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Can one photon be anywhere in Universe?

  1. Jun 18, 2008 #1
    Lets use a perfectly sinusoidal oscillator to produce one photon of
    an exact momentum which we know.

    Now, Heisenberg Uncertaintity Principle says that if we know the momemtum
    exactly then the position of the photon is completely unknown. So, said photon
    has a probability of being ANYWHERE in the Universe.

    Say the photon which goes from A to B (emitter and screen)
    could have gone any path in its entangled, or unobsereved travel
    - even round the entire Universe to get to B. Quantum experiments
    seem to show that if we can say the path of a photon then its a particle,
    but if we never have the possibility of knowing its path (ie a trip round
    the Universe perhaps) then its a wave.
  2. jcsd
  3. Jun 18, 2008 #2
    Frankly I find that pretty odd, don't you? How did I measure the momentum if its on the other side of the universe?

  4. Jun 18, 2008 #3
    If we release it with a certain velocity that we know almost exactly, why would that velocity significantly change?
  5. Jun 18, 2008 #4


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    There's no way to produce a photon in a momentum eigenstate. (You can get arbitrarily close though, at least in a theory that describes space-time as Minkowski space). If you're thinking that I'm wrong because the wavelength of a photon that's emitted by an atom is completely specified by the difference between the energy levels, and that this wavelength corresponds to a particular momentum, then you're probably overlooking that there's some uncertainty in both the position and momentum of the atom that emitted the photon.
  6. Jun 18, 2008 #5
    Fredrik is correct. Furthermore, consider this, that one cannot simply say that one 'perfectly measures' the momentum, instead consider that in order to be completely correct one has to consider the 'system of interest' to not just be the particle but the particles of your detector as well and all of those particles together (and any possible entanglements between them) make up your system. Therefore, there is no way to do some measurement with your apparatus and get a read out and read it and have it correctly tell you the exact momentum of your particle. (i.e. say you measure the position of a charged particle by having a grid of wires and 'triangulating' its position by measure the current induced in the grid of wires (by noticing the magnitude of the current in each individual gridline one can calculate in which grid section and where in that grid section the particle passed through). However, one could never be EXACT because there is an uncertainty associated with the position of the gridlines and the momentum of the electrons creating the current. I mean tihnk about it, ultimately a momentum measurement might be made by comparing the position of a particle at two different times however, there is also an uncertainty relation between time and energy ([tex]\Delta t \Delta E \ge hbar/2[/tex]) thus in order to have infinite time resolution one must allow infinite energy uncertainty.

    So long story short the universe saves us from a paradox/contradition/violation by the fact that no such apparatus could be built.
    Last edited: Jun 18, 2008
  7. Jun 18, 2008 #6
    Wouldn't it be the same for an electron?
  8. Jun 18, 2008 #7
    Yes; anyway a photon created by decay of a metastable state has a smaller indetermination in its energy/momentum. In theory there could be much more stable...metastable states and so much more "dispersed" photons (but certainly not dispersed in the entire universe).
  9. Jun 19, 2008 #8
    Uncertainty position faster than light speed?

    hmmm... fascinating, lets say the uncertainty position of the photon managed to be very large - say 3 light years - then the photon's position could be, say, 3 light years from the position of the nearly exact momemtum measurement experiment. Well... did the photon take 3 light years to get there or did it 'appear' there instantly - ie faster then light speed?
    If so, it 'went' 3 light years in no time or what?
  10. Jun 19, 2008 #9


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    Another way to look at this is to consider the Fourier transform that is the basis of the uncertainty principle. By the properties of the Fourier transform we know that a perfect sinusoid (i.e. one single exact frequency) would take an infinite amount of time to generate. Therefore you cannot know when your photon was released so it could be anywhere.
  11. Jun 19, 2008 #10


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    The statement that the photon "is" at a position 3 light-years away implies that its wave function is sharply peaked around that location, and that implies that the uncertainty in its position is small. So you're describing an impossible (self-contradicting) scenario.

    If the uncertainty in its position is large, then the particle is smeared out over a large region. Note that a position measurement is not about "finding out" where the particle is. A position measurement is an interaction that changes the state of the particle so that the region over which it is smeared out becomes smaller. There's no way to predict where that smaller region is going to be, but (as always in QM) we can calculate the probabilities of the possibilities. What we "find out" when we measure a particle's position isn't where it was before the measurement. Instead we find the (smaller) region over which the particle is smeared out after the measurement.
  12. Jun 19, 2008 #11
    To give further motivation here for this assertion, consider that the uncertainty principle (as Griffiths claims) is something that is commonly missused. This is one of those instances in that the OP missuses the uncertainty principle. Heisenberg's uncertainty principle says that, AT BEST:

