The Hall effect can be used to determine the sign of the charge carriers, as a positive particle drifting along the wire and a negative particle drifting the other direction get deflected the same (as $F = q \vec{v}\times\vec{B} = (-q) (-\vec{v})\times\vec{B}$). But I don't understand how positive charge carriers are ever possible.

As I've understood, a positive hole is nothing more than the absence of an electron. As all the electrons in the valence band are still negatively charged, why would this hole behave in a magnetic field as if it were positive?

Also, a hole is created if an electron is excited into the conductance band. If there is always the same numbers of holes as electrons, how can any Hall effect ever occur?

Thank you very much

  • 1
    $\begingroup$ "But I don't understand how positive charge carriers are ever possible." Electrons and holes are not the only charge carriers. Positive ions and bare protons can flow, too. physics.stackexchange.com/a/17131/176 $\endgroup$
    – endolith
    Commented Mar 23, 2012 at 17:32

3 Answers 3


There are two essential facts that make a hole a hole: Fact (1) The valence band is almost full of electrons (unlike the conduction band which is almost empty); Fact (2) The dispersion relation near the valence band maximum curves in the opposite direction to a normal electron or a conduction-band electron. Fact (2) is often omitted in simplistic explanations, but it's crucial, so I'll elaborate.

STEP 1: Dispersion relation determines how electrons respond to forces (via the concept of effective mass)

EXPLANATION: A dispersion relation is the relationship between wavevector (k-vector) and energy in a band, part of the band structure. Remember, in quantum mechanics, the electrons are waves, and energy is the wave frequency. A localized electron is a wavepacket, and the motion of an electron is given by the formula for the group velocity of a wave. An electric field affects an electron by gradually shifting all the wavevectors in the wavepacket, and the electron moves because its wave group velocity changes. Again, the way an electron responds to forces is entirely determined by its dispersion relation. A free electron has the dispersion relation $E=\frac{\hbar^2k^2}{2m}$, where m is the (real) electron mass. In the conduction band, the dispersion relation is $E=\frac{\hbar^2k^2}{2m^*}$ ($m^*$ is the "effective mass"), so the electron responds to forces as if it had the mass $m^*$.

STEP 2: Electrons near the top of the valence band behave like they have negative mass.

EXPLANATION: The dispersion relation near the top of the valence band is $E=\frac{\hbar^2k^2}{2m^*}$ with negative effective mass. So electrons near the top of the valence band behave like they have negative mass. When a force pulls the electrons to the right, these electrons actually move left!! I want to emphasize again that this is solely due to Fact (2) above, not Fact (1). If you could somehow empty out the valence band and just put one electron near the valence band maximum (an unstable situation of course), this electron would really move the "wrong way" in response to forces.

STEP 3: What is a hole, and why does it carry positive charge?

EXPLANATION: Here we're finally invoking Fact (1). A hole is a state without an electron in an otherwise-almost-full valence band. Since a full valence band doesn't do anything (can't carry current), we can calculate currents by starting with a full valence band and subtracting the motion of the electrons that would be in the hole state if it wasn't a hole. Subtracting the current from a negative charge moving is the same as adding the current from a positive charge moving on the same path.

STEP 4: A hole near the top of the valence band move the same way as an electron near the top of the valence band would move.

EXPLANATION: This is blindingly obvious from the definition of a hole. But many people deny it anyway, with the "parking lot example". In a parking lot, it is true, when a car moves right, an empty space moves left. But electrons are not in a parking lot. A better analogy is a bubble underwater in a river: The bubble moves the same direction as the water, not opposite.

STEP 5: Put it all together. From Steps 2 and 4, a hole responds to electromagnetic forces in the exact opposite direction that a normal electron would. But wait, that's the same response as it would have if it were a normal particle with positive charge. Also, from Step 3, a hole in fact carries a positive charge. So to sum up, holes (A) carry a positive charge, and (B) respond to electric and magnetic fields as if they have a positive charge. That explains why we can completely treat them as real mobile positive charges in their response to both electric and magnetic fields. So it's no surprise that the Hall effect can show the signs of mobile positive charges.

  • 4
    $\begingroup$ That is brilliant. Most of the explanations I have seen are thus completely wrong and it always intrigued me because it gave incorrect results for the Hall effect. Now I understand it's a completely quantum-mechanical effect that can't be easily explained with the parking lot metaphor. Your explanation should be in my textbook (Kittel). $\endgroup$
    – Kasper
    Commented Jun 7, 2011 at 9:50
  • 1
    $\begingroup$ Great explanation. I still do not understand why electrons and holes move the same way at the top of the conduction band. I can see how this puts a negative sign in front of the Hall Coefficient for P type materials, but I do not understand dispersion. $\endgroup$
    – Bill Slugg
    Commented Jun 12, 2011 at 0:47
  • 1
    $\begingroup$ Step 4 doubt, sorry, I am parking lot confused: why is that in a full valence band electrons can't move but when there's a single hole around, all of them move like the water in the river? And, if those electrons were actually flowing with the hole like the buble, wouldn't that imply a big current (that I believe doesn't happen)? $\endgroup$
    – Rojo
    Commented Sep 3, 2012 at 2:41
  • 3
    $\begingroup$ @Rojo: Good question! An electric field affects all the electrons in the valence band, both the top of the band (where the action happens) and the bottom of the band (which is always full). The field pushes the electrons near the top of the band in one direction, and pushes the electrons near the bottom (where the band curves the opposite way) in the opposite direction. In a FULL band, the average motion is zero. (This is related to the fact that the band is a periodic function of k-space, so it can only curve upward in some places if it curves downward in other places.) $\endgroup$ Commented Sep 3, 2012 at 12:09
  • $\begingroup$ @ Steve B I noticed this interesting post, I agree with you electrons are moving in the same directions as holes, however, than the net current cancel in the P-type region? this is quite wierd..physics.stackexchange.com/questions/86097/… $\endgroup$
    – Lorniper
    Commented Nov 11, 2013 at 7:31

"...why would this hole behave in a magnetic field as if it were positive?"

