory.

This theory is corpuscular and assumes that the charge carriers act like gaseous atoms in the kinetic theory.

Effects such as Hall and conductivity are dependent upon the number/density of carriers, regarded as individual corpuscles and independent of the effects of the lattice and wave mechanics.

 

This theory is only really successful for alkali metals and unsuitable for other materials.

 

Better theories from wave/quantum mechanics show that the De Broglie wave suffers interference in the lattice such that, although its main component is in the direction of the corpuscular vector, the wave also has components in other directions.

 

The main component is perpendicular to the imposed magnetic field, but the others are not and so are affected by it differently, leading to a skewing / diminishing / reversal of the Hall voltage.

 

This is the best easy explanation I can offer.

[Enthalpy]

 

I have doubt in it. A positive Hall coefficient could be often found in pure metals. For example in Bismuth. Does it mean electrons have negative effective mass in Bismuth? As well as in Zinc, Cadmium etc. I have large doubts in it.

 

It's the good and only reason. But feel free to doubt it: your problem, not mine.

[studiot]

 

It's the good and only reason. But feel free to doubt it: your problem, not mine.

 

Negative effective mass is not the reason it's the consequence.

[Enthalpy]

Hall voltage as a consequence of the effective mass sounds reasonable. But as a cause? Do you claim that a Hall voltage is necessary to observe other consequences of the effective mass?

[studiot]

I would say that Moreno has been chased off by your condescending attitude, when you could have offered explanation.

 

I will take the first part of your last post as a reasonable question and try to expand the explanation I gave before.

 

 

 

But as a cause?

 

The mechanics of the 'free electron model' is Newtonian in that the response of a particle is proportional to an externally applied force.

The constant of proportionality is called the mass and is the same for all sorts of applied forces in all sorts of situations.

 

Electrons in a lattice are not actually free, but subject to constraints due to the lattice.

As a result of these constraints the electron in a lattice does not respond in the same proportionality to an applied electric and or magnetic force.

To a first approximation we can replace the free electron constant (mass) by a new constant and we call this the effective mass to distinguish.

 

But it should be stressed this is a measured or observed value not a calculated or derived one.

 

This is why I say the effective mass is a consequence, not a cause.

 

To fully explain this phenomenon we need to delve into quantum band theory, but here is a simple explanation that might help.

 

If we accept the idea of bands, without too much detail, we can observe that if a band is full the electrons cannot move in response to an applied force so no drift or driven current can occur.

This is because an electron can only move into a vacancy. If there are none available it cannot move.

This of course describes a perfect insulator.

 

Conversely if the band is sparsely populated with electrons then these can move quite freely about making the material a good conductor, since there are many vacancies available.

 

Now consider what happen is the band is almost full and short of exactly one electron.

We still cannot have an electron migration current, but we can have a 'hole' migration current, which is, of course, of the opposite sign to the normal electron current.

This is essentially the situation for the 'anomalies' asked about, but life is, as always, really much more complicated.

Edited July 29, 2016 by studiot

[Enthalpy]

This is a common description of holes, but some details are usually imperfect.

 

If one imagines electrons have a positive mass even where the band's curvature is inverted, then in the Hall effect, the electrons will move to the side deduced from positive electron mass, and the holes move to the opposite direction, and since the holes and electrons have opposite charges, the sign of the Hall voltage would remain the same.

 

The very reason for opposed Hall voltage is the negative electron mass, for instance near the maximum of a valence band, because this negative mass sends the electrons to the opposite direction.

 

The negative mass is a consequence of the crystal, sure.

 

And when the electrons have a negative mass, it is useful to introduce the idea of hole, because the hole has then a positive mass.

 

A hole current is just an electron current in the opposite direction. In a valence band too.

 

Now in a metal, whenever a band has an inverted curvature near the Fermi level, the electrons there have a negative mass, and we prefer to speak of holes. This happens at many metals and it doesn't need an almost-full band. The situation is often more complicated because metals can have several bands that intercept the Fermi level, and with varied curvatures.

