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I have a question that's intrigued me for awhile. How do Heisenberg Uncertainty and Quantum Decoherence coexist? I know in the early days of QM, a common reaction to the measurement problem was to say that wave function collapse resulted from the consciousness of the measuring observer a la Schrodinger's cat. However, I understand that most modern physicists believe that wave function collapse occurs independently of an observers' consciousness and that quantum decoherence into classical states results from particles effectively measuring each other and arriving at "pointer states" with superposition and quantum information reduced in proportion to the particles' degree of entanglement. Assuming this is true, wouldn't decoherent classical objects (e.g., dense matter) be made up of particles that are constantly measuring each others momenta and position. I also understand that momentum and position are Fourier transforms and therefore cannot be simultaneously measured. Does this mean that the constituent particles of a classical object are alternating between positional and momentum measurements? Or am I missing something?

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Note In this answer, technical asides are in italics.

There is a false dichotomy in your question, which I think comes from the way quantum mechanics is often presented at a popular science level. The uncertainty principle is often stated as "you can only measure position or momentum, but not both." In fact this is a simplification. The full statement is that (in some state $\psi$), that \begin{equation} \Delta x \Delta p \geq \frac{\hbar}{2}, \end{equation} where $\Delta x$ is the uncertainty in position in the state $\psi$ (more precisely, the standard deviation of measurements of the position of many particles prepared in the same state $\psi$), and $\Delta p$ is the uncertainty in the momentum.

What this means is that you can in fact make a measurement of $x$ and $p$ simultaneously, so long as you have some uncertainty in both quantities which satisfies the limit implied by the uncertainty principle. What you can't have, is a perfect measurement of both quantities simultaneously.

It's useful to imagine a particle in phase space, which is the space of the position and momentum coordinates of a particle. For a particle moving in one dimension, phase space is two dimensional (a plane); that will be sufficient for this question. In classical physics, a particle is represented as a point in phase space, corresponding to having a definite position and momentum. In quantum mechanics, because of the uncertainty principle, a particle is represented as a "blob" in phase space. (This is a bit of a simplification; more rigorously this blob is a Wigner quasiprobability distribution) The area of this blob must be at least $\hbar/2$ (but can be larger). The key point is that this blob can involve roughly equal spread in both position and momentum; it need not be extremely well localized in position and correspondingly unlocalized in momentum, or vice-versa.

In a "typical" state, the blob for a particle in phase space has some uncertainty in both position and momentum. It's not a terrible approximation to think of a particle as being a circle in phase space, with an area that saturates the bound from the uncertainty principle, and having "equal" spread in position and momentum (in some units). The area of the "blob" taken up by one particle for a classical object is very small (since $\hbar$ is a small number).

Now let's think about a large, classical object, with many particles. Each has a position and momentum, and so each can be represented as a blob in phase space. The separation between different particles in phase space, is very large compared to the area of individual blobs (again, because $\hbar$ is small). Therefore, when we reason about a large, classical collection of particles, the "classical statistics" in the positions and momenta of each particle (corresponding to the distance between blobs), is much more important than the "quantum statistics" in the position and momentum of a single particle (corresponding to the area of one blob).

For highly quantum collections of particles (well, really for bosons), the "blobs" of different particles tend to overlap in phase space; the "quantum statistics" becomes as important or even more important than the "classical statistics". For fermions, in highly quantum states of matter, the Pauli exclusion principle means the blobs don't overlap, but the key point is that the particles are close enough in phase space that we actually have to worry about the exclusion principle.


A comment asked me to comment on collapse and entanglement.

I don't have any comment on "objective collapse." In the Copenhagen interpretation, a collapse into a particular eigenstate would mean the blob would instantaneously change from some arbitrary state, to a state where the width in the direction of measurement was equal to the measurement uncertainty, and the width in the orthogonal direction is at at least large enough that the area of the blob is larger than $\hbar/2$. After this, the blob would tend to spread out so there would be large uncertainty in both directions. Heuristically, I would visualize "collapse" as the process of a roughly circular blob instantaneously becoming a highly elongated ellipse that is narrow in the direction that was measured, followed by normal quantum evolution where the state will tend to become more spread out and circular and with increasing area.

Phase space also enables a very nice picture for entanglement. For multi-particle systems with $N$ particles in 1 spatial dimension, phase space is really $2^N$ dimensional, since each particle has $2$ phase-space dimensions (a position and momentum). Now, we often will simplify this picture by talking about $N$ dots on a 2 dimensional space. But for entanglement, we cannot make this simplification.

In this larger $2^N$ dimensional space, one point actually represents a configuration of the whole system of $N$ particles (that is a lot of information for one point!). Let's just take $N=2$ particles for simplicity. An entangled state like \begin{equation} |\psi\rangle = \frac{1}{\sqrt{2}} \left(|0\rangle|1\rangle - |1\rangle |0\rangle\right) \end{equation} would correspond to something like having two "blobs", one blob near point $A$, where the momentum and position of particle $1$ are $p,x$ and particle $2$ are $P,X$, and another blob near point $B$, where the momentum and position of particle $1$ are $P,X$ and particle $2$ are $p,x$ (normally the states $|0\rangle$ and $|1\rangle$ refer to the spin of an electron along some axis or polarization of a photon, but here I am using them to refer to a 1-particle state localized near $(x,p)$ and another localized near $(X,P)$, respectively). If we know that the position of particle $1$ is near $p$, then we know we are near point $A$ and not point $B$, so we know particle $2$ has momentum $P$. This is the idea of entanglement: knowing something about particle $1$ tells you about particle $2$.

You may have noticed nothing in the above description requires $x$ and $X$ to be close by to each other; the EPR "paradox" corresponds to a situation where $x$ and $X$ are very widely separated in space, we measure something about particle $1$ to learn we are near either point $A$ or point $B$ in phase space, and therefore immediately learn about particle $2$.

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    $\begingroup$ It says im not supposed to waste our tine complimenting, but that was a pleasure to read and I understood every word, without any re-read, although I didnt previously know it. Thanks. Actually you know what, if you get bored you could add some text about how they objectively cause collapse (which I guess youre saying is really just the blob flattening into a “line” of equal “area”?) if thats even doable in a similar way, maybe objective collapse is too complex for that. Or anything about collapse in that case. Or about entanglement maybe (cant be blobs overlapping), if ya want. Thanks again $\endgroup$
    – Al Brown
    Jul 28 at 23:11
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    $\begingroup$ @AlBrown I certainly don't mind compliments :) I added a few comments. $\endgroup$
    – Andrew
    Jul 28 at 23:38

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