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An eigenstate, or determinate state, is a state where the measurement of some observable always yields the same result. This means that the standard deviation of the observable is zero. If a distribution has a standard deviation of zero, this means that every value is the same value. There is only one value. Does this mean that an eigenstate, represented by its eigenfunction, looks like a collapsed wavefunction?

This doesn't seem right to me because the eigenfunctions of the hamiltonian for the infinite square well are sine or cosine functions (they don't look like a spike).

Maybe I'm getting a little lost with what an eigenfunction means/represents. The state of a system in Quantum Mechanics is represented by a vector, usually an infinite dimensional vector. Wavefunctions give you the probability amplitude that a particle is at a specific location at a specific time. When I solve the time-independent Schrodinger equation, I'm solving for the eigenfunctions of the energy operator, the hamiltonian. So when the hamiltonian acts on a system in one of those eigenfunctions, that system always collapses to the same