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In basic quantum mechanics courses, one describes the evolution of quantum mechanics chronologically. Interference experiments with particles showed that particles should have a wave character; on the other hand, the photo-electric effect showed that electromagnetic waves should have a particle character. So there's a particle-wave duality.

The course continues with postulating the Schrödinger equation, and solves the problems for the wave character of classical particles (e.g. with respect to harmonic oscillators to solve the UV-catastrophe in the description of blackbody radiation). On the other hand, one describes the photo-electric effect extensively with perturbation theory, as the electromagnetic wave acts as a periodical force on an electron. But the electromagnetic wave is still described with the Maxwell equations, and not by Schrödinger.

Suppose history went a little differently: the corpuscular character would still be the generally accepted behaviour of light (as Newton preferred), and one discovered the wave character of photons and electrons simultaneously by doing interference experiments. The laws of Maxwell, Gauss, Faraday and Ampère weren't discovered yet. Schrödinger comes up with his equation, and one describes the wave character of electrons as well as photons with this equation.

Would this equation give a exhaustive description of the electromagnetic (and vector like) behaviour of photons? In other words: is the Schrödinger theory, in some way, equivalent to the Maxwell theory for the description of photons? Or are the Maxwell equations some kind of limit for greater dimensions (like the Newton equations for mechanics)? What is the link between these two "wave character" descriptions of photons?

BNJMNDDNN
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    The Schrodinger Equation is not relativistically covariant so it doesn't have a chance of describing photons. – Alfred Centauri Jun 17 '13 at 22:01
  • So the wave character of the "classical particles" and the "classical waves" (neglecting the fact that Maxwell's equations were in fact "modern" already) aren't actually comparable, apart from the fact that they are waves? Is there a relativistically covariant variant of the Schrödinger equation (as there's the Lorentz-transformation for the Galileo-transformation)? Some equation that would describe all wave-like behaviour in nature? – BNJMNDDNN Jun 17 '13 at 22:31
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    Interestingly, Schrodinger worked with a covariant equation first and then gave up on it. That equation is now know as the Klein-Gordon equation. – Alfred Centauri Jun 17 '13 at 22:36
  • The Weyl Equation describes photons, it is a special case of the Dirac Equation where the particle is massless. – Abhimanyu Pallavi Sudhir Jun 18 '13 at 04:57
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    @dimension10 Doesn't the Weyl equation rather describe neutrinos, since it describes a system with spin 1/2? – twistor59 Jun 18 '13 at 09:02
  • @twistor59: It definitely doesn't describe neutrinos (in reality, but in the standard model, they do) but the Weyl Equation for Rank 2 tensors does describe photons because the photon field is a massless dirac field. – Abhimanyu Pallavi Sudhir Jun 18 '13 at 09:13
  • @dimension10 The term "Weyl Equation" is generally reserved for the equation satisfied by spin 1/2 fields, e.g. section 4.4.1 here – twistor59 Jun 18 '13 at 09:24
  • related:What equation describes the wavefunction of a single photon? – Larry Harson Jun 18 '13 at 14:15
  • "The concept of E and B as ordinary fields is a classical notion ... In the domain of macroscopic phenomena and even in some atomic phenomena the discrete aspect of the electromagnetic field can usually be ignored" J.D. Jackson –  Aug 16 '13 at 01:31

4 Answers4

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Are the Maxwell equations a correct description of the wave character of photons?

Yes, Maxwell's equations are the wave equation for a photon, just as the Schrodinger equation is the wave equation for a nonrelativistic electron.

In other words: is the Schrödinger theory, in some way, equivalent to the Maxwell theory for the description of photons?

As others have pointed out, you can't apply the Schrodinger equation $i\hbar d\Psi/dt=-(\hbar^2/2m)\nabla^2\Psi$ to photons, because the Schrodinger equation is nonrelativistic, and a photon is never nonrelativistic.

