# Can one photon be anywhere in Universe?

1. Jun 18, 2008

### wawenspop

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. Jun 18, 2008

### pmb_phy

Frankly I find that pretty odd, don't you? How did I measure the momentum if its on the other side of the universe?

Pete

3. Jun 18, 2008

### gamesguru

If we release it with a certain velocity that we know almost exactly, why would that velocity significantly change?

4. Jun 18, 2008

### Fredrik

Staff Emeritus
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.

5. Jun 18, 2008

### maverick_starstrider

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 ($$\Delta t \Delta E \ge hbar/2$$) 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
6. Jun 18, 2008

### lightarrow

Wouldn't it be the same for an electron?

7. Jun 18, 2008

### lightarrow

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).

8. Jun 19, 2008

### LaserMind

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?

9. Jun 19, 2008

### Staff: Mentor

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.

10. Jun 19, 2008

### Fredrik

Staff Emeritus
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.

11. Jun 19, 2008

### cmos

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:

$$\Delta x \Delta p=\hbar / 2$$

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

$$\Delta x \Delta p=0 \times \infty$$

Therefore the uncertainty principle does not even allow for the act of a perfect measurement to occur.

Last edited: Jun 19, 2008
12. Jun 19, 2008

### DaveC426913

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?

13. Jun 19, 2008

### pmb_phy

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.
Exactly.
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.

Pete

14. Jun 19, 2008

### Phrak

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
15. Jun 19, 2008

### Count Iblis

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:

|atom>|photon>

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
16. Jun 19, 2008

### pmb_phy

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

|$\Psi(t_0)$> = c1|$\phi_1$> + c2|$\phi_2$>

Then

|$\Psi(t)$> = c1e-iE1(t - t0)|$\phi_1$> + c2e-iE2(t - t0)|$\phi_2$>

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

$\Delta E$ ~ |E2 - E1|

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

$\Delta t$ ~ h/$\Delta E$

This is the same equation as

$\Delta E$ $\Delta t$ ~ 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.

Pete

17. Jun 19, 2008

### pmb_phy

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.

Pete

18. Jun 19, 2008

### pmb_phy

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

|$\Psi$> = e-Ent/hbar|$\phi_n$>

It is know that these states are unstable. If the initial state is |$\phi_n$> 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 |$\phi_n$>, 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 $\tau$ be the lifetime of the state. If one prepares the system at t = 0 in the state |$\phi_n$> one observes that the probability P(t) of it still being excited at a later time t is

P(t) = e-t/hbar$\tau$

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

E'n = En - i*bar*$\gamma_n$/2

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

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

The uncertainty $\Delta E$ 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.

Pete

Last edited: Jun 19, 2008
19. Jun 19, 2008

### pmb_phy

Actually you are misusing the uncertainty principle too. Seems like most people do though so don't feel bad.

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 $\Delta N$ = 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.

Pete

20. Jun 19, 2008

### Fredrik

Staff Emeritus
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 $|\psi|^2$ 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.

21. Jun 19, 2008

### Fredrik

Staff Emeritus
About the atom that emits a photon...

Everyone agrees that the laws of QM implies that an isolated system in an energy eigenstate will remain in that state. However, I think an excited atom would eventually emit a photon even if it was the only matter in the universe. The reason is that the "energy eigenstate" is a state vector in a Hilbert space of an approximate model of the atom.

The real atom is composed of interacting subsystems. It's likely that the interactions between subsystems will be able to change the state of the "complicated" system from a state that can be approximated as an energy eigenstate in the simplified model to a state that is better approximated as a superposition of several energy eigenstates in the simplified model.

So the energy of the emitted photon isn't going to be exactly the difference between two energy levels, because even the existence of those energy levels is an approximation. (The Hamiltonian of the "complicated" system has eigenstates too, but it's possible that the differences between energy eigenvalues are much smaller. Its spectrum can even be continuous).

But suppose there exists a system that really does have exact energy eigenstates, and that sends out photons when it makes a transition from a higher energy level to a lower. Would this photon be in a momentum eigenstate? No, it wouldn't. I believe that what I said in #4 is sufficient to explain that.

22. Jun 20, 2008

### cmos

This is exactly what I was saying. The only "mishap" in my post was relating a zero uncertainty to an "exact measurement." I will have to think about exactly what that term implicates.

Regardless though, the point that I was trying to make is that assuming an uncertainty of zero leads to an indeterminant form. Therefor it is not physically possible to make that type of measurement; in other words, there must always be some finite uncertainty. So, I do think that I invoked the uncertainty principle correctly.

