# What exactly is a hole?`

I_am_learning
I never actually understood what exactly is a Hole? Could you explain it clearly.

## Answers and Replies

Homework Helper
It's simply the absence of an electron.
Because of historical accident, before the discovery of the electron, current was assumed to flow from positive to negative.
It's a bit error prone to have electrons moving one way and current moving the opposite way - so we imagine the movement of the absence of electrons, which go in the same direction as current

I_am_learning
But that don't seem to explain the whole story. If movement of holes were actually movement of negatively charged electron in the opposite direction then the Hall effect shouldn't have shown positive hall voltage. But as far as I know, the holes behave totally like positive charges, in every respect. This is what bothering me for years!

Modey3
I never actually understood what exactly is a Hole? Could you explain it clearly.

Consider a string of hydrogen atoms. If you take away one electron on a particular hydrogen atom you would get a string of something like: ... H, H, H+, H, H ...

If the current moves in the right-direction the positive-charge (not the hydrogen cation!) moves in the left-direction. As you can see the hole is the positive charge induced by the un-screened nucleus.

modey3

vin300
But that don't seem to explain the whole story. If movement of holes were actually movement of negatively charged electron in the opposite direction then the Hall effect shouldn't have shown positive hall voltage.
If B is along +ve z and either velocity of holes is in +ve x or velocity of electrons is in -ve x, Hall voltage is +ve.
If B is along +ve z and either velocity of electrons is in +ve x or holes in -ve x, Hall voltage is -ve

I_am_learning
If B is along +ve z and either velocity of holes is in +ve x or velocity of electrons is in -ve x, Hall voltage is +ve.
If B is along +ve z and either velocity of electrons is in +ve x or holes in -ve x, Hall voltage is -ve

I don't think thats correct. Look at the following illustration I sketched If the movement of holes were actually movement of electrons, then should always find -ve hall voltage (Right Figure) . Shouldn't we?

Phrak
If the movement of holes were actually movement of electrons, then should always find -ve hall voltage (Right Figure) . Shouldn't we?

I understand your problem but not the solution. Apparently the others are unfamiliar with the the Hall effect in P-type materials in violation of classical elecrtromagnetism. Perhaps someone can enlighten us both.

Maybe you should take into account that electrons at the upper end of the band have a negative effective mass. Hence a hole (the absence of one of these electrons) has a positive mass.

josephajain
I agree with the critic.

I_am_learning
Maybe you should take into account that electrons at the upper end of the band have a negative effective mass. Hence a hole (the absence of one of these electrons) has a positive mass.

O.K. I guess you mean that the holes are not just the absence of electron but absence of electron with negative effective mass (Someone somewhere has already told me this). But if that is the case then; when the electric Field is applied in upward direction (the right side figure of my sketch some posts above), then electrons having negative mass should have moved in the same direction (upward). O.K yes, That could produce and hence explain the +ve hall voltage, but the absurd thing is that, that would result in current in downward direction (opposite to direction of applied field).
But we always see the current in the direction of applied field, don't we?

Studiot
In post#6 you have a contradiction.

This is because you are trying to show the current in three different ways
That is not a good idea.

You have shown a conventional current. This current is in one direction only, regardess of the movement of its carriers and is one of the determining vectors of the Hall voltage.

This conventional current arrow may be made of all holes, all electrons or a mixture.
Regardless the arrow is still the same.

So the two important vectors in your diagram are the magnetice field and the current (with its associated electric field of course).

This leads to build up of negative charge on the left and positive on the right which in turn makes the polarity Hall voltage, developed across the section negative on the left and positive on the right.

Are you also aware that this voltage develops its own electric field so that the final electric field is no longer parallel to the current direction but the vector sum of the two fields?

I think this is a fine example of trying to work in terms of charge carriers rather than a (ficticious) entity called conventional current brings difficulties. All the other equations of physics are adjusted to be in sync with this.

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I_am_learning
In post#6 you have a contradiction.
The derived force vector is correct in the right hand diagram but should be reversed in the left one as positive and negative charges are deflected in opposite directions.

