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The phase difference can be measured but is it possible to measure the instantaneous phase of the electric field in light wave at one position?

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    $\begingroup$ If you can define it, you can measure it (in principle). However, to actually define phase, you need a temporal reference, which one can argue ultimately means that you're just defining the phase difference between your system of interest and your reference oscillator. So: define "phase" without that caveat, or you're just running in circles. $\endgroup$ – Emilio Pisanty Jun 1 '17 at 11:24
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Yes, for a classical field (e.g. a large photon number, coherent state), in principle the phase can be measured in the sense that one can in principle observe that the electric (magnetic) field at a given point oscillates sinusoidally with time and one can measure at what times the field is zero and when it reaches its maximums.

If you don't believe this for light, simply imagine the experiment with microwaves, or radiowaves. Phase locked loops have been synchronizing oscillator circuits with incoming radio waves for the best part of a hundred years.

So at optical frequencies, the difference is one of technology only, not physics. Optical phase locked loops now exist and realize coherent modulation schemes over optical fiber communication links.

If you mean by your question the phase of a quantum state of light, then, of course, the state is a ray in projective Hilbert space and is invariant with respect to multiplication by a global phase. No meaning can therefore be given to the absolute phase of a quantum state. Phase differences between the complex amplitudes of base states in superposition are meaningful, and these determine interference effects and the evolution of the quantum state.

Classical and quantum phase are quite distinct, however, and don't simply merge through a "correspondence principle" in a large photon number limit. To understand the difference between them, and the lack of a "correspondence principle", consider a photon number coherent state $\psi(\alpha,\,t)$ of the quantum harmonic oscillator:

$$\begin{array}{lcl}\psi(\alpha,\,t) &=& e^{-i\,\frac{\hbar}{2}\,\omega_0\,t}\,\exp\left(\alpha\,e^{-i\,\hbar\,\omega_0\,t}\,a^\dagger - \alpha^\ast\,e^{+i\,\hbar\,\omega_0\,t}\,a\right)\,|0\rangle \\&=& e^{-i\,\frac{\hbar}{2}\,\omega_0\,t}\,e^{-\frac{|\alpha|^2}{2}}\,\sum\limits_{k=0}^\infty\,\frac{\alpha^k}{\sqrt{k!}}\,e^{-i\,k\,\hbar\,\omega_0\,t}\,|k\rangle\end{array}\tag{1}$$

which is a superposition of Fock (number) states beating together as their relative phases oscillate at different frequencies. Suppose we measure the photon number with the number observable $\hat{n}=a^\dagger\,a$; its mean $\langle\psi|\hat{n}|\psi\rangle$ is constant at $|\alpha|^2$ (indeed, so are all the moments $\langle\psi|\hat{n}^k|\psi\rangle;\,k\in\mathbb{N}$ and by calculating them, one can show that the photon number is Poisson distributed with mean $|\alpha|^2$). However, suppose we measure the position with the position observable $\hat{x}= \sqrt{\frac{\hbar}{2}\frac{1}{m\omega}}\,(a^\dagger + a)$ or the momentum with observable $\hat {p}= i\sqrt{\frac{\hbar}{2}m\omega}\,(a^\dagger - a)$; the means of these measurements $\langle \psi|\hat{x}|\psi\rangle$, $\langle \psi|\hat{p}|\psi\rangle$ are phase-quadrature, time harmonic functions:

$$ \langle \hat{x}(t) \rangle = |\alpha| \sqrt{\frac{2\,\hbar}{m\,\omega}} \cos (\arg(\alpha) - \omega_0 t)$$ $$ \langle \hat{p}(t) \rangle = |\alpha| \sqrt{\frac{2\,m}{\hbar\,\omega_0}} \sin (\arg(\alpha) - \omega_0 t)\tag{2}$$

and the phase of these quantites, i.e. $\arg(\alpha)$, is the classical phase. Take heed that we can multiply the state $\psi$ in (1) by any wildly time-varying quantum phase $e^{i\,\varphi(t)}$ we like and the results of the above calculations will be unchanged; the quantum phase disappears under the multiplication of the state and its complex conjugate in the calculation $\langle \psi|\hat{X}^k|\psi\rangle$ of any moment of any observable $\hat{X}$. Note that the classical phase is perfectly well defined for any, arbitrarily small mean photon number coherent state (it's simply that for $\alpha$ very small, you'll need many measurements on the same state to measure it, whereas a single measurement will do for laser-sized values of $\alpha$).