[studiot]

 

then in the Hall effect, the electrons will move to the side deduced from positive electron mass, and the holes move to the opposite direction,

 

I'm having some difficulty with this statement.

 

As I understand the Hall effect, positive and negative charge carriers are swept in the same direction by the Lorenz force.

This results in either a positive or negative Hall voltage depending upon which carrier predominates.

 

My calculations to support this are as follows

 

Establish a right handed xyz coordinate system.

Charge flows along the y axis with velocity V; electrons flow -y direction and holes flow in the positive y direction.

and the applied magnetic field (B) is directed along the negative Z axis.

 

This results in Lorenz force, F, directed along the negative x axis as follows

 

The resultant Lorenz force on charge q is given by F = q(V x B) - a vector equation - and resolves as follows

 

F_x = q(V_yB_z - V_zB_y)

F_y = q(V_zB_x - V_xB_z)

F_z = q(V_xB_y - V_yB_x)

 

Applying values

 

For holes

 

q = +e ; B_z = -B ; B_y = B_x = 0

V_y = +V ; V_x = V_z = 0

 

F_x = +e{(+V)(-B) - 0) = -eVB

F_y = +e(0 - 0) = 0

F_z = +e(0 - 0) = 0

 

For electrons

 

q = -e ; B_z = -B ; B_y = B_x = 0

V_y = -V ; V_x = V_z = 0

 

F_x = -e{(-V)(-B) - 0) = -eVB

F_y = -e(0 - 0) = 0

F_z = -e(0 - 0) = 0

 

It can be seen that once all the signs are taken into account both holes and electrons are driven in the same direction.

 

I look forward to your calculations supporting the opposite view.

 

Edit

 

Of course this describes the standard Hall effect. It does not explain the anomalous situations moreno asked about.

These cannot be satisfactorily explained by simple free electron theory.

Edited August 1, 2016 by studiot

[Enthalpy]

As I understand the Hall effect, positive and negative charge carriers are swept in the same direction by the Lorenz force.

This results in either a positive or negative Hall voltage depending upon which carrier predominates.

 

I need to clarify the logic, yes.

 

If electrons conduct the main current, the Hall effect sweeps them to one side.

If holes conduct the main current, the Hall effect sweeps them to the same side.

And since their charges are opposite, the sign of the Hall effect depends on the charge of the (main) carriers.

 

But holes conceived only as the absence of electrons do not suffice:

Holes are just a simplification to help reasoning. It's still electrons that move, whatever their density, the band filling, and their mass.

A hole moving in one direction means that an electron moves in the opposite direction.

When the Lorenz force sweeps an electron to the right, the hole sweeps to the left.

So if the electrons still behaved normally in, say, a P semiconductor, we would observe the same sign for the Hall voltage as in an N semiconductor.

 

It is the electron's negative mass, in situations where the band's curvature is inverted like in a P semiconductor, that gives the hole a positive mass, and lets it behave as a normal particle but with a positive charge, explaining the sign of the Hall voltage.

 

This also tells that the hole is a useful idea where the band curvature in inverted, rather than when a band is almost full. And indeed, holes are commonly defined in metals, where they explain the sign of the Hall effect when needed, exactly as they do in semiconductors.

 

I haven't found within a reasonable time a real band diagram for a metal with mainly hole conductivity. But at least the (heavily simplified!) sketch there helps understand:

http://www.doitpoms.ac.uk/tlplib/semiconductors/energy_band_intro.php

The band overlap tells it's a a metal (there would be many mode bands, and the Fermi level very far from their minima and extrema).

Some bands have a normal curvature near the Fermi level, and the electrons have a positive mass there, usually not their mass in vacuum.

Other bands have an inverted curvature near the Fermi level, the electrons there have a negative mass, so holes are more useful.

Both electrons and holes conduct is most metals. The main current as well as the Hall voltage results from both, with varied contributions.

