- #31
Maaneli
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<< Can you tell me what's the cause of the HUP if it has nothing to do with measurement? >>
For particles, it does have to do with measurement. Zapper's example of using a slit to locate a "photon" is confusing because he hasn't defined what he means by a "photon". Presumably he's thinking of a "photon" as a point particle of some kind. But then he should be more specific about the dynamics of that photon, i.e., whether it moves according to a path integral or a de Broglie-Bohm guidance equation, or a Wigner function.
I think there is lot's of confusion on this board because no one has really stated and distinguished the 3 different formulations of the HUP, which relate to
1) the structure of the wavefunction/packet.
2) a property of an ensemble of particles.
3) a dynamical property of measuring a single particle.
The Heisenberg microscope is an example of 3), and it isn't a very valid example because it is actually a semiclassical thought experiment, not a quantum mechanical one. But the examples of 1) and 2) have not yet been specified here.
An example of 1) is the general property of a Fourier analysis on waves of any kind, classical or quantum mechanical. In the case of a general wave packet, it is manifested as a reciprocity relation between wavenumber, dk, and spatial pulse width, dx, given by d(k)*d(x) = 1, or between frequency band width, domega, and time width, dt, given by domega*dt = 1. Quantum mechanics differs only in that you multiply both sides of these reciprocity relations by hbar (or just apply the de Broglie relation dp = hbar*dk), implying the magnitude of the relations is very small (or on quantum mechanical lengthscales), and this is basically the HUP for quantum wave packets. It says nothing about point particles, which is ultimately all we see in physical experiments.
An example of 2) is the operator algebra that Heisenberg historically formulated and applied to experiments. Starting from the commutation relation
[\hat{x},\hat{p}] = i*hbar, where \hat{x} = x and \hat{p} = -i*hbar*d/dx,
one can then use these position and momentum operators to compute the expectation values of observables by
< x > = int[\bar{psi}*\hat{x}*psi]dx
< p > = int[\bar{psi}*\hat{p}*psi]dx
and then
dx = Sqrt[< x^2 > - < x >^2]
dp = Sqrt[< p^2 > - < p >^2]
Physically, what the example of 2) means is that suppose you have an initial wavefunction psi_0(x) so that the particles have a distribution of rho(x) = R^{2}_{0}(x). Now consider the ith particle in the ensemble and at time t, measure its x-coordinate, x_i(t), N times. Then evaluate from these N position measurements the average and mean positions described above. Then, starting from the same initial wavefunction, perform now a series of momentum measurements in the x-direction. Let the ith particle in the ensemble at time t be found to have momentum p_x,i(t). Then evaluate from these N momentum measurements the average and mean momenta described above. The produce of dx and dp turns out to be
dx*dp \geq hbar/2.
Notice that 1 position and momentum measurement is not sufficient for this definition to work. It requires an *ensemble* of measurements. This tells us that the HUP is not a property of any individual particle, but rather of an *ensemble* of particle measurements.
There are of course uncertainty relations for other conjugate quantities in QM, such as for angular momentum, and the same general rules above apply. One comment however about the energy-time uncertainty relation. Wikipedia on the uncertainty relation gets it exactly right, so I will actually just quote them:
"Since energy has the same relation to time as momentum does to space in special relativity, it was clear to many early founders that the following relation holds:
dE*dt \geq hbar/2.
but it was not clear what dt is, because the time at which the particle has a given state is not an operator belonging to the particle, it is a parameter describing the evolution of the system. Lev Landau once joked "To violate the time-energy uncertainty relation all I have to do is measure the energy very precisely and then look at my watch!"
Einstein and Bohr however understood the meaning of the principle. A state which only exists for a short time cannot have a definite energy. In order to have a definite energy, the frequency of the state needs to be accurately defined, and this requires the state to hang around for many cycles, the reciprocal of the required accuracy.
For example, in spectroscopy, excited states have a finite lifetime. By the time-energy uncertainty principle, they do not have a definite energy, and each time they decay the energy they release is slightly different. The average energy of the outgoing photon has a peak at the theoretical energy of the state, but the distribution has a finite width called the natural linewidth. Fast-decaying states have a broad linewidth, while slow decaying states have a narrow linewidth.
One false formulation of the energy-time uncertainty principle says that measuring the energy of a quantum system to an accuracy ΔE requires a time interval Δt > h / ΔE. This formulation is similar to the one alluded to in Landau's joke, and was explicitly invalidated by Y. Aharonov and D. Bohm in 1961. The time Δt in the uncertainty relation is the time during which the system exists unperturbed, not the time during which the experimental equipment is turned on."
The question that one should now ask is how can one relate the uncertainty relation for wave packets to the empirically deduced uncertainty relation for particles in experiments. The pilot wave theory of de Broglie and Bohm gives an exact answer to this question which I can go into if anyone is interested.