You are correct that a positively charged hole is nothing but a lack of an electron. It is positively charged because there is a proton in that atom that does not have an electron to balance its charge. Holes move around by having an electron cross the hole to get to the other side. For example. A hole is located at 200 Elm Street. There is no house there. Just an empty lot. The house from 198 Elm street is moved onto the 200 lot. A house has moved to the right and a hole has moved to the left. This is how a hole behaves in a magnetic field as if it were positive.

  • 2
    $\begingroup$ I don't think this explains how a hole acts as a positive charge with respect to the Hall effect. In this picture, one would still expect the Hall effect to produce results corresponding to those of moving negative charges. To me, the strange thing is not that a hole acts like a positive charge, but that it also acts like a positive mass. Say the hole is moving in the $x$ direction and the magnetic field is in the $y$ direction. Why does the hole move in the $+z$ direction? The "real" force is on the electrons and points in the $+z$ direction, so the hole "should" go the other way. $\endgroup$
    – Ted Bunn
    Commented Jun 6, 2011 at 13:32
  • $\begingroup$ It does go the other way. Holes act as mirror images of electrons. A hole is a spot where a p type dopant atom has an outer orbital one electron short of being filled. It will take an electron from a nearby semiconductor atom and fill that spot. That semiconductor atom now is missing an electron and is positively charged. For that hole to move, it must pull an electron out of the outer orbital from the atom on the far side of it thus moving the hole one step over. Electron carriers of charge are different, they stay in the conduction band all the time and thus are more mobile. $\endgroup$
    – Bill Slugg
    Commented Jun 6, 2011 at 23:38
  • $\begingroup$ By "It does go the other way," you mean that the Hall voltage has the opposite sign for $p$-type semiconductors than it does for metals, right? At least, that's how I've always understood it. The classical picture of holes predicts the opposite: it predicts that the electrons feel a force in the $+z$ direction, in the geometry I describe, leading to a buildup of negative charge on top. But as I understand it, the opposite actually occurs (e.g., journal.lapen.org.mx/jan09/…) $\endgroup$
    – Ted Bunn
    Commented Jun 7, 2011 at 0:58
  • $\begingroup$ No, it is incorrect to say that the effect has opposite signs depending on the material involved. If you set up a Hall effect test stand and swap out samples but do not reorient the magnetic field and do not change the amount or direction of the current that flows, you will always have the same sign (although not the same magnitude) on the meter that is reading out the Hall effect. What IS different in semiconductors is that there are two types of charge carriers, each with a different mobility, each going in opposite directions to the other, but in the end producing the same voltage sign. $\endgroup$
    – Bill Slugg
    Commented Jun 7, 2011 at 2:17
  • $\begingroup$ It's been a long time since I studied this stuff, but that sounds very different from what I learned once upon a time. For instance, is this incorrect: "For most metals, the Hall coefficient is negative, as expected if the charge carriers are electrons. In beryllium, cadmium and tungsten, however, the coefficient is positive. In these metals, the charge carriers are holes, which act like positive charges. In a semiconductor, the Hall coefficient can be positive or negative, depending on whether it is P or N type." (mysite.du.edu/~etuttle/electron/elect19.htm) $\endgroup$
    – Ted Bunn
    Commented Jun 7, 2011 at 2:56

Steve Byrnes has already explained very clearly in his great answer why we can treat holes as if they were real positive charge carriers. From this it is clear that if we have holes near the top of the valence band and an empty conductance band, we get a Hall voltage which is opposite to normal metals. However, what still left me puzzled for a moment is that in this scenario, microscopically, it is in fact only electrons in the valence band which flow. So we should be able to explain the positive Hall voltage without referring to the "positive hole" picture. Since $\vec{F} = q \vec{v} \times \vec{B} = (-q) (-\vec{v}) \times \vec{B}$ has the same sign for right-moving positive holes as it has for left-moving electrons, it seems like they should get deflected in the same direction, which would give contradicting results for the sign of the Hall voltage.

The solution to this is that, as highlighted by Steve Byrnes, electrons near the top of the valence band have negative effective mass. So while the force $\vec{F}$ is the same for the holes and for the electrons, this is not the case for the acceleration $\vec{a} = \vec{F} / m_\text{eff}$. Since $m_\text{eff}$ is positive for the holes, while it is negative for the electrons, they move into opposite directions so that the resulting Hall voltage is the same in both descriptions as it should be.

Also, a hole is created if an electron is excited into the conductance band. If there is always the same numbers of holes as electrons, how can any Hall effect ever occur?

This is only true for an intrinsic semiconductor. In this case, the Hall voltage would indeed always vanish if electrons and holes had the same mobility. However, in real systems they usually have different mobilities, so that one measures a finite Hall voltage also for intrinsic semiconductors. For a more elaborate answer see Hall effect with similar positive and negative carriers?.


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