Yes, I have considered that positive and negative charges are deflected in opposite directions; but since their direction of motion is also opposite; they are finally deflected in the same same direction, as shown in the figure (post 6).
Either you missed it or I am in serious trouble with the Left-Hand Rule.

Studiot
But your current vector is the same in both diagrams.

The LH rule expects this vector (as you have it) irrespective of the polarity of the carriers.

This is a sign convention thing. You also have to start introducing -ve signs into the vector cross product if you must use charge carrier direction. Don't forget that the current vector also defines a direction in space, ahich you have chosen by using conventional current (wisely in my opinion).

Modey3
Folks,

This subject is being over-analyzed. IMO the Hall Effect experiment doesn't adequately describe what a hole is. A hole is not a particle. Yes, I said it! A hole is an absence of a particle. In semiconductors the absence of an electron within (not at the top) the DOS is manifested locally on an atom. Thus, it may seem like a hole is a particle, but in reality it is not. The positive charge of the hole comes from the nucleus. Holes are created by adding localized p-type defect states near the VBM, which pulls electrons out of the VB and this creates holes in the VB. The p-type defect gets neutralized while an atom in the parent lattice gets ionized . This is why holes are localized around an atom. In metals, a hole has a different meaning because as the temperature is raised you are technically creating holes in the VB, but the holes are not localized to a particular atom, which is why it's impossible to describe current in terms of hole-flow in metals.

I hope this helps

modey3

Studiot
Yes you have shown the electron and hole flow in opposite directions.

Your diagram in post 6 is both correct and compatible with vin300 statement in post 5.

The Lorenz force pushes both carriers to the same side. This is fairly easy to show mathematically.

As a result the Hall voltage across the section will be either positive or negative on the left depending upon whether the flow is holes or electrons.

This situation when there are both carriers present (semiconductors) is much more complicated. It is not true to say that just because there are more holes or electrons the left side will end up more positive or more negative. This is because the Lorenz force equation has to be modified to take into account the diffeent mobilities of holes v electrons. This can lead to the minority carrier determining the Hall voltage polarity.

You can have the equations if you want them, but it is a lot of algebra.

I_am_learning
Your diagram in post 6 is both correct
Its a great relief to hear that.
Now, I am at point at asking what I intended to ask.

In some material Hall Voltage is found to be positive. This is explained by saying that the current is constituted by holes (Left figure, Post 6).

But the contradiction arises when we study about the nature of holes. If we are to say that holes are actually absence of electrons (and there is no such thing as positively charged hole) and that the motion of holes is accomplished by transfer of electrons in the opposite direction, then we are forced to model the current flow in the material by Rightside Figure in Post 6. And now to the arousal of great contradiction the Hall Voltage is Found to be -ve.

Please for gods sake, let be get through this. I am dying.

I_am_learning
Your diagram in post 6 is both correct
Its a great relief to hear that.
Now, I am at point at asking what I intended to ask.

In some material Hall Voltage is found to be positive. This is explained by saying that the current is constituted by holes (Left figure, Post 6).

But the contradiction arises when we study about the nature of holes. If we are to say that holes are actually absence of electrons (and there is no such thing as positively charged hole) and that the motion of holes is accomplished by transfer of electrons in the opposite direction, then we are forced to model the current flow in the material by Rightside Figure in Post 6. And now to the arousal of great contradiction the Hall Voltage is Found to be -ve.

Please for gods sake, let be get through this. I am dying.

Studiot
First let us look at the model you are employing in post#6.

The lattice is electrically neutral. So if we take a negative electron from point A and shove it over to point B, point B becomes –1 negative and this leaves a corresponding +1 positive charge at point A, supplied by the positive charges in the area that were formerly balancing the -1 charge on the electron we have moved.

However, once the hole and electron have separated (ie the pair has been 'created') there is nothing in the model to continue to link a particular hole to a particular electron. If the electron moved on again to point C it would not leave another hole at B, just a neutral point.
Equally the hole could move by ‘displacing’ a different electron, or if you like a different electron could occupy point A which would again be neutral, leaving a +1 hole somewhere else in the lattice.

