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I'm an amateur in quantum mechanics. I am confused after reading the following in the wikipedia article about quantum superposition:

If the operators corresponding to two observables do not commute, they have no simultaneous eigenstates and they obey the uncertainty principle. A state where one observable has a definite value corresponds to a superposition of many states for the other observable.

This seems to state that the properties obeying the uncertainty principle (position, momentum, etc.) are quantum observables. This would imply that each value we can measure for such property is orthogonal to all others. I've always thought such properties are classical and with continuous range of possibile values.

I couldn't find any other information about the relationship between superposition and uncertainty.

Am I misinterpreting something? Am I wrong or is the article wrong?

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    $\begingroup$ Can you be a bit more clear in your question. Are you asking if quantum observables are continuous, or are you asking if you can know more than one observable at a time, or are you asking a different question? $\endgroup$ – Jonathan Wheeler Feb 29 '16 at 23:38
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    $\begingroup$ Non-commuting observables are not restricted to quantum phenomena. They are occurring for perfectly classical waves. The width of wave packets in position and in frequency (momentum) space are non-commuting and the uncertainty relation is also known as a general property of Fourier transforms. We are using that a lot in electrical engineering where it imposes crucial restrictions on the design of digital communication systems. $\endgroup$ – CuriousOne Feb 29 '16 at 23:39
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    $\begingroup$ Start with this: The quote from Wikipedia is flat out wrong. Non-commuting observables can certainly have simultaneous eigenstates. Next: Yes, position and momentum have continuous spectra. You seem to think this is inconsistent with either (or both) having a complete set of orthogonal eigenvectors, but I don't understand why you think this. It is, in any event, not correct. (There are issues with the "eigenvectors" of, say, position, not actually belonging to the naively defined state space, but I think these issues are not what's bothering you.) $\endgroup$ – WillO Mar 1 '16 at 1:45
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As WillO points out in the comments, the quote is actually wrong: non-commuting observables can have (some) simultaneous eigenstates--but a better statement would be that not all of their eigenstates are simultaneous. In other words, if observable $\hat{A}$ has an eigenstate that's a non-trivial superposition of eigenstates of observable $\hat{B}$, then they're non-commuting, and vice versa.

The position-space and momentum-space wavefunctions are Fourier transforms of one another. In position-space representation, a state of definite position $x'$ would be a Dirac delta, $\psi_{x'}(x) = \delta(x-x')$, and the and a state of definite momentum $p'$ would be $$\psi_{p'}(x) = \exp(ip'x/\hbar)/\sqrt{2\pi\hbar}\text{.}$$ It is therefore by default a superposition of position eigenstates.

In fact, when you're writing a wavefunction in position-space, you are automatically representing it as a superposition of position eigenstates, with the function value $\psi(x)$ as the coefficient of the eigenstate of definite position $x$ in that superposition. That's what a wavefunction means: for arbitrary state $|\psi\rangle$, the position-space wavefunction is $\psi(x) = \langle x|\psi\rangle$, where $|x\rangle$ is a position eigenstate.

A similar statement would be correct for momentum-space: momentum eigenstates would be Dirac deltas in momentum-space representation, etc. The only difficultly is that that states like $\psi_p'(x)$ is not normalizable--but that's not a big deal; it's completely dual to the problem of the Dirac delta not actually being a function. They still make sense for the things they're practically used for, as distributions or various other more mathematically sophisticated ways of dealing with that issue.

I've always thought such properties are classical and with continuous range of possibile values.

You seem to be interpreting 'quantum' as implying 'discrete'. This would be very mistaken, and observables in quantum mechanics can and often do have continuous spectra--i.e., a continuous region of allowed results of measurements (eigenvalues). Typically, quantum-mechanical position and momentum observables are continuous. The difference between classical mechanics is that the operators that represent those observables are non-commutative.

This seems to state that the properties obeying the uncertainty principle (position, momentum, etc.) are quantum observables. This would imply that each value we can measure for such property is orthogonal to all others.

It's completely the opposite of this. If you have two observables $\hat{A}$ and $\hat{B}$, with eigenstates (states of definite values) of one that are all orthogonal to eigenstates of the other, then they don't have any non-trivial uncertainty relation between them.

I couldn't find any other information about the relationship between superposition and uncertainty.

One thing that might be helpful to realize is that the observables-as-operators formalism of quantum mechanics in Hilbert space applies just as well to classical mechanics, too--the difference is entirely in what operators correspond to which physical observables, including which ones you are allowed to measure.

For example, if you have an observable $\hat{z}$ with two eigenstates $|z_+\rangle$ and $|z_-\rangle$, an arbitrary superposition is always a valid state, e.g. the following two particular ones $$|x_\pm\rangle = \frac{1}{\sqrt{2}}\left(|z_+\rangle \pm |z_-\rangle\right)\text{.}$$ Now, in quantum mechanics, you could easily have an operator such as $$\hat{x} = |z_+\rangle\langle z_+| + |z_-\rangle\langle z_-|$$ that has exactly those two superpositions as its eigenstates. This operator would not commute with $\hat{z}$, and there will be a non-trivial uncertainty relationship between them. The difference is that in standard classical mechanics, something like $\hat{x}$ would not correspond to a valid physical observable. You would simply be forbidden from measuring it.

Instead, if you're only limited to measuring $\hat{z}$ and observables that commute with it, the states $|x_\pm\rangle$ would be indistinguishable from the mixed state of $|z_+\rangle$ and $|z_-\rangle$ with probability $1/2$ each. But quantum mechanics allows you more freedom.

An even easier example: if $|\psi\rangle$ is a non-trivial superposition of multiple eigenstates of some some observable $\hat{A}$, then the projector $\hat\pi = |\psi\rangle\langle\psi|$ is a valid observable in quantum mechanics that doesn't commute with $\hat{A}$ and with a single eigenstate of $|\psi\rangle$.

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