Suppose history went a little differently: the corpuscular character would still be the generally accepted behaviour of light (as Newton preferred), and one discovered the wave character of photons and electrons simultaneously by doing interference experiments.

This is a very good question. There is a very nice discussion of this sort of thing in Peierls, Surprises in Theoretical Physics, section 1.3:

One of the most basic ideas of quantum mehcniacs is the analogy between light and matter [...] From this, it might appear an accident of history that physicists originally encountered only the wave aspects of light and only the corpuscular aspects of particles with mass, such as electrons. It therefore comes as a surprise to discover that this is no accident at all, and that the analogy between light and matter has very severe limitations.

I won't try to provide an analysis as complete as Peierls' in a physics.SE answer, but a crucial point is that because photons are bosons, you can have a coherent superposition of photons in which a large number of photons are packed into a volume equal to a cubic wavelength. Such a superposition can have a well-defined amplitude and phase that can be measured by classical measuring devices such as antennas.

But you can't do this with electrons, because they're fermions. This is why the electron wavefunction isn't a classical field that can be measured directly.

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    I feel it's a bit dangerous not to be using the term "Schrodinger equation" for the "time-dependent" Schrodinger equation for time evolution or at least make it clear that you're referring to the eigenvalue problem for a particular Hamiltonian as opposed to the time-evolution equation. I say this because since the Schrodinger equation governs time-evolution of every physical system, even relativistic ones such as those involving photons, and some might get confused. – joshphysics Jun 18 '13 at 04:03
  • @joshphysics: so according to your comment, it is possible to describe electrons as well as photons with the time dependant (relativistic variant if needed) of the Schrödinger equation? And does that lead in some way to the same conclusions as the Maxwell equations? – BNJMNDDNN Jun 18 '13 at 09:34
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    @BNJMNDDNN The Schrodinger equation ("time-dependent") describes the time evolution of every physical system. In order to describe photons quantum mechanically, however, one needs to additionally use quantum field theory (QFT) in order to properly model the quantum states of photons which are interpreted as arising from quantized electromagnetic fields. If you want to learn more, I'd recommend reading about quantum electrodynamics (QED). – joshphysics Jun 18 '13 at 09:44
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    This is the standard confusion between the Schrodinger equation = the fundamental and always exact equation of quantum mechanics, vs the Schrodinger equation which happens to be the non-relativistic approximation to the Dirac equation for the electron field. The equations look identical but have totally different interpretations. I wish this was drilled up front and explicitly in every quantum mechanics course, but unless the teacher has done enough qft themselves they probably don't even know the difference. – Michael Jun 18 '13 at 12:49
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    @BNJMNDDNN, the distinction joshphysics is making is the distinction between the abstract Schrodinger equation (which is the statement that the Hamiltonian operator is the generator of infinitesimal time evolution), and the partial differential equation that bears his name. In your original question, to which were you referring? – Alfred Centauri Jun 18 '13 at 13:45
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    @joshphysics: Thanks for the comment. I've clarified in the answer which meaning of "Schrodinger equation" I was assuming the OP had in mind. –  Jun 18 '13 at 14:21
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    @BenCrowell I have 2 Questions, First how is the probability defined , if you use maxwells equation as wave equation for a photon(My guess is it will be E^2+B^2). Second what does fermionic nature have to being unable to use classical apparatus to detect the field like nature of the Electron. – Prathyush Jun 19 '13 at 01:13
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    Re the first question, see this http://physics.stackexchange.com/questions/437/what-equation-describes-the-wavefunction-of-a-single-photon . The Peirls book has a good discussion of this; he points out that $E^2+B^2$ has the wrong transformation properties to be a number density. Re the second question, the idea is that to get the classical-field behavior, you need a large number of particles so that you can do classical measurements in which the random fluctuations average out. –  Jun 19 '13 at 02:28
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    @AlfredCentauri: I haven't heard of those distinctions between different Schrödinger equations. I've had 2 quantum courses where we just used time-dependant and mostly time-independant Schrödinger equation (to solve potential problems e.a.). I never had QED of QFT (yet), so a lot of the answers here are a bit difficult to understand for me. But thank you all for your answers, it's a little more clear for me now! – BNJMNDDNN Jun 20 '13 at 17:25
  • The book from Peierls, Surprises in Theoretical Physics is really a gem and available for free from https://archive.org/details/SuprisesInTheoreticalPhysics – asmaier Nov 20 '16 at 22:02
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Would this equation give a exhaustive description of the electromagnetic (and vector like) behaviour of photons? In other words: is the Schrödinger theory, in some way, equivalent to the Maxwell theory for the description of photons? Or are the Maxwell equations some kind of limit for greater dimensions (like the Newton equations for mechanics)? What is the link between these two "wave character" descriptions of photons?

