## Can one photon be anywhere in Universe?

 Quote by 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.
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

 Quote by 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.
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

Mentor
 Quote by pmb_phy If I knew of a postulate or theorem that states that then I'd agree. However I know of none.
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.
 Mentor 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.

 Quote by 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
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.

 Quote by 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.
What is 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.
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.
 So, I do think that I invoked the uncertainty principle correctly.
I disagree. You wrote
 This says that, AT BEST, the combined uncertainty in the two measurements must multiply to $$\hbar/2$$.
This indicates that you are refering to individual measurements. You also wrote
 However, by having an exact measurement in one variable and no clue in the other variable, we have an indeterminant form:..
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_wor...auss_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

 Quote by pmb_phy I don't understand.......
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.

Mentor
 Quote by cmos This is exactly what I was saying. The only "mishap" in my post was relating a zero uncertainty to an "exact measurement."
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.
 Quote by cmos 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.
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.

 Quote by cmos pmb_phy, it appears that you are taking some of what I say out of context.
That could very well be true. My only intention was to make it clear that the value of the uncertainty of an observable is determined by the state that a system is in and is unrelated to the error in a measurement. You seemed to be speaking about the error in measurement.
 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.
Heisenberg uncertainty relations only relate the uncertainties in physical observables, not uncertainties in measurements. In fact there is no meaning whatsoever to uncertainty in measurements. The Heisenberg uncertainty relations say nothing about measurements. When you speak of measurements it makes it seem that the uncertainty is not intrinsic to the system.
 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.
I disagee. The value of the uncertainty has nothing to due with the quality of your instruments.
 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.
There is nothing wrong with having a zero uncertainty in a variable and it certainly can't be viewed as a misuse. In fact many good QM textbooks address just this kind of situation.
 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.
The OP probably meant what everyone else means by that. I.e. when the uncertainty of one observable is zero then the probability density of the conjugate observable will be a constant. However I see no problem with what the obsever said. The uncertainty is a calculated value. It is intrinsic to the system. As such it can be zero or infinite because the mathematical expression can be.
 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.
Ah. I see. Thanks for clarifying that.
 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.
I see. But I don't see this as a problem. But don't get me wrong. I fully understand your concern and what you mean by this.

I guess the only objection that I have with what you posted is your use of the term "measurement" since the error in measurement of an observable has nothing to do with the uncertainty of the observable. You referred to your instruments and how good or bad they are. You can have the best intruments and that will not change the value of the uncertainty in an observable. The uncertainty is a property of the state itself, regardless of how good your instruments are. Likewise when the equality holds in the uncertainty relation it is not because you have the best of all instruments. It is because the state itself has that property. That is the importance of the Gaussian state. If a free particle is in a Gaussian state then the equality in the uncertainty principle holds.

Pete

 Quote by cmos 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.
I've been thinking about this more and I guess I have no objection to what you've said above since an uncertainty of exactly zero corresponds to an eigenstate of position which is not a square integrable function and therefore does not represent a physical state. The uncertainty therefore has zero as a lower limit, i.e. it can be as small as you'd like but cannot equal zero.

Pete

 Quote by pmb_phy I've been thinking about this more and I guess I have no objection to what you've said above since an uncertainty of exactly zero corresponds to an eigenstate of position which is not a square integrable function and therefore does not represent a physical state. The uncertainty therefore has zero as a lower limit, i.e. it can be as small as you'd like but cannot equal zero. Pete
I will say though, my use of explanation of measurements was careless in my previous posts; thank you for pointing that out. While the uncertainty is intrinsic to the system, it must be finite and non-zero; I think we both agree to that now.

 Tags momemtum, path, photon, single, uncertaintity