    [tex]\Delta x \Delta p=\hbar / 2[/tex]

    This says that, AT BEST, the combined uncertainty in the two measurements must multiply to [tex]\hbar/2[/tex]. However, by having an exact measurement in one variable and no clue in the other variable, we have an indeterminant form:

    [tex]\Delta x \Delta p=0 \times \infty[/tex]

    Therefore the uncertainty principle does not even allow for the act of a perfect measurement to occur.
    Last edited: Jun 19, 2008
  13. Jun 19, 2008 #12


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    Fair enough but it simply modifies the question: how accurate would one have to measure the momentum of a particle so that its position is indeterminate within the volume of the known universe?
  14. Jun 19, 2008 #13
    There is no lower bound to the precision in which the momentum of a particle can be measured. In principle the momentum can be measured exactly. If the momentum of a particle, which is initially in an arbitrary quantum state, is measured then one can measure its value exactly. The state will then be in an eigenstate corresponding to that value of the momentum. If this process is repeated then the measured value of the momentum will be different in each case. There will therefore, in general, be an uncertainty in the momentum. This is the ideal situation. In practice this is never the case. Note that the energy of a particle can be measured exactly when the particle is initially in an energy eigenstate. However in such cases the uncertainty in position will be infinite.
    If I knew of a postulate or theorem that states that then I'd agree. However I know of none. And just because I don't know how to measure the exact value of momentum it doesn't mean that it doesn't have that value or that it can't be determined exactly. There is a major difference between measuring the exact value and the particle possessing an exact value. Let me give you an example. Suppose an atom is initially in an energy eigenstate. Let the atom fall into a lower eigenstate. These energy levels and quantize. Therefore an atom in an eigenstate possesses a definite value of energy. When the atom falls into a lower level eigenstate of energy the atom will emit a photon. Since the energy of the two levels are known then the photon will have a precise value. Suppose you have instruments have a finite precision to them and as such the exact value of the photons energy cannot be determine. However it could certainly be determined what transition occured from the energy (even though it is not precisely known) and we can deduce what the energy of the photon must be. Then both the energy and momentum of the photon is precisely determined.

    I understand your concern though. I also believe that the same idea applies to all physical observables which have a continuos spectrum.
    The uncertainty in momentum is independant of how you measure it. The uncertainty is completely determined by the state that particle is in.

  15. Jun 19, 2008 #14
    With the momentum well defined, the time of emmition is undefined, thus the position, as well.

    In fact, the quanta never really stops being emmited, nor begins to be emmited, but is emmited for all time.

    Edit: I see Dalespam has already said as much.
    Last edited: Jun 19, 2008
  16. Jun 19, 2008 #15
    I don't agree. If the electron (plus rest of the universe) were in an exact eigenstate of the full Hamiltonian then it would stay there forever. However, the very fact that the electron makes a transition to a lower energy level tells you that the electron was not in an exact eigenstate.

    The full Hamiltonian involves also the coupling of the electron to the electromagnetic field. If we ignore the coupling and treat this in perturbation theory, then the unperturbed Hamiltonian has eigenstates like:


    Where |atom> is some eigenstate of the Hamiltonian of the atom which only takes into account the interactions of the electrons with each other and the nucleus. |photon> are the momentum states of the photon.

    Now, if the initial state is:

    |atom in excited state>|no photons pesent>

    Then, the coupling between electrons and the elctromagnetic field couples such a state to the state

    |atom in ground state>|one photon present>

    The matrix element of the perturbation between the two states determines the decay rate. This is related to the spread of the photon's energy.

    You can also turn this around: The fact that the matrix element of the full Hamiltonian between the two states is nonzero leads to the nonzero transition probablity. That being the case, you can conclude that the two states are not eigenstates of the full Hamlitonian and therefore do not have sharply defined energies.
    Last edited: Jun 19, 2008
  17. Jun 19, 2008 #16
    That is not the case. One must use caution when attempting to use the time-energy uncertainty relation. There is no such thing as a time operator and as such there term "uncertainty" in the phrase uncertainty in time has a very different meaning than in does in "uncertainty in momentum." In fact the value [itex]\Delta t[/itex] is called the time of evolution.
    That is impossible since the very meaning of the term quanta is that it comes in finite discrete quantites.
    And I disagree with him too.

