No it does not.

If it somehow did, it would have to go through zones where its presence would be quantum physically impossible, and besides it would take a nonzero time to move.

You must think in terms of creation and annihilation operators, which is the only coherent way to understand quantum leaps.

The excitation of an electron to an energetically higher state is not connected with a local removement either in a isolated atom or in a solid state material. But in a solid state the possibility for a local displacement is created by stimulation. The electron can be moved away in the conduction band.

I often have used the term  "the electron is lifted to" in order to give some idea of the change of the state, and therefore heard the same question as Vijay,

Younger students today can probably identify better with the term "altered state" in order to dispel the picture book concepts of classical mechanics when describing the electron and its properties.

I'm afraid to say it, but I've got to disagree with the answers here. An electron certainly does move when absorbing a photon. It obviously moves in many molecular excitations, such as an n to pi* transition, where the initial and final states have drastically different wavefunctions centered in different areas with respect to the molecular center of mass. But I'll argue that it moves during atomic transitions as well, and even in solid state transitions.

If we consider a 1s -> 2p excitation in hydrogen atom, and work through some time-dependent perturbation theory, treating the electric field classically and the atomic system quantum mechanically, we find that the perturbation mixes in some small amount of 2p character into an initially ground-state hydrogen. The in-phase portion of the 2p orbital interferes constructively with 1s, and the out-of-phase portion interferes destructively, causing the electron distribution to be centered slightly to one side of the nucleus. This is a superposition state, which will naturally oscillate at the difference frequency of 1s and 2p, and after a half cycle the electron is centered on the other side of the nucleus. This produces an oscillating electric field, which gives rise to a free induction decay, and is exactly how the electron is able to absorb a photon under the dipole operator.  The details of this quantum description differs quite a bit from the classical, but the main picture of an electron interacting with a light field by oscillating side-to-side about its equilibrium position is preserved quite well, even when both initial and final states have probability densities centered at zero.

While this is a semiclassical argument, the same picture can be derived from a fully quantized description (both atom and light field treated quantum mechanically) by using time-dependent density operators and a position basis. I'm not sure how one would attempt to describe the details of many photon echo experiments without using some concept of a material polarization, which implies that charge distributions are oscillating at optical frequencies.

This should also work in the solid state, although I'm only really conversant with LCAO methods (I believe you solid-state guys call this tight-binding). We write a wavefunction as the product between some plane wave (wavelength on the order of a unit cell up to much longer) and some local wavefunction which is otherwise the same at every lattice point, which I'll take to be atomic or molecular orbitals. Solid state people like to focus on the long-wavelength portion, but the unit-cell part of the wavefunction will behave exactly like individual atoms, and will have the electron oscillating side-to-side in order to absorb a photon.

One could argue that if the wavefunction really extends over an entire infinite crystal that such an oscillation won't matter, but in real experiments an electron is localized to some wavepacket during the absorption process. The size can be calculated by taking your laser bandwidth (assuming the coherence time is at least as long as the laser pulse) and using the band structure to convert to a momentum width; heisenberg uncertainty can allow you to convert that to a position width. When I was a graduate student we did an experiment with 25fs laser pulses in PbS that produced wavepackets 15-20 nm wide, if I remember correctly. I don't know if that value is typical or not, but I would bet most experiments produce less localization. Note of course that the electronic oscillation at each lattice point is very small compared to a lattice constant, and the wavepacket is large, so the electronic motion needed to absorb a photon is very small (but not zero).

William Peters gives a beautiful and interesting answer, but I'm personally inclined not to identify the electron with its wavefunction -- the first is a particle and the second is a piece of math -- so that changes in the wavefunction are not ipso facto changes in where the electron is at any particular instant.  I would tend to answer the question by saying that, as far as we know, absorption or emission of a photon by an electron is an event that happens instantaneously and at one point in space and time, since both the photon and electron are point particles (when they behave as particles).  That precludes any kind of motion during the actual absorption event itself.

But of course, since the state the electron ends up in is by assumption different from that in which it started, the probability distribution of its position before and after the absorption will generally be different, so in that sense, of course the electron generally has to "move" as a consequence of absorption.  Move how and where -- this depends on the exact nature of the before and after wavefunctions.

I don't think that the absorption and emission of photon by an electron (Compton scattering) is happening at the same space-time point. A real electron can absorb a real photon (1st vertex at a given space-time point ) and "propagates" as a virtual electron  until it (the electron) undergoes another interaction  (e,g. interaction with vacuum will be accompanied by the emission of a real photon ) at different space-time point (second vertex)  and becomes a real electron. The propagation between the two vertices is understood as being  virtual. 

Christophers explanation meets my thoughts. Absorption and fixing of an electron later on by some interactions processes are not the same. The convincing spatial change of a molecular transition to a different orbital is not the absorption process itself. It´s just a following change of the state and the orbitals from energetic and some other atomar reasons.

Your question I find needs three answers.  It's terribly complex, even for one of those answers. 

For the main question, which is an excited electron in an atom, it does change 'orbitals', or 'shell.'  Depending on the application this can be perceived/measured as a movement of the electron.  For example a metastable state, the orbital can be considerable elliptical, where the major axis is five times larger than the minor.  The electron spends considerable time 'away' from the nucleus.  I call that moving.  But it is not conduction band related.

By "application" I mean what you are using the electron for, and even how you are measuring it.  Measure it as if it's a particle, then it moves.  Measure it as if it's a wave, then it does not 'have' to move, but it might.  Recent experiments have measured a particle in both states, both a wave and particle, previously thought not possible, by the main stream.