A photon is an excitation of a "mode", i.e. a solution of Maxwell's equations satisfying the appropriate boundary conditions. For example, a field constrained to be within a cavity has to satisfy the boundary conditions determined by the cavity. A field in free space might be a spherically symmetric solution, depending on the source properties etc. Once you have chosen your solution you can, in principle (in practice this may be very tricky - unless you're doing particle physics rather than quantum optics!), put a single excitation into it, creating a one-photon state.

Now, although the mode is a solution of Maxwell's equations, the state (at least in the Schroedinger picture), satisfies a Schroedinger equation. This is just equivalent to saying that it evolves unitarily in time.

This Schroedinger equation, however, isn't the "wave equation for the photon" in the same way that in single-particle quantum mechanics the Schroedinger equation is the wave equation for the particle. Rather, it's the time evolution equation for the state, which takes place in Hilbert space, not in spacetime. Trying to emulate the single particle QM description by constructing a wavefunction for the photon is difficult:

The wavefunction would be the inner product of the state with position eigenstates $|x\rangle$ $$ \Psi(x)=\langle x|\Psi\rangle$$ The difficulty comes about because there isn't a(n undisputed!) Lorentz invariant position operator $\bf{\hat{x}}$ for photons.

However, we can create single photon states. These, however, are not really "like" the classical field which corresponds to the mode you excited. See Lubos' answer here for discussion of the electric field in a single photon state for example. If you want something which looks like the classical field, you need to construct the corresponding coherent state. This does not have a definite number of photons.

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twistor59
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  • Excellent answer. I really like to think about the photon as being only $\left|\Psi\right\rangle $ or as you said a state in an abstract Hilbert's space, then the photon can be seen as $\left\langle n|\Psi\right\rangle $, whereas the wave would be $\left\langle x|\Psi\right\rangle $. Thanks for sharing your point of view. – FraSchelle Jun 18 '13 at 13:03
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    Equivalent of the standard quantum oscillator $X(t)$, for the quantized Maxwell field, is, for a given momentum and polarization, $\Phi_ {\vec k, \lambda}(t)$. So you can define a "Schrodinger equation" by replacing $x$ by $\Phi_ {\vec k, \lambda}$ (idem for differential operators). It is not a very used formalism, it is perfectly correct, and you have a wave function $\Psi(\Phi_ {\vec k, \lambda}, t)$. The equations for different momentum and polarizations being independent (linearity), we may think of a global wave function as $\prod_{\vec k, \lambda} (\Psi(\Phi_ {\vec k, \lambda}, t))$ – Trimok Jun 18 '13 at 17:13
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Following the comment by Alfred Centauri, let me suppose that you discussed the Dirac theory of electron, and not the Schrödinger one. I will come back later on the possibility to describe matter-light interaction using the Schrödinger equation.

The Dirac theory describes the (special) relativistic behaviour of a particle. When complemented by the principle of gauge invariance (in particular the substitution of the normal derivative by the covariant one), it gives the basic playground for the simultaneous descriptions of the electromagnetic field (Faraday law and absence of magnetic monopole), the charge associated to the relativistic particle (the equation which replaces the Newton equation with the Lorentz force if you wish, but this has to be though with care) and their coupling (equations similar to the Maxwell-Ampère and Gauß, but there the current and charge densities have full quantum meaning, no more fluid interpretation as for the classical electromagnetism).