[studiot]

Most of what you have said in post#11 I agree with but this bit needs some thought.

 

But holes conceived only as the absence of electrons do not suffice:

Holes are just a simplification to help reasoning. It's still electrons that move, whatever their density, the band filling, and their mass.

A hole moving in one direction means that an electron moves in the opposite direction.

When the Lorenz force sweeps an electron to the right, the hole sweeps to the left.

So if the electrons still behaved normally in, say, a P semiconductor, we would observe the same sign for the Hall voltage as in an N semiconductor.

 

Not exactly.

 

If hole movement was the same as electron movement then current could be described in terms of electrons only.

 

But the holes and electrons are often in different bands.

 

Anyway that does not explain negative effective mass.

 

This is because mass is a non-negative isotropic scalar.

Effective mass is much more complicated and is in fact a tensor.

Effective mass in two or more dimensions can be positive or negative and vary in value.

 

I don't know if either moreno or anyone else is still interested but I am going to start at a very basic level and expand some things normally just stated as granted in textbooks.

 

So here is a gentle introduction to the implications of this.

First let us consider speed and velocity as a simple example to set the scene.

 

Speed is a non-negative isotropic scalar.

A body travelling at a speed of V travels at V, whatever direction it is heading.

Velocity is a vector.

Velocity in two or more dimensions has positive or negative components that vary in value, depending upon the direction.

So a velocity V due East has a component Vcos (45) in the North Easterly direction and -Vcos(45) in the South Westerly direction.

 

And yet both speed and velocity have the same units or dimensions - metres per second or LT^-1

 

Sometimes speed and velocity can be used interchangeably, sometimes they can't, as in the case of the calculation of momentum, which is always speed times mass, but you need to take all the velocity components into account if you calculate mass times velocity.

 

Speed and velocity are generally well known and understood.

 

Mass, reduced mass and effective mass less so.

Especially as some authors in quantum theory call 'reduced mass' a form of effective mass and there are yet other definitions in other circumstances.

 

So what definition of effective mass should we use in the Hall effect and what are the consequences?

 

Well I will just outline the definition in this post and fill in derivation and detail later if anyone is interested.

 

We define effective mass as the ratio of the square of Plank's constant to the second derivative of the particle's energy with respect to its De Broglie wave vector.

[math]{\rm{EffectiveMass}}\quad {\rm{ = }}\quad {{\rm{m}}^*} = \quad \frac{{{{\left( \hbar \right)}^2}}}{{\left( {\frac{{{\partial ^2}E}}{{\partial {k^2}}}} \right)}}[/math]

To see why it is called effective mass look at the units/dimensions.

Energy has units Joules or L²MT^-2

[math]{\left( \hbar \right)}[/math] has units Joule-seconds or L²MT^-1

the wave vector, k has units: per metre or L^-1

So the second differential has units energy per metre per metre or L²MT^-2LL or L⁴MT^-2

Putting this together with the square of the reduced Plank's constant we have

[math]{\rm{Dimensions}}\;{\rm{of}}\;{\rm{effective}}\;{\rm{mass}}\;{\rm{are}}\quad \frac{{\left( {{L^2}M{T^{ - 1}}} \right)\left( {{L^2}M{T^{ - 1}}} \right)}}{{{L^4}M{T^{ - 2}}}} = M[/math]

This has the same dimensions as the scalar inertial mass.

So why is this a tensor?

Well look at the bottom of the fraction. In the wave vector we have the cartesian product of two vectors, ie a tensor.

Just like the simple example with a vector velocity the tensor has different components in different direction.

These components are conditioned by the environment, not the electron but that is the subject of another post.

Edited August 8, 2016 by studiot

[studiot]

I haven't used any diagrams yet so here is one that I think is clear.

