Voltages are often thought of as continuous physical quantities. I was wondering whether by zooming in a lot, they are discrete.

I feel like the answer to the above question is yes as voltages in the real world are generated by actions of electrons. Can someone give me a more formal proof or a disproof?

Whether voltages are discrete of continuous can have some impact the correctness of devices such as the analog to digital converter. For example, if voltages in the real world are continuous, then the Buridan principle[1] says that there cannot be a perfect analog to digital converter because such a device makes a discrete decision from continuous inputs.

[1] : Lamport, L. (2012). Buridan’s Principle. Found Phys 42, 1056–1066. http://link.springer.com/article/10.1007/s10701-012-9647-7

(It would be great if someone could also answer a related question https://electronics.stackexchange.com/questions/126091/is-there-an-adc-with-a-finite-bound-on-measurement-time)

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    $\begingroup$ Since $e$ is a charge and $k$ is dimensionless (natural number), $k\cdot e$ is still a charge, not an electric potential (voltage). $\endgroup$
    – Jan Hudec
    Commented Aug 20, 2014 at 6:01
  • $\begingroup$ Thanks for pointing that embarassing mistake of mine. I removed that line. $\endgroup$ Commented Aug 21, 2014 at 16:54

7 Answers 7


For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So even though the charge term is quantized, the capacitance — and therefore, the voltage — is not.

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    $\begingroup$ Building on Dave's response... You might think, "Aha, but for a given capacitance, voltage must be discrete because charge is discrete." Not even that is true, at least, not always. Usually we can continuously vary voltage because voltage only determines the average number of charges. Or in another sense, the number of charges varies all the time so the voltage fluctuates and its average is continuous. This sort of subtlety (about what is "voltage") appears regularly in statistical mechanics and has real effects in nanoelectronic devices which only contain a few charges. $\endgroup$
    – Nanite
    Commented Aug 19, 2014 at 18:58
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    $\begingroup$ Actually, atomic spacing/distance is quantized: en.wikipedia.org/wiki/Planck_length $\endgroup$
    – sbell
    Commented Aug 20, 2014 at 13:55
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    $\begingroup$ @sbell: Yes, I was wondering if someone would bring that up. But the Planck length doesn't imply a quantization; it simply puts an ultimate limit on our ability to measure position. $\endgroup$
    – Dave Tweed
    Commented Aug 20, 2014 at 14:30
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    $\begingroup$ @sbell: The article also says, "There is currently no proven physical significance of the Planck length". $\endgroup$
    – Dave Tweed
    Commented Aug 20, 2014 at 14:44
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    $\begingroup$ @user22180: Note that I carefully didn't say "defined", I said "relationship". The OP's question is whether the quantization of charge implies a quantization of voltage, and my example shows that it doesn't. $\endgroup$
    – Dave Tweed
    Commented Aug 20, 2014 at 19:01

Voltage is a continuous function. If you are a certain distance from a (point) charge $q$, the potential is

$$V=\frac{q}{4\pi\epsilon_0 r}$$

By adjusting the value of $r$ to anything you want (not quantized), you can get any potential you want. And so yes, when you do any analog-to-digital conversion, you will "destroy" a certain amount of information.

The question is always "is that of practical significance"? If it is, you need to build yourself a higher resolution converter...

  • $\begingroup$ The problem with ADC's is more than just destruction of information. If voltages are continuous, it means that we cannot build an ADC that is guaranteed to output a logically valid digital value after any fixed (finite) amount of time (assuming the device does not trivially output the same value at all time and that the device's output is a function over the voltage (same input voltage => same output value)). $\endgroup$ Commented Aug 19, 2014 at 17:40
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    $\begingroup$ Noise on any device ensures that for some values of the input, there will be some variation on the output. In my mind that is destruction of information (because you are adding noise to the signal, you lose information about the signal). In other words, there is no device that will give you the same (non trivial) output for a given input, for all inputs - if it has more than one state, then there is a transition region where the "same" input can lead to one of two output states. Even hysteresis won't save you. $\endgroup$
    – Floris
    Commented Aug 19, 2014 at 17:46

Voltage doesn't come directly from the charge of the electron. It's the energy per charge. The charge carriers may be discrete, but the energy is not.

We can easily generate a potential by moving a wire through a magnetic field. The potential is proportional to the speed of the wire, which is a continuous value. $$V = vBL\sin{\theta}$$

  • $\begingroup$ Is that formula just a macroscopic approximation of a discrete phenomena? I guess we will never know! $\endgroup$ Commented Aug 19, 2014 at 17:35
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    $\begingroup$ @Abhishek QED (Quantumelectrodynamics) explains much complexer scenarios as this as well. It is the macroscopic approximation of a continous quantum phenomena. $\endgroup$
    – peterh
    Commented Aug 20, 2014 at 9:19

It's not a fundamental feature of electrical potential, but:

If you have a polycrystalline metal and you cut and polish a smooth surface, the differently-oriented regions will present a different lattice plane to the outside. Crystals cut along different planes may have slightly different work functions, and so the electric potential very close to such a conductor will vary randomly at the level of a few millivolts. This is sometime called the "patch effect" and it can be comparable to the Casimir force (see e.g.) and other small electrostatic effects.


I'm actually rather surprised none of the other existing answers (at time of writing) mention shot noise. So, I'd like to talk a little bit about this, but since there is also a bounty asking specifically about the how quantum field theory plays with voltages, I'll have a couple of things to say about that as well.