Obviously everything get more complicated when you try to quantise the electromagnetic field. The previous discussion didn't discussed the appearance of the photon.

I would say that the Wikipedia page related to the Dirac equation is not so helpful for understanding this point, but you could try to open the book by A. Messiah Quantum mechanics (volume II if not in an edition with the two volumes in one book), which contains all the pedagogical details you need, including the quantisation of the electromagnetic field in term of photon.

Schrödinger vs. Dirac description of matter

One can also describe the interaction between matter and light using the Schrödinger equation for the atom. This is the main study of the book

C. Cohen-Tannoudji, J. Dupont-Roc and G. Grynberg Photons and Atoms: Introduction to Quantum Electrodynamics, Wiley (1992)

that I suggest you to read. In short, when the magnetic-like interaction is weak, the description using the Schrödinger equation is sufficient. You can understand this with the pictorial idea: when the (continuous laser) light field does not interact too much with the (gas) atoms, the effect can be describe by the first order term in interaction, which is already given by the Schrödinger prescription.

Historical perpectives

Now, regarding your historical perspective, it seems highly not probable that Dirac would have described the coupling between electrons and photons if he were unaware of the Maxwell's equations. This is once again because the gauge invariance is crucial in deriving the coupling. You may find more details about the history of gauge theory in the excellent collection of historical articles by

L. O'Raifertaigh The dawning of gauge theory, Princeton series in Physics (1997).

The same reasoning apply to the Schrödinger equation, because all these physicists were deeply influenced by the notion of field, that Maxwell really invented half a century before.

In short, the gauge invariance is the main ingredient of matter-field interaction, not the equation you're using to include it.

To be also noted:

  1. The particle behaviour of light was not the generally accepted behaviour of light (as you said) at the time of Maxwell's equations. Indeed, the Young two-slits experiment was already known by the end of the 18-th century.

  2. You do not really need to quantise the photon field to understand the photoelectric effect. This is discussed in a paper

    Lamb, W. E., & Scully, M. O. The photoelectric effect without photon, in Polarisation, matière et rayonnement (pp. 363–369). Presses Universitaires de France (1969).

where they calculate the photoelectric effect quantising only the electron / detector behaviour.

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FraSchelle
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  • Please see also the complementary answer by @twistor59 below: http://physics.stackexchange.com/a/68450/16689 – FraSchelle Jun 18 '13 at 13:03
  • I just realised that @BenCrowell was discussing the paper by Lamb and Scully in a separate post http://physics.stackexchange.com/q/68147/16689 he gave. – FraSchelle Jun 19 '13 at 05:17
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The photon is an elementary particle.

Elementary particles are described by their rest mass, spin and quantum numbers as seen in the table. They then have attributes that are variable, like energy and momentum, which are states measured with the corresponding quantum mechanical operator.

The photon has spin 1, zero mass and zero charge. Its energy is given by

E=h*nu, where nu is the frequency characterizing it .

The frequency is the one common point with the classical electromagnetic light description. This is verified with the double slit experiment which displays the probabilistic nature from quantum mechanics when the photons are projected one by one and the interference pattern builds up . It is exactly the interference pattern given by the solutions of the classical Maxwell equations of electromagnetic waves.

Somehow, the large ensemble of photons contained in the light wave are coherently producing the electromagnetic wave studied macroscopically. Of course this is necessary for a self consistent physical theory. The macroscopic mathematical forms have to emerge from the microscopic theories.

So yes, Maxwell's equations do describe photons since they give a light wave with frequency nu, the defining property of each photon in the ensemble.

@LubošMotl has an article on his blog that shows how from the second quantization description of photons the classical electromagnetic wave emerges, similar to how thermodynamic quantities emerge from statistical mechanics.

The Schrodinger equation in this sense is not equivalent to the Maxwell's equations.

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