 

 

 

[Enthalpy]

I don't see a strong relationship between negative mass and mass tensor. In most semiconductors with indirect gap, including silicon, the valence electrons near the band edge have an isotropic negative mass (or rather two masses, light and heavy holes) while the conduction electrons near the band edge have a positive mass that isn't isotropic, at least in individual valleys. The electrons' mass can look isotropic if all valleys contribute as much to the conduction, especially if the crystal isn't stressed.

 

Yes, the current is explained by electrons only. It's the only mobile particle in silicon anyway. It's only because the electrons' mass is negative near the valence band's edge that we introduce holes.

 

The mass is a tensor because the material does that. In each direction of a minimum of the conduction band, the Fermi surfaces are ellipsoids rather than spheres. In gallium aresenide for instance, the conduction band minimum is centered in the Brillouin zone, the Fermi surfaces are spheres, and the electron mass is isotropic.

[studiot]

A few calculation and references to support your statements would be nice.

 

 

Looking at my diagram here is something to ponder.

 

We are discussing conductors and semiconductors, that is conductivity current as opposed to displacement current, not the special states that exist in superconductors.

All conductors have some resistance (conductivity) and require an electric field to drive the current (even if small).

 

Any explanation that states

 

'It's due to negative mass guv, honest'

 

must also explain why the main current proceeds in the normal expected direction and only the hall current/deflection is reversed.

 

Edit

If you are going to offer the negative effective mass explanation, your explanation is insufficient, particularly if you wish to reject the deifinition I gave which is entirely conventional.

 

Your statement is basically that of Dr Steve B here

 

http://physics.stackexchange.com/questions/10800/how-can-the-hall-effect-ever-show-positive-charge-carriers

 

Whilst I have great respect for Steve, his explanation was as lacking as yours as to why there should be 'curvature of the bands' or what that might mean or signify.

 

There is something more fundamental going on that I was leading up to.

Edited August 9, 2016 by studiot

[Enthalpy]

Any explanation that states

'It's due to negative mass guv, honest'

must also explain why the main current proceeds in the normal expected direction and only the hall current/deflection is reversed.

 

I know that problem. I do not know the answer. I had hoped you wouldn't notice it and bring it in the discussion. Gasp.

 

The situation as I see it:

• Just "positive charge carriers" is the least pleasant common answer, because the only mobile particles are the electrons, so holes behaviour must be explained in terms of electrons even if the reasoning is less direct. I got only that explanation from my professor, and he couldn't answer why in P semiconductors holes drift to the same side as electrons would do in N semiconductor.
• Later I got the less bad explanation of "electrons have a negative mass hence holes behave normally" and was satisfied with it until recently.
• I haven't seen a better one, despite both are flawed. "Positive charge carriers" is a simpler one but isn't right.

I know an other objection (or an other formulation of the objection) to the "holes have a positive mass" explanation:

• We know the flow direction of the main current, it's the same whether the electrons have a positive or negative mass.
• The Lorenz force doesn't depend on the electron's mass, and with no current taken from the transverse electrodes, the electric field corresponding to the Hall voltage compensates the Lorenz force. As the force by the electric field doesn't depend on the electron's mass neither, the Hall voltage shouldn't reverse where electrons have a negative mass.

So both common explanations are flawed, or at least incomplete.

 

Maybe it's a matter of speed versus acceleration. Or of velocity near the conduction band minimum compared with the speed around the minimum. Or of aliasing to the Brillouin zone. While the issue is interesting, I don't plan to invest time in it.

 

This is not the only flaw in the usual models about electron conduction in semiconductors and metals. The heat capacity of metals is commonly explained by lattice vibrations plus a negligible contributions by the electrons because only a few ones are near the Fermi level hence can move and store kinetic energy, BUT Hall measures tell that about one electron per atom is mobile in a metal.

Whilst I have great respect for Steve, his explanation was as lacking as yours as to why there should be 'curvature of the bands' or what that might mean or signify.

 

The bands are not the free electrons' parabola because of interaction with the lattice, that's clear and even unavoidable.