Now we know that all charges, holes and electrons, moving as part of the current, are subject to the leftwards Lorenz force.
But all other electrons in the lattice are not subject to this force as they are not moving as part of the current. And there are plenty available.
So a leftwards movement of the hole at point A can be accompanied by a rightwards movement of electrons without violating any laws.
This, of course, is movement at right angles to the current flow so no work is required from the electric field causing the current ie there is no interference with it.

Note also my earlier comment about the different ‘mobilities’ of holes and electrons.

A simple explanation of holes is positional. This is because electrons are regarded as point particles. As such they have a location in space and it makes sense to talk about electrons moving one way and 'holes' moving the other.

However when an electron moves it does not leave behind ‘nothing = absence of electron’
It is rather like pigeon holes in a letter rack and letters or pages in a book and writing. Just because there are no letters in a particular pigeonhole or no writing on a particular page does not mean that the pigeon hole or page does not exist. They are still there, available for letters or writing.
So if you imagine the solid block as the rack or book and the pigeonholes or pages as the wave states and the letters or writing as the electron perhaps you can see the analogy.
Remember the analogy is not perfect, just an aid to visualisation.

Further electrons and holes are not point particles. They have no location in space. In solids the wavefunctions extend across the entire block of material.
Some are filled some are not, but all are present.
This is how the electron can actually ‘move’ – It jumps from state to state. This is a quantum thing and the quantum Hall effect is much more complicated than the classical smooth one.

rootX
I don't think thats correct. Look at the following illustration I sketched
...

If the movement of holes were actually movement of electrons, then should always find -ve hall voltage (Right Figure) . Shouldn't we?

1. Could the confusion be that you mixing n-type materials with p-type? I believe current is solely due to the particles in the conduction band, so it might depend whether you have n or p majority.

I did not read through the whole thread .... I would have to review the Hall's effect before coming with a valid answer

I_am_learning
I read and re-read your reply for some hints, but please forgive me, I couldn't still get it.

The lattice is electrically neutral. So if we take a negative electron from point A and shove it over to point B, point B becomes –1 negative and this leaves a corresponding +1 positive charge at point A, supplied by the positive charges in the area that were formerly balancing the -1 charge on the electron we have moved.
This is explaining how conduction begins, Right?
However, once the hole and electron have separated (ie the pair has been 'created') there is nothing in the model to continue to link a particular hole to a particular electron. If the electron moved on again to point C it would not leave another hole at B, just a neutral point.
Equally the hole could move by ‘displacing’ a different electron, or if you like a different electron could occupy point A which would again be neutral, leaving a +1 hole somewhere else in the lattice.
This is explaining how conduction continues.

Now we know that all charges, holes and electrons, moving as part of the current, are subject to the leftwards Lorenz force.
So a leftwards movement of the hole at point A can be accompanied by a rightwards movement of electrons without violating any laws.

Same story repeats, I think. How can electrons be made to move rightwards?
I would like to put it this way
Be it n-type or p-type material,
1. the only thing that can move is electrons, right? (Am considering Classical Model)
2. If the current is in upward direction, No matter how you try to explain the phenomenum, The net electron flow must be in downward direction, Right?
3. If the elctron flows in downward direction, then Lorentz Force pushes it left, and Hence
-ve Hall Volatge, Right?

4.I must be wrong at least Somewhere, Right? But where (1,2,3,4)?

Gold Member
A hole is NOT merely an absence of an electron. By that definition, no charge would be a hole.

A hole IS the absence of an electron in a formerly charge-balanced material, giving the material a net positive charge.

Studiot
If the current is in upward direction, No matter how you try to explain the phenomenum, The net electron flow must be in downward direction, Right?
3. If the elctron flows in downward direction, then Lorentz Force pushes it left, and Hence
-ve Hall Volatge, Right?

Your problem is in number 2/3

The crux of it is here

Now we know that all charges, holes and electrons, moving as part of the current, are subject to the leftwards Lorenz force.
But all other electrons in the lattice are not subject to this force as they are not moving as part of the current. And there are plenty available.
So a leftwards movement of the hole at point A can be accompanied by a rightwards movement of electrons without violating any laws.

etc

The electrons that need to move rightwards are not the electrons that take part in the upward flowing current.