23. Jun 20, 2008

### pmb_phy

What is an indeterminant form.
I don't understand why you are saying that. Can you please elaborate and explain why "there must be" a finite uncertainty? The uncertainty is unrelated to a measurement. It is meaningless to speak of the uncertainty of a single measurement. The uncertainty is determined by the values measured in a large number of measurements. As I explained above, just because you make exact measurements in no way implies that there is zero uncertainty. And you can't make the uncertainty smaller by using instruments which allow you to make more precise measurements.
I disagree. You wrote
This indicates that you are refering to individual measurements. You also wrote
This indicates that you believe that if, in a single measurement, you measured the exact value of the observable (i.e. the error in measurement was zero) that you would be unable to measure the value of the other and have a finite value for the error in that measurement. These two ideas are not what the uncertainty principle is all about.

I think you may be confusing the error in the measurement of a particular observable with the uncertainty in that same observable. Such a notion is wrong. Lets say that we have imperfect instruments and as such there will always be a finite error in the measured values of the observables. Then these values are not related to the uncertainties. For this reason I recommend that we use the notation $\delta A$ to refer to the error in the measurement of the observable A. This will keep us straight regarding whether we are speaking of errors in measurements as to uncertainties in an observable. As such there is no relation akin to the uncertainty principles for the quanties $\delta A$. It is the uncertainties which obey the uncertainty principle and not the error in the variables.

Let me give an example for clarification. Suppose we have a free particle (i.e. one for which the potential energy function is zero) constrained to move in a straight line. Call this the x-axis. Let the state be such that the amplitude of the wave function $\Psi$(x) be a Gaussian function. The function that I'm referring to is given in Eq. (1) in this web page

http://www.geocities.com/physics_world/qm/gauss_state.htm

(Note: I'm not supposed to post links to my web site but I'm hoping the moderator won't mind this time since I'm only using it to refer to a function)

If you were to plot the function $\psi$(p) it to would be a Gaussian function. The the width of $\Psi$(x) is x0. The width of $\psi$(p) is p0. These values are inversely proportional to each other. A fact which follows from Fourier Analysis. This does not mean that you can't measure the position of the particle with an error of less than x0. It also doesn't mean that you can't measure the momentum of the particle with an error of less that p0. The values x0 and p0 are completely independant of the instruments that you use nor does it depend on the error in your measurements. There is absolutely nothing you can do to decrease the uncertainty of a particular quantum state. Better instruments (i.e. which have smaller errors) will get you nowhere.

Pete

24. Jun 20, 2008

### cmos

pmb_phy, it appears that you are taking some of what I say out of context. My first post has two aims, 1) further elaborate on the correct assertions of maverick_starstrider and 2) to show that the question raised by the OP has no physical meaning (which, BTW, you seem to agree with; just that you do not yet clearly understand my mathematical arguments).

Heisenberg:
$$\Delta x \Delta p \geq \frac{\hbar}{2 \pi}$$

So even if you have the BEST instruments available, you will still have uncertainty in your measurements. Combined, this will equal the RHS of the equation. However, if you have crappy instruments, then you will have a combined uncertainty greater than $$\frac{\hbar}{2 \pi}$$; this occurs everyday in every lab.

The OP asserted the question, what if $$\Delta p = 0$$? The OP goes on to divide the RHS of the equation by 0 to get infinity in the uncertainty of the position measurement. This is where I claim that the OP has misused the uncertainty principle.

The uncertainty principle states that when you multiply the uncertainties you will get a finite non-zero quantity. This implies that you cannot have an uncertainty of zero nor can you have one of infinity.

Furthermore, if we entertain the idea of what I just claimed you cannot have, then you would have:

$$\Delta x \Delta p = 0 \times \infty$$

This is an example of an indeterminant form. Furthermore, because it is an indeterminant form we throw it out as physically implausible. And again, a fortiori, we must satisfy the very first equation I wrote down.

25. Jun 20, 2008

### Fredrik

Staff Emeritus
Your post was fine. The only thing that it makes sense to disagree with is your choice of words when you said "uncertainty in the two measurements". It suggests that the uncertainty is a property of the measurement, which is what set pmb_phy off, because he likes to point out that the uncertainty is a property of the state the system is in.

However, he's neglecting the fact that if you make a position measurement that shows that the particle is definitely between x-a and x+a, where a is some small number, then you have put the particle in a state that's represented by a wave function that's zero outside of that interval. Such a wave function has a $\Delta x$ that's small if a is small and goes to zero when a goes to zero.

What this shows is that even though $\Delta x$ isn't a function of a (which is a property of the measurement), the two aren't unrelated. So it definitely doesn't make sense to attack everyone who suggests that the uncertainty has something to do with measurements. The uncertainty is a function of the state the system is in, but if we have recently performed a measurement, then the measurement is a part of the reason why the system is in the state it's in.
I think the argument is valid, at least if we interpret it as representing some kind of limit procedure. The problem is that an exact position measurement would by definition leave the system in a "state" that's zero everywhere except at the point where the particle is now located. The uncertainty of such a "state" isn't zero, it's undefined. (Actually it isn't even a state). But if we instead consider a measurement that confines the particle to an interval of a certain length and have that length go to zero, $\Delta x$ will go to zero and $\Delta p$ will go to infinity.

Last edited: Jun 20, 2008