For particles, it does have to do with measurement. Zapper's example of using a slit to locate a "photon" is confusing because he hasn't defined what he means by a "photon". Presumably he's thinking of a "photon" as a point particle of some kind. But then he should be more specific about the dynamics of that photon, i.e., whether it moves according to a path integral or a de Broglie-Bohm guidance equation, or a Wigner function.
I think there is lot's of confusion on this board because no one has really stated and distinguished the 3 different formulations of the HUP, which relate to
1) the structure of the wavefunction/packet.
2) a property of an ensemble of particles.
3) a dynamical property of measuring a single particle.
The Heisenberg microscope is an example of 3), and it isn't a very valid example because it is actually a semiclassical thought experiment, not a quantum mechanical one. But the examples of 1) and 2) have not yet been specified here.
An example of 1) is the general property of a Fourier analysis on waves of any kind, classical or quantum mechanical. In the case of a general wave packet, it is manifested as a reciprocity relation between wavenumber, dk, and spatial pulse width, dx, given by d(k)*d(x) = 1, or between frequency band width, domega, and time width, dt, given by domega*dt = 1. Quantum mechanics differs only in that you multiply both sides of these reciprocity relations by hbar (or just apply the de Broglie relation dp = hbar*dk), implying the magnitude of the relations is very small (or on quantum mechanical lengthscales), and this is basically the HUP for quantum wave packets. It says nothing about point particles, which is ultimately all we see in physical experiments.
An example of 2) is the operator algebra that Heisenberg historically formulated and applied to experiments. Starting from the commutation relation
[\hat{x},\hat{p}] = i*hbar, where \hat{x} = x and \hat{p} = -i*hbar*d/dx,
one can then use these position and momentum operators to compute the expectation values of observables by
< x > = int[\bar{psi}*\hat{x}*psi]dx
< p > = int[\bar{psi}*\hat{p}*psi]dx
and then
dx = Sqrt[< x^2 > - < x >^2]
dp = Sqrt[< p^2 > - < p >^2]
Physically, what the example of 2) means is that suppose you have an initial wavefunction psi_0(x) so that the particles have a distribution of rho(x) = R^{2}_{0}(x). Now consider the ith particle in the ensemble and at time t, measure its x-coordinate, x_i(t), N times. Then evaluate from these N position measurements the average and mean positions described above. Then, starting from the same initial wavefunction, perform now a series of momentum measurements in the x-direction. Let the ith particle in the ensemble at time t be found to have momentum p_x,i(t). Then evaluate from these N momentum measurements the average and mean momenta described above. The produce of dx and dp turns out to be
dx*dp \geq hbar/2.
Notice that 1 position and momentum measurement is not sufficient for this definition to work. It requires an *ensemble* of measurements. This tells us that the HUP is not a property of any individual particle, but rather of an *ensemble* of particle measurements.
There are of course uncertainty relations for other conjugate quantities in QM, such as for angular momentum, and the same general rules above apply. One comment however about the energy-time uncertainty relation. Wikipedia on the uncertainty relation gets it exactly right, so I will actually just quote them:
"Since energy has the same relation to time as momentum does to space in special relativity, it was clear to many early founders that the following relation holds:
dE*dt \geq hbar/2.
but it was not clear what dt is, because the time at which the particle has a given state is not an operator belonging to the particle, it is a parameter describing the evolution of the system. Lev Landau once joked "To violate the time-energy uncertainty relation all I have to do is measure the energy very precisely and then look at my watch!"
Einstein and Bohr however understood the meaning of the principle. A state which only exists for a short time cannot have a definite energy. In order to have a definite energy, the frequency of the state needs to be accurately defined, and this requires the state to hang around for many cycles, the reciprocal of the required accuracy.
For example, in spectroscopy, excited states have a finite lifetime. By the time-energy uncertainty principle, they do not have a definite energy, and each time they decay the energy they release is slightly different. The average energy of the outgoing photon has a peak at the theoretical energy of the state, but the distribution has a finite width called the natural linewidth. Fast-decaying states have a broad linewidth, while slow decaying states have a narrow linewidth.
One false formulation of the energy-time uncertainty principle says that measuring the energy of a quantum system to an accuracy ΔE requires a time interval Δt > h / ΔE. This formulation is similar to the one alluded to in Landau's joke, and was explicitly invalidated by Y. Aharonov and D. Bohm in 1961. The time Δt in the uncertainty relation is the time during which the system exists unperturbed, not the time during which the experimental equipment is turned on."
The question that one should now ask is how can one relate the uncertainty relation for wave packets to the empirically deduced uncertainty relation for particles in experiments. The pilot wave theory of de Broglie and Bohm gives an exact answer to this question which I can go into if anyone is interested.