Consequently they are not the ones being forced leftwards by the Lorenz force.
(You are aware that the force you have labelled 'Thrust' is called the Lorenz Force?)

Don't forget there are no electrons moving directly leftwards. These are the ones whose trajectory has two components, viz the upward flow of current and the leftwards push, resulting in a path curving towards the left.

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I_am_learning
(You are aware that the force you have labelled 'Thrust' is called the Lorenz Force?)
Yes, I am aware of that.

Don't forget there are no electrons moving directly leftwards. These are the ones whose trajectory has two components, viz the upward flow of current and the leftwards push, resulting in a path curving towards the left.
I am aware of that too.

The electrons that need to move rightwards are not the electrons that take part in the upward flowing current.
Consequently they are not the ones being forced leftwards by the Lorenz force.
Then which electrons are they? And which force causes them to move rightwards?
(I think I have reached to the root point, now. Wow!)

Ok, here's my view:

ma=qE+q vxB
is the equation of motion. Due to friction, the acceleration will lead to an equilibrium velocity v that will show in the same direction as the acceleration a.
Case a: m>0, q<0 (electrons at the lower end of the band)
v is antiparallel E, j =q v is parallel E. If E is in x-direction and B in z-direction, then the electrons will be deflected in direction of -y.
Case b: m<0, q<0 (electrons at the upper end of the band with negative effective mass)
v is parallel E, j is antiparallel E. If E is in x-direction and B in z-direction, then the
electrons will be deflected in direction of -y, as in case a. However, as the direction of the current is reversed in comparison with case a, the hall coefficient will be of alternate sign.
case c) m>0, q>0 (holes):
v is parallel E, j is parallel E and holes are deflected in direction of -y. The Hall coefficient equals the one in case b.

In the band picture, the situation in a nearly full band is complicated by the fact that most levels are filled and "Pauli block" the mutual motion. In a completely full band, the electrons cannot react at all to the applied field and acceleration a=0. In a nearly full band, the mean acceleration of the electrons will be very small, but will on the mean follow case b. Alternatively, the situation can be described in terms of some holes of case c.

Phrak
Thanks for that DrDu. I was hoping you'd come up with more. You are the one who has done the work, not I, so I'm a little embarassed to mention this. In case b) where electron mass is less than zero, j is antiparallel with E. The positive current is moving counter to the applied field! So we get perpetual motion due to negative resistance (which would be awfully nice to have...)

I think we might consider the equilibrium condition where an electric charge gradiant has already accumulated in the ±Y-direction so that there is an additional Ey field due to this charge.

The drift velocity, vx = (½d)max = k-1max, where d is a some nominal positive distance between collisions, so that k is positive and

sign(vx) = sign(ax). (I'm just repeating what you’ve established in this.)

kvx = max = qEx + q(vyBz)

vy=0 by conservation of charge in the equilibrium condition, so that

1) kvx = qEx. Now the electrons are not giving us free energy. Bummer.

kvy = 0 = may = qEy – vxBz so that

2) qEy = vxBz

Equations 1 and 2 are as far as I've gotten. Maybe the negative electron mass model works. I have no idea. ---or did I make a fundamental error somewhere?

I_am_learning
Ok, here's my view:

ma=qE+q vxB
is the equation of motion. Due to friction, the acceleration will lead to an equilibrium velocity v that will show in the same direction as the acceleration a.
Case a: m>0, q<0 (electrons at the lower end of the band)
v is antiparallel E, j =q v is parallel E. If E is in x-direction and B in z-direction, then the electrons will be deflected in direction of -y.
Case b: m<0, q<0 (electrons at the upper end of the band with negative effective mass)
v is parallel E, j is antiparallel E. If E is in x-direction and B in z-direction, then the
electrons will be deflected in direction of -y, as in case a. However, as the direction of the current is reversed in comparison with case a, the hall coefficient will be of alternate sign.
case c) m>0, q>0 (holes):
v is parallel E, j is parallel E and holes are deflected in direction of -y. The Hall coefficient equals the one in case b.