There are many forms of noise in electronics. For example there is (approximately) white thermal noise (also called Johnson-Nyquist noise which is the voltage fluctuations caused by the thermal motion of electrons in resistors throughout the device. There's also $1/f$ noise which is often called pink or flicker noise since it was originally observed in vacuum tubes (a low-frequency visual flickering), though I believe the origins of flicker noise are still not very well understood.

Shot noise is another type of noise which is caused by the discretization of electric charges into electrons. Basically the idea is that as current flows through a junction, like a diode, vacuum tube, or anything else really. The effect is, in most applications, smaller than the other types of noise I mentioned, so it sometimes gets overlooked, but it's there nonetheless. The idea is that the current flowing through a junction tells us the mean number of electrons per second passing through, but the number of electrons which actually pass through the junction in a fixed interval of time is a Poisson process, and so there are some fluctuations about the mean in the number of electrons, and then those fluctuations imply fluctuations in the current (and hence voltage) in the circuit.

Since we have discrete numbers of electrons jumping across our junction, we will also have (approximately) discrete jumps in the current (and hence voltage) to do with it, which is why shot noise came to mind when I saw this question. All of this, of course, is really just a semi-classical approximation making use only of the discretization of charge into electrons and does not represent a fundamental discretization of the potential.

So then, let me comment on how quantum field theory (QFT) plays into this since the bounty asks about it. The short answer, which might be a little disappointing, is that even in QFT there is no discretization to the potential, but let me explain a little rather than just leave it there.

The name and language surrounding QFT would seem to imply that fields and everything else are "quantized" in the sense of being "discretized," but these are actually distinct things: Just because something is quantized does not mean it is discretized. So let's just look at what quantization, at least roughly, means. As a disclaimer, everything I say here is going to be a bit informal. To get a full picture of what's going on one would really need to know a little QFT, to which many books are devoted.

In a classical (field) theory, we have a bunch of fields and we specify the "state" of the system by specifying a particular field configuration. So for example, in electrodynamics our fields are the potential $V$ and the vector potential $\vec A$, which we usually put together into a 4-vector $A_\mu$ which we refer to as "the" vector potential. To specify what's going on, we would just need to specify a vector potential which obeys Maxwell's equations.

In a quantum (field) theory, we "promote" all our fields to operators and to specify the state of our system, we specify a vector in a Hilbert space (a vector space with a notion of inner product and some other technical niceties) instead of specifying a field configuration which obeys the equations of motion. So for example, in electrodynamics we would now have 4 operators $\hat A_\mu$ which act on vectors in our Hilbert space.

Ignoring a bunch of subtleties which aren't important to the idea I want to get at, we can actually do something in QFT which makes things more similar to classical mechanics. See, since the Hilbert space is a vector space, we are free to choose a basis of vectors however we like. The "standard" and "nice" way to choose a basis is to use the eigenvectors of some operators, such as $\hat A_\mu$.

There are some technical issues with this involving gauge invariance, but here's the picture we should have in mind. Imagine specifying some field configuration everywhere in space and time, and then just say $|A(everywhere)\rangle$ is the state which corresponds to this configuration we have imagined. Then, by definition, $\hat A_\mu(x)|A(everywhere)\rangle = A_\mu(x)|A(everywhere)\rangle$ where $\hat A_\mu(x)$ is the vector potential operator at the location $x$ in space and time while $A_\mu(x)$ is the value of the vector potential at $x$ which we have imagined.

So, whereas in classical mechanics we could specify the state of the system by a field configuration which obeyed Maxwell's equations, in QFT we can specify the state of the system by (a linear combination of) field configurations, which may or may not obey Maxwell's equations.

This is the main idea I wanted to get at: even in QFT we can still import some of our thinking about field configurations from classical mechanics, just with a few new bells and whistles. More to the point, at no stage here do we encounter a notion of quantization of the field configuration itself. So, while some things are quantized in QFT, not everything is, and in particular the electric potential is not. At least, not any more than it is classically (which is approximately is in the case of shot noise I mentioned earlier).


Voltages on microscale
Voltage is a concept of macroscopic electrodynamics (also electrodynamics of continuous media) - that is, it is a concept applicable to volumes that contain macroscopic number of atoms, although they can be treated infinitesimally small for other practical purposes (e.g., for writing Maxwell equations as differential equations). On a microscale voltage cannot be unambiguously defined, since some of its properties do not hold: e.g., the lumped-circuit description does not apply to microscopic circuits, i.e., due to quantum coherence we cannot view a voltage across a device/circuit as a sum of voltages on its consequtive elements. Another important point at the microscopic level is that voltage is better identified not with the potential difference, but with the electrochemical potential.

Landauer formula
In the standard approach to transport in nanostructures, the Landauer-Büttiker formalism, voltages are veiwed as chemical potentials of the leads (electrodes), which connect the device to the outside world and maintain the non-equilibrium current flow. This view is further generalized to non-interactuing situations, e.g., using the Mair-Wingreen formalism.

Coulomb blockade
Some of the answers have mentioned discrete nature of electrons - indeed, when electrons tunnel through a small region of a space - a quantum dot or even a simple semiconductor junction of high quality - the discreteness of charge starts to play a role and results in the phenomenon, known as Coulomb blockade, where the current through the device is possible only at certain values of voltage across the junction, determined by the capacitance of the latter. The voltage here is understood in Landauer-Büttiker sense, and is still continuous.



I interprete voltage as the static difference in chemical potential between two conductors. I assume that the geometry is constant, as is reasonable for a device. With these assumptions voltage is discrete as it depends on the discrete number of charge carriers on each conductor.


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