If the change in orbital is from one spherical shell to another, then it's waveform has spread out.  Does that mean it has 'moved' to you?  It's 'center' has remained the same.  Just it can be measured further from the atom, more often.

Regarding the conduction band, as I understand them, they do not exist unless current/voltage is present.  Thus, the electron moving from being captured by it's atom, into the 'current', why yes, then it must be moving, as it's now current.  Is that what you mean?  Exciting a solid to have a conduction band, without current (just heat it up), is possible.  "Measuring" the electron in question likely will create current?  Thus, moving your electron?

I'm not trying to be contrary, just pointing out the second question you pose is not easily answered, yes or no.  Again, it depends on the application/measuring method.  Now, to your third question.

Trying to isolate an atom, one atom, and it's electrons, outside of a 'solid state' and asking the question concerning 'conduction band' is poorly understood, at least by me, as there is no conduction band in a single isolated atom.  Conduction band exists in what physics calls 'solid state.'   That is, current flows in the solid/liquid/gas.

So, you can now tell, from the above simplistic views I have presented, of any QM Interpretation, Copenhagen or those less popular, or from the Classical viewpoint, it's very complex, depending on the "application."

What I think you have found by everyone's answer, is QM plays a huge role these days in your questions.  What I am answering, is until you 'measure' it, you can not know.  That's what basic QM states (Copenhagen).  And using Classical physics, the answer would be yes, likely always, as measured relative to it's atom nucleus.  At a minimum, it has 'moved away' from the nucleus.  Been "lifted."  Further from the positive attraction of the protons, so it can participate in the current.

My last words on this issue, it sounds like you are explaining to students.  They are taught white lies throughout most of the curriculum.  Later, when exposed to underlying advanced theories, they get lost, easily, due to having a 'poor' intuition built upon the preceding years of white lies.  Good luck with them.  I suspect my answer has not assisted you much.

This reply I will focus exclusively on "Conduction Band".  It is not a classical concept.  It is just about a pure QM mathematical construct.  With advances in the last decade, the QM math predictions have become experimentally verified.  This verification means the QM math is a good modelling tool, for predictions of new alloys/minerals and their conductivity.  Many new superconductors were predicted by their molecular makeup and verified by experiment.  So, the QM definition of "conduction band" is now widely accepted. 

There is no equivalent "classical viewpoint" of this conduction band.  One could ask why.  Briefly, your question hits the nail on the head.  Electron "moves" ... just is not a question that the QM math can answer.  And there is no classical math/model for this type of "conduction band."  Not like shells, orbitals, Bohr style, or other approach.

Thus, if you are teaching students, about conduction bands, and attempting to related that classical electron velocity and position, I suggest you not make this attempt.  Instead, teach that classical model does not exist for conduction band theory.  Instead, one must use the QM approach.

Another way of talking about this, is from a side approach.  In the conduction band can an individual electron be "identified?"  Does it belong to that nucleus?  Or does the electron in the band get 'shared' between two or more nuclei?  Can that electron be distinct/distinguished with a second electron that is shared with the same nuclei?  These questions focus on classical approach, mapping it from the QM conduction band model. 

I do not have answers to these questions.  I do not know if in the last 5 years, since I 'caught' up with current conduction band theories, if anyone can answer these questions.  I asked these questions to demonstrate that the accurate QM modelling math does not provide classical answers.

And that's a key point as to why QM approach to reality is likely to 'beat' the classical approach for conduction bands.  QM makes good predictions, where there is no classical prediction, as there is no classical model worthy of mention (please post if there is such known to you).

When an electron enters the conduction band, one can no longer talk about individual electrons, in that band.  The electron has 'morphed' into a 'cloud' or waveform that has no solution for individual particles.  Thus, the QM waveform makes accurate predictions (more than 3-4 decimal places) for resistivity of the solid state, over a range of temperatures.  I could be outdated in this opinion, as the last I updated myself was over five years.  I find it very interesting.

Now, to clarify "lifting" an electron away from it's nucleus, exciting it into the conduction band, making it more 'isolated' from the nucleus proton positive and attractive charge.  The 'distance' of the excited electron from the nucleus does not change during this "lifting".  Say what?  The direction of lift would have to be radially outward from the nucleus, right?  And in a solid state, that means this electron moves towards a neighbouring nucleus, or two.  But to do that it must overcome those nucleus' electrons' repulsive force.  And it can not. 

Thus, when an electron in it's ground state is excited into the conduction band, the electron does not "move" in any classical sense, of outward radially motion from the nucleus. 

What does happen is the electron 'transitions' from the ground state 'band', through the forbidden band (no electrons in this energy range exist), and into the higher energy conduction band.  Thus, it becomes available as 'current.'  Without moving.

Now, being available as current, has it's own set of questions, quite similar to the classical questions I posed above.  Do individual electrons actually 'move' to create the current?  Certainly, there is an electron flux leaving and entering the solid state component (i.e. resistor, etc).  That does not explicitly imply the entering electrons push excited electrons to exit the medium.  Again, I do not believe QM modelling has any prediction for this question.  I'd like to know.

Do conduction band electrons exist as individual electrons?  Not according to the QM math.  So, asking if they move, is not answerable by QM math.  Nor by classical math, except as a gross approximation, in the limit of QM math, that is, electron flux entering and exiting the volume where the conduction band exists.  What happens to electrons between the entry and exit surfaces, is classically unknown, and perhaps unknowable as human knowledge.