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Can quantum operators be pictured in terms of wave packets?

  1. Apr 22, 2013 #1
    I've found the wave-packet picture quite useful as I work my way through the very basics of quantum mechanics. But I'm having trouble finding a wave-mechanical picture of operators. For example, at least in terms of a free particle, using the wave mechanics treatment (as opposed to the matrix mechanics treatment) I think I can very roughly picture:
    • pure states or eigenstates/vectors, following the form e (at least for momentum states) as helical sine waves or phase waves in the complex plane
    • eigenvalues as the amplitudes, moduli or radii of those phase waves in the complex plane
    • wave functions as wave packets or group waves comprised of the interfering component phase waves
    • probabilities as vectors along the real axis of the cross-section of the wave packet in the complex plane

    But I'm having trouble picturing an operator in this context. Is it possible to picture it at all? My best guess is that it would be a subset of the overall wave packet, and so in some sense a wave packet in its own right, but not sure. Any help would be greatly appreciated.
  2. jcsd
  3. Apr 22, 2013 #2

    Simon Bridge

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    No. Operators "operate on" wavepackets to give a number.
    You can always expand a wave-packet into a superposition of eigenfunctions of an operator.
  4. Apr 24, 2013 #3
    we could say that an operator transforms a wavepacket into another.
  5. Apr 25, 2013 #4

    Simon Bridge

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    If you really pushed it, sure... i.e. if ##\psi## is an eigenfunction of operator A, with eigenvalue a, then ##A\psi=a\psi## appears to have scales the wavefunction by a.

    A operator is a mathematical representation of an operation.
    An example of an operation, in this sense, could be "addition" or "multiplication".

    OP does not seem to feel a need to describe The "+" operator in terms of wavepackets and yet:

    What seems to be key here is that operators can only be described by their effect on the things they operate on.
  6. Apr 26, 2013 #5
    Thanks, that certainly gives me some direction.

    Thanks Simon. To explore the point a bit further, suppose you took one wave packet (A), for example, and had it interfere with another wavepacket (B), resulting in a new packet or state. In this context, would wave packet A be functioning as an operator? If not, how about a measuring device such as a slit in the double slit experiment. Could that be characterized in some sense as an operator?
  7. Apr 26, 2013 #6

    Simon Bridge

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    Nice example - in this case, if we were to say that packet A was an operator then cannot the same be said of packet B? So which the operator pray and which the operand?
    In this case, the operation in question is "addition" (superposition?). The operator is binary - taking two wavepackets and outputting one.

    There is a lot of discussion about that: be prepared for someone to contradict or clarify something about what I am about to write ;)

    Narrow slits, in effect, measure the position of the particle - as such, their operation can be described by a position operator - as in Marcella (2009). The result of the measurement is to prepare the wavefunction in a superposition of delta functions (for very narrow slits) which time-evolve to interfere. Subsequent measurements accumulate into the famous diffraction pattern.

    Operators do not have to look like the things they operate on - they can be totally abstract.
    i.e. If I have a $1 coin, and so do you, and you give yours to me, then I now have $2.
    The operation of interest is "addition". I can take the two $1 coins and exchange tham for a $2 coin - that would also involve an operation. You will notice that there is nothing about the operations themselves that resemble a coin although they intimately involve coins.

    Read Marcella (2009) with an eye on Rothman (2010).
    I don't think it affects the example re: slits as operator - but it does limit how far to run with Marcella's argument.
    Last edited: Apr 26, 2013
  8. Apr 29, 2013 #7
    Thanks for helping me out here Simon, and I see your point. At the same time, it does seem, at least in normal circumstances, that anything that operates on X is by definition an operator with respect to X. A concrete example might be a hammer driving a nail into the wall. Of course, one could rightfully argue from basic action-reaction principles that the nail is operating on the hammer, and that would be valid, but doesn't subtract from the fact that the hammer, in this context, is an operator performing the operation of hammering on the nail.

    I'd agree, with one small qualification. Although the hammer, for example, doesn't have to look anything like the nail, it does have to share a common property with it, in this case hardness corresponding to the surface of the nail head, necessary to make contact and perform the function at hand. I think one could reasonably argue that the same holds for one wave packet performing the operation of interfering with another.

    I'd agree that an operation that an operator performs is abstract, but the operator itself seems to be something relatively concrete, an object that can be pictured. The hammer is certainly operating on the nail, just as the measuring device's wave packet is operating on the wave packet being measured, and in that sense would seem to clearly be an operator. Would you agree?

    Regarding the references to Marcella and Rothman, thank you, I'm looking into them now.
  9. Apr 29, 2013 #8

    Simon Bridge

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    When you want to think about operators, it helps to ask, "what is the operation here?"
    In your example, the operation is "hammering a nail".
    The operation involves three objects - the nail, the hammer, and the wall.
    So the operator needed to describe it will have three operands.

    You'll also see that the operator has no physical existence independent of the operands.

    You could use the same operator to hammer a candle into a cake if you liked - in which case you'd put your hand (say) as the second operand. Of course, pushing would be better for this, but hammering also works. What I want to draw your attention to is that it is the same operator even though the operands are different.

    The way you described the process was in terms of actions - you'd think of the hammer as acting on the nail to some result. In normal English it is reasonable to describe this action as the hammer operating on the nail - but in math, and in physics, the word "operator" is jargon, and has a more narrow range of uses.

    "Oh but," I hear your mind start to say, "perhaps the person holding the hammer is the concrete form of the operator - that person takes all three..." well yes, like the "operator" of heavy machinery. This is the use of the word that does not apply here. If you included the hammerer in the process, then the operation includes the hammerer, and so the operator must be expanded too. The operator should not depend on the operands - so the operator for a person with a hammer knocking a nail into a wall should be the same for an ape holding a lump of hard stone knocking a wooden stake into a sand bank.

    That's why the operator has to be abstract.

    A slit (prev example above) can be described as having an effect involving a position operator.
    I gave you some reading material on this : Marcella (2009).

    However, the slit is not an operator.
    The operation is just isolating the position - it does not matter if it is a hole in a sheet of cardboard or a crystal lattice or something else. It's the same operation. The operation actually involves an interaction between the physical objects involved - a point that is made by Rothman (2010), discussing Marcella.

    I know it is tempting to try to fit every mathematical process into some concrete physical one.
    As you progress you will run into more and more difficulties with this. Some things exist only as concepts in our imaginations.

    I don't understand why you are restricting your descriptions to wave-packets.
    But I think I've taken this discussion as far as I reasonably can.
    Good luck ;)
    Last edited: Apr 29, 2013
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