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I Laser vs photodiodes and spectra

  1. May 11, 2017 #1
    Hi all!

    I would have two questions, related to laser and photodiodes spectra.

    1) We know that lasers produce a very monochromatic radiation, even if they are low to moderatly expensive. That is because the emitted frequency light is dependend on Ec - Ev = Eg which is the bandgap. So electrons are stimulated to "jump" from the higher level to the lower level, emitting a photon. Photodiodes works similarly; they absorb a photon and an electron jump from the lower level to the higher one.
    The question is, if the bandgap is the same, why emitting spectrum of laser is much narrower than absorption spectrum of photodiodes?

    2) Now suppose that you have a perfect monochromatic laser. They emit only at a certain frequency, so the light is a pure sine in time domain. The Fourier Transform is than a Dirac Delta (actually two deltas), and it's right because it emits only at one frequency.
    Now suppose you don't turn the light on permanently but you emit a single short pulse. In the time graph we could model the function in many ways like a sine times a gaussian, or a sine times a raised-cosine. If we make the fourier transform of the signal we would obtain not a delta but some broader curve. And this is mathematically right.
    But what happens physically? If the laser is perfect and emits at only one wavelength, if I pulse it, it should still emits at that frequency, isn't it? But the spectrum mathematically broadens, so I cannot figure out what happens

    Thank you
     
  2. jcsd
  3. May 11, 2017 #2

    Drakkith

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    If the energy of the absorbed photon is greater than the bandgap it can still be absorbed. The electrons are excited into the conduction band of the material, which has a huge number of closely spaced states that effectively forms a continuum. The excess energy usually goes into heat I believe.
    In lasers, there is a dominant transition that the laser works on and each state that the electrons transition from/to is at a set energy level, unlike the conduction band of a photodiode.

    You cannot make a perfectly monochrome laser unless it has always been in operation and will forever be in operation. Turning it off or on introduces other frequencies into this spectrum and so in reality a perfectly monochrome laser is a impossible as a perfect plane wave.
     
  4. May 11, 2017 #3

    Henryk

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    The answer to your question 1 is the following:
    Both, conduction and valence band are quite wide. A photodiode will absorb a photon whose energy is at least equal to the energy gap but if the photon has a higher energy, it will be absorbed by an electron transition from below the top of the valence band to above the bottom of the conduction band. The upper limit is essentially the energy gap plus the width of the valence band plus the width of the conduction band.
    Laser requires a transition of electron from the upper to the lower state. A semiconductor laser is essentially a p-n junction. In the n-region, you have electrons at the bottom of the conduction band up to around kBT. In the p-region, there are holes only at the top of the valence band down to c.
    Thus, there available transitions for a semiconductor laser is limited to where you have carriers and that's around 2kBT.
    Then, comes the laser's resonance cavity. It allows only a discrete set of wavelength. That narrows the wavelength of the emitted light even further.

    As for question no. 2. The math is right. A short pulse can be represented by a superposition of sine wave of a range of frequencies. and this is, mathematically, exactly the Heisenberg uncertainty principle:
    For a perfectly monochromatic light, the energy uncertainty (##\Delta E = h\Delta \nu ##) is zero, that means it time uncertainty has to be infinite, i.e. the sine wave has to continue forever.
    If you localize the wave to a finite time interval, you also introduce uncertainty in energy, hence, a spread of frequencies.
     
  5. May 12, 2017 #4
    1) Thank you both, now it's make more sense to me!

    2) So it's just a matter of uncertainty? If we had a gas laser which does not have a band of allowed levels, but only two dicrete levels, would that still hold? If the energy of emitted photons depends on the difference between the two energy levels, and in this case they are precisily two, a part from the non idealities of the hypothetical device, how may the frequencies be different from that one?
     
  6. May 12, 2017 #5

    f95toli

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    Yes, the math is still correct:smile:

    The fact that you can't turn a signal on and off without changing the frequency content of that signal is somewhat difficult to wrap your head around; but it is indeed always the case.

    One way of thinking about it is to remember that frequencies are not "real"; when we are talking about the frequency of something we are -implicitly- referring to the Fourier transform of a signal in the time domain. However, this transform is just one of many possible way to analyze the signal; it happens to be the most useful one in this case because there is a simple relationship between the energy of a photon and its frequency. However, whenever we deal with real systems we have to remember that this is an idealization. Hence, in the case of a the laser we are indeed modifying the waveform (in the time domain) it is sending out by "chopping" it.

    Note that this is also the basic principle behind many types of spectroscopy: by creating pulses (in the simplest case using switches) we can get spectra over a wide range of frequencies even though our main source is monochromatic.
     
  7. May 13, 2017 #6
    Well, yes and no, as far as I understood. I got that for a generic signal talking about frequencies is just a math abstraction. As regards light, and EM radiation in general, frequencies are somehwat real. Take a prism or a grating that spread the white light to its colors. It's a continuum (generally) but each line can be considered monochromatic if taken by itself. An almost infinitely on gas laser won't have only one frequency because of radiation may interact with other atoms and produce different frequency.
    The dual of a photon with a certain energy is a wave with a very precise frequency.

    I mean, I cannot understand physically what happens to make the spectrum broaden.
     
  8. May 13, 2017 #7

    PeterDonis

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    This is not actually true. Laser light is in a coherent state of the electromagnetic field. It is not an eigenstate of either the energy (frequency) operator or the number operator, which is what "a pure sine in the time domain" would be.

    A decent introduction to coherent states is here:

    https://en.wikipedia.org/wiki/Coherent_states

    The correct Fourier transform is actually a Gaussian, because a coherent state is a Gaussian.

    "Broadening" in this context just means increasing the spread of the Gaussian in the frequency domain, which corresponds to decreasing its spread in the time domain. In other words, the shorter you want to make the laser pulse, the more frequency broadening you get.

    Strictly speaking, the above is still an idealized description, which does not take into account fluctuations in the source--for example, thermal motions of the atoms in the lasing medium. These can also contribute to broadening (and in fact the term is often used specifically to describe increase in the frequency spread due to these causes).
     
  9. May 14, 2017 #8

    m k

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    Transform the mathematical abstraction to physical reality.

    Say a photon hits your meter.
    What next?

    A photon is one but the definition of a curve needs at least two points.
     
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