 

At the valence band maximum, the (valence) electron's wavefunction [but I like to say shortly "the electron"] changes its phase by 180° from one atom position to the next. Near the maximum, the phase differs a bit from 180°, so the wavefunction changes less abruptly, and the electron's energy is smaller. Smaller energy from more momentum means an inverted curvature and a negative mass.

 

Real life is in 3D, one should distinguish the directions of the crystal which indeed change the shape of the bands. Also nasty, at least one semiconductor has the maximum of its valence band at nonzero wavenumber - excentered in the Brillouin zone.

 

Other explanations exist, one being that we have folded the free-electron's parabola over the Brillouin zone, so what was a normal increase of energy when nearing the 180°-per-atomic-distance wavenumber is now around the zero wavenumber. Depending on one's background, call it aliasing (signal theory) or stroboscopic effect. Not wrong, but too short: just above that wavenumber, the energy of a free electron continues to increase, while after folding we should have obtained a flat top at the valence band. Because of that, I prefer a wavefunction resulting from orbitals interactions rather than from free electrons; software models for bands also combine orbitals.

 

What stands firm is that the bands can be observed (more or less), they fit rather well what computer models tell, and their (often inverted) curvature also fits the measured Hall voltage in semiconductors and metals.

[studiot]

Thank you for the positive and frank discussion so far.

I have tried to encourage others (who have read this judging by the viewing numbers) to participate, not least the OP, moreno.

 

I'm sorry you have limited your interest since I was preparing the next stage of my discussion, viz the transition from the 'little balls theory' to quantum theories.

 

I will again post a few summary observations in case there is further interest.

 

When we consider the bonding in alkenes we discus the pi bond and these carry over to polymer dienes.

These pi bonds, which are formed as hybrids from the bonding atomic obitals have a definite shape and locality in space.

This describes where we can expect to find the electron and where we can't.

 

When we consider the same process in metallic bonds the bands that result are not like this and the band structure is not equivalent to a series of pipes through the lattice. In particular they exist in phase space, not xyz physical space. So the 'curvature' you describe exists on graph paper, not in real space, unlike the curvature of p orbitals and pi bonds.

 

An electron in a band does not move about like a little ball but once it enters a particular band it can be anywhere within the range of that band.

The electron acts a wave that spans the entire band but can only appear in one particular phase space sub orbital according to Pauli rules.

 

I was preparing a comparative table of orbitals here are a few.

We can discuss the implications if you wish.

 

Note that Lithium Sodium, Potassium, Rubidium, Caesium, Copper, Sliver and Gold have a full shell plus an outer S orbital with a singleton electron.

They all have 'normal' Hall coefficients.

 

Beryllium, Cadmium, Tungsten and Zinc have 'anomalous' coefficients and have that outer S orbital fully occupied.

 

Borderline cases, magnesium, indium and aluminium have the full S orbital, but also have a singleton outer p electron.

Edited August 10, 2016 by studiot

[Enthalpy]

Yes, I have to put time in other activities.

 

I believe the analogy with alkenes and polyenes holds - in fact, I understand (or mis- maybe) polyenes by analogy with semiconductors, insulators and metals. The analogy is so deep that long polyenes need doping to become good conductors, and the absorption spectrum of dyes refects the multiple energy levels.

 

Yes, the curvature is in the (h, E) space. Nothing observable easily in xyzt.

 

What's the state of an electron, and what does Pauli's exclusion mean here? It gets complicated as soon as there are several electrons.

 

The electronic "states" in a solid usually refer to a definite energy and momentum. By definition, such states extend to the whole solid. When the solid is the wire of a high voltage line, the states are 1000km long and 10cm wide. Definite momentum means indefinite position.