    As I mentioned above the quantity [itex]\Delta t[/itex] is called the time of evolution. Let me explain what this term means. Assume that the initial state of a system, |[itex]\Psi(t_0)[/itex]>, is the linear superposition of two eigenstates of H, |[itex]\phi_1[/itex]> and |[itex]\phi_2[/itex]> with different eigenvalues E1 and E2;

    |[itex]\Psi(t_0)[/itex]> = c1|[itex]\phi_1[/itex]> + c2|[itex]\phi_2[/itex]>


    |[itex]\Psi(t)[/itex]> = c1e-iE1(t - t0)|[itex]\phi_1[/itex]> + c2e-iE2(t - t0)|[itex]\phi_2[/itex]>

    If we measure the energy we get either E1 or E2, precisely. However this does not mean that the uncertainty in energy is zero. Far from it. In fact the uncertainty in energy is given by

    [itex]\Delta E[/itex] ~ |E2 - E1|

    The above equation for the state shows that the system oscillates between two values with the Bohr frrequency [itex]\nu[/itex] = [itex]\Delta E[/itex]/h. The characteristic time of evoltion of the system is therefore

    [itex]\Delta t[/itex] ~ h/[itex]\Delta E[/itex]

    This is the same equation as

    [itex]\Delta E[/itex] [itex]\Delta t[/itex] ~ h

    which is where the time-energy uncertainty relation comes from.

    For some reason people have gotten the idea that a non-zero uncertainty in a value means that the value can't be measured precisely. That is wrong.

  18. Jun 19, 2008 #17
    Forget that post please. It was a mistake. I just wrote a post stating the meaning of the time-energy uncertainty relation. Please see that post. Thanks.

  19. Jun 19, 2008 #18
    I disagree. Let the initial state of the system (a hydrogen atom for example) be an eigenstate the Hamiltonian. The system then evolves according to

    |[itex]\Psi[/itex]> = e-Ent/hbar|[itex]\phi_n[/itex]>

    It is know that these states are unstable. If the initial state is |[itex]\phi_n[/itex]> corresponding to the energy En greater than the ground state it generall falls back into the ground state by emitting one or more photons. Therefore the state |[itex]\phi_n[/itex]>, while being an eigenstate, is not really a stationary state. As you indicated the atom is really in constant interaction with the electromagnetic field. The approximation consisting of completely ignoring the electromagnetic field is very good, except, of course, if we are interested in the instability of the states.

    The instability of a state can be taken into account phenomenoligically. Let [itex]\tau[/itex] be the lifetime of the state. If one prepares the system at t = 0 in the state |[itex]\phi_n[/itex]> one observes that the probability P(t) of it still being excited at a later time t is

    P(t) = e-t/hbar[itex]\tau[/itex]

    The mean value of the time that the system remains in the state |[itex]\phi_n[/itex]> is [itex]\tau[/itex]. This can be accompished if one defines an imaginary value of the energy as

    E'n = En - i*bar*[itex]\gamma_n[/itex]/2

    Therefore, to take into account phenomenologically the instability of the state |[itex]\phi_n[/itex]> whose lifetime is [itex]\tau_n[/itex] it suffices to add an imaginary part to its energy by setting

    [itex]\gamma_n[/tex] = 1/[itex]\tau_ni[/itex]

    The uncertainty [itex]\Delta E[/itex] has the value which is the difference between the initial state and the ground state. So you see it is not a problem to measure the energy of the photon since the value of each energy level is a known number.

    Last edited: Jun 19, 2008
  20. Jun 19, 2008 #19
    Actually you are misusing the uncertainty principle too. Seems like most people do though so don't feel bad. :smile:

    The uncertainty of a quantity can be non-zero even when the measurement of that quantity is non-zero. The uncertainty of a physical observable has a statistical meaning. It is not related to the precision of a single measurement. Uncertainty refers to the range of values that can be measured in identical experiments even when each measurement is exact.

    Consider the uncertainty in the number, N, that is rolled on a die. You can readily see that the number rolled on a die can be "measured" exactly. It can only take on certain well defined values, i.e. n = 1, 2, ... , 6. The uncertainty in the number of die turns out to be [itex]\Delta N[/itex] = 3.14!

    The uncertainty in a physical observable is inherant in the state itself and is independant on the precision in which one is able to measure that observable in a single measurement. That is to say that one can determine the uncertainty in a physical observable by merely knowing what the quantum state is. Thus the uncertainty of an observable depends only on the state the the system is in when the observable is measured and has nothing to do with how precise you can measure the observable.

  21. Jun 19, 2008 #20


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    The relevant postulates are:
    1. States are represented by vectors (actually rays) in a Hilbert space.
    2. The probability interpretation of the wave function.

    An exact measurement of position would make the wave function zero everywhere except at one point in space. That means the integral of [itex]|\psi|^2[/itex] isn't going to be 1 (or even well-defined).

    Also, if you take some kind of limit to squeeze the wave function into a delta-function shape, it's Fourier transform goes to 0 everywhere. So in that limit the momentum-space wave function becomes the 0 vector, which doesn't represent the state of a physical system.
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