Are you sure case b, exist? That E field and j can sometimes be anti-parallel?
And for case c, I need to repeat the same story, If we in fact do believe that holes are actually absence of electrons (in formerly charge neutral place) and that holes are abstract things and that they are really no particles and that really no force acts on them; can't we always talk in terms of electrons (at least for the sake of this discussion)? Then Case C should be modeled with case b, I think?

Yes, I do believe that case b exists and that E and j can be anti-parallel.
Of course holes (case c) are abstract things, but they precisely model situation b.

Staff Emeritus
There's a similar discussion on this already in another thread:

For anyone claiming that "holes" aren't as "real" as "electrons", keep in mind that these are done in the context of many-body interaction. If you think the electrons you get in such a system are "real" while holes are not, then you need to convince me why the Landau's Fermi Liquid theory is wrong, why these are actually "quasiparticles" and not bare electrons (look at the mass renormalization of these things), etc.. etc.

Zz.

Studiot
You have to show consistency in a model.

The critic wanted to discuss the hall effect with a point charge model of current.
This is considered difficult to impossible to use to explain all the observed phenomena.

However you cannot both say that a hole is a convenient fiction or non real entity and that the Lorenz force acts on it.

So if we discuss the diagrams in post#6 we are tacitly giving substance to holes.

Allowing this argument the point charge theory goes as follows

1) Point charges do not interfere with each other.

2) All point charges positive or negative, are swept in the same direction by the Lorenz force.

3) At equilibrium the net buildup of charge to one side is manifest as +ve or -ve voltage at right angles to the direction of flow, depending upon whether there are more +ve or negative charges diverted.

With regard to the existance of holes or otherwise, the molecular orbital exists whether it is occupied or not.

I finally found the time to have a look to a book (how old fashioned!), Ashcroft Mermin, solid state physics.
They discuss the situation in detail, backing up my rather hand-waving argumentation.
The most difficult part is the justification of the equation of motion. Ashcroft and Mermin refer only to the literature on that point for the general formula (especially for the Lorentz term) although it is rather easy to prove for the reaction to a static electric field. That the current must be opposite for some electron levels compared to what one would expect for a free electron follows from the fact, that the total current from a full band must vanish.

Studiot
My copy of Ashcroft and Thingy expounds a quantum explanation which quantitatively accounts for various values of the Hall voltage.

But they do not provide a motivation for an answer to the OP's question.

Here is a mathematical instance of the same question.

When multiplying two ordinary numbers
+ times + makes +
- times + or + times - makes -

But why does - times - make +?

The best answer I have seen to this (due to Gelfand)

3x5 = 15
Getting 5$three times is gaining 15$

3x(-5) = -15
paying 5$three times is losing 15$

(-3)x5 = -15
not getting 5$three times is losing 15$

(-3)x(-5) = 15
not paying 5$three times is gaining 15$

The question about the hole is similar to the last one.

Phrak
Yes, I do believe that case b exists and that E and j can be anti-parallel.

I don't see how this is justifiable. With E and j antiparallel the P type Hall effect device is a source of emf; energy is being put into the system--but from where? This is a perpeptual motion machine of the first kind.

And more, it is not equivalent to case c) which behaves like a resistor.

The point is that without taking scattering which leads to resistance into account, an electron of initial crystal momentum k_0 would move trough the whole Brillouin zone (1 dimensional case). During that motion, its velocity would also change sign so that finally it would end where it started, so that no energy is deposed in the system on the average.

Naty1
Here's an interesting description:

The electron–hole pair is the fundamental unit of generation and recombination, corresponding to an electron transitioning between the valence band and the conduction band. ......

the valence band is so nearly full, its electrons are not mobile, and cannot flow as electrical current....However, if an electron in the valence band acquires enough energy to reach the conduction band, it can flow freely among the nearly empty conduction band energy states. Furthermore it will also leave behind an electron hole that can flow as current exactly like a physical charged particle. Carrier generation describes processes by which electrons gain energy and move from the valence band to the conduction band, producing two mobile carriers; while recombination describes processes by which a conduction band electron loses energy and re-occupies the energy state of an electron hole in the valence band.

http://en.wikipedia.org/wiki/Electron-hole_pair

Studiot
That's exactly what I said before.

If you allow quantum explanations, the pigeon hole is already there, whether occupied or not.