For measurement of classical phase of light, see Emilio Pisanty's answer here describing the relevant experimental technology in the Physics SE thread Have we directly observed the electric component to EM waves?.

Also, one can indeed define an observable for the classical phase of a quantum state, although it's tricky; an influential quantum optics researcher has written a whole book on the subject:

Stephen M. Barnett & John A. Vaccaro, "The Quantum Phase Operator: A Review"

(Stephen M. Barnett, along with Bruno Huttner, was one of the first people to work out a fully quantized theory of the electomagnetic field in a dielectric (in this paper here)).

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  • $\begingroup$ On the technological side, attosecond streaking (as I explained here) can directly measure the waveform in the IR regime, though you'd struggle if you wanted to trace out more than ten-twenty cycles or so. However, it is still, arguably, an interference experiment, as you always need a phase-locked reference - including the microwave and radio experiments you describe. $\endgroup$ – Emilio Pisanty Jun 1 '17 at 9:48
  • $\begingroup$ On the quantum side, that depends on what states you're measuring. For Fock states the phase is obviously completely undefined, but not all states are Fock states - and there are indeed phase 'eigenstates' (modulo some strong difficulties in defining them). Phase is (almost) as legitimate an observable as anything else - or, at the very least, there is a perfectly workable POVM for its measurement. (More details here.) $\endgroup$ – Emilio Pisanty Jun 1 '17 at 9:51
  • $\begingroup$ ... though re-reading your post, you're talking about Hilbert-space phases, which are not the correct analogs to the classical phase of an oscillator. Instead, the quantity of interest is the quantum phase as described in my last link. $\endgroup$ – Emilio Pisanty Jun 1 '17 at 11:22
  • $\begingroup$ @EmilioPisanty could you possibly summarize for those of us who don't have access to the book? Maybe you could write your own answer? $\endgroup$ – Brian Bi Jun 1 '17 at 19:25
  • $\begingroup$ @BrianBi To be honest, the quantum perspective is a good deal of a sidetrack w.r.t. the OP, so it wouldn't be all that appropriate to add an answer here. On-site there's this question but it's not a great fit and I'm not thrilled with the answers; it would make a good separate question - though you'd need to demarcate it clearly from the existing one, maybe along the lines of "given that the naive view won't work, what can you do?". $\endgroup$ – Emilio Pisanty Jun 1 '17 at 19:33
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Yes, of course it is possible to measure phase. Holography works by combining a source wave with a reflected-light wave and where the phases match (zero degrees) the light intensity is high (and darkens the film), and where the phases mismatch (180 degrees) the light intensity is minimum, and the film stays transparent. That hologram image, then, captures the reflected-light phase (and reconstruction of the image afterward is therefore possible by illuminating the film with a copy of that original source wave).

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    $\begingroup$ "combining a source wave with a reflected-light wave" is probably the kind of thing OP meant when they said they know "the phase difference can be measured", so this doesn't really answer their question. $\endgroup$ – The Photon Jun 1 '17 at 14:35
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    $\begingroup$ The meaning of 'phase' that makes most sense to me in this context, is the value (phi) in the oscillatory expression "sin(omega *t + phi)" for the wave at a point in space. So, an arbitrary choice of the zero of the time elapsed is intrinsically required, i.e. there must be a reference. So, the physical situation of creating a hologram defines that reference with the 'reference beam'. The reference beam can take another absolute phase and the hologram will reconstruct because only phase differences are important. $\endgroup$ – Whit3rd Jun 1 '17 at 20:58
  • $\begingroup$ I agree with you that that's the definition that makes sense. But based on the question asked, I'd guess that OP doesn't understand that, or maybe even why it's an issue. So a complete answer should explain why what OP asked for is not sensible. $\endgroup$ – The Photon Jun 1 '17 at 21:49

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