 

These states, with definite energy, are useful because the Fermi-Dirac statistics tells that, in a band about 1eV tall and at a temperature of 26meV, and depending on where the Fermi level is relative to the band, some states are (almost) certainly occupied, other empty, and just a fraction of them are sometimes ocupied - this is where movement happens, for instance if an electric field picks an electron from a state flowing to the left and puts it in a state flowing to the right. It explains nicely why a full band in an insulator doesn't conduct. As an other advantage, we know from Fourier transform or series that any waveform is a linear combination of these states, so they're a good base.

 

Though, these states are not the only possible base. One simple example: take the exp(+ikx) and exp(-ikx) states [*exp(iwt) implicitly]. You can write them as weighed sums of cos(kx) and sin(kx), so the complete collection of sin and cos for all permitted k is a base too, and not even a bad one as it's orthogonal too, so it's just equally correct to have one electron pair in each of these standing waves rather than a pair in each propagating wave. Pauli rule means that electron (pairs at most) occupy states that are orthogonal to an other.

 

So not only is the choice of the base subjective to describe the waveform of an electron. It's worse: we have no other reason than human-rated simplicity to claim that the set of electrons fills one kind of waveform base rather than an other. The set of electrons below the Fermi level fills the available states, but we don't know what kind of states. Only after a successful interaction can we tell "the absorbing electron had such a state and now it has such".

 

So "delocalized to the whole solid" applies to the available states which have a definite momentum. For other states, say orthogonal wavelets, the momentum would be less defined and the position better defined. And the electrons? Well, there is no reason to attribute them a set of waveforms rather than an other.

 

In some cases - and this doesn't differ much from vacuum - we do have information about an electron's position. Say, if it has been injected in the conduction band from a deep dopant level: for some time we know it's near to the dopant atom. Then, the adequate wavefunction to describe it isn't one plane wave with definite momentum - but the wave can be described as a weighed sum of plane waves if this is advantageous. Or as a Gaussian or a sum of Gaussians, which is interesting because it keeps its shape during a diffusion process.

 

----------

 

As a sidenote, in the BCS theory for supraconductivity, two electronic states that propagate in opposite direction with definite momentum make a series of nodes and antinodes which, as the momenta are not exactly opposite, propagate, and the nodes interact with the charge wave of the lattice compressed or expanded by a phonon. Since the two electrons that can be put in this state are delocalized to the size of the solid, there is no need to "explain" how they should couple despite their repulsion. They aren't near to an other.

 

----------

 

Band structure from orbitals... Yes they do result from atomic orbitals, but that's accessible to software rather than qualitative or algebraic understanding.

 

While electrons in a band don't result directly from atomic orbitals (and transition metals like Cu rearrange the electrons a lot) the orthogonality of all electron (pairs) still holds, including that delocalized electron waveforms are orthogonal to the electrons local to the atoms. Being already orthogonal, the available atomic orbitals are interesting candidates to define mathematically the delocalized states, but that's all. Heavy weighting and summing may be necessary, just like you don't find the original 2s and 2p orbitals of the carbon atom in a CH₄ molecule.

 

A striking example that the deep local atomic orbitals influence the band structure: C, Si and Ge have the same crystal structure and the same number of valence electrons, but the conduction band minimum has a different momentum for each material. This tells that simplistic models like the periodic potential are insufficient.

[Moreno]

So, there are holes in metals which could be regarded just like holes in semiconductors?

[studiot]

So, there are holes in metals which could be regarded just like holes in semiconductors?

 

 

You started this thread asking about exceptions to the norm.

 

I suggest you draw a line under the exceptions in thinking about this new question.

 

The usual theory classifies outer electrons as being either bonding (also wrongly called valence) electrons or electrons which are available to carry conduction current.

 

A hole is then formed when a bond is broken and a bonding electron moves into the conduction block of electrons.

The hole contributes to the conduction when another bonding electron breaks its bond but moves into the vacant bonding position.

In this way the hole moves and thus contributes to the total current.

 

It should also be noted that in this version of the theory the electric fields to drive current within a conductor are much weaker than those required for the same current in a semiconductor.

Edited September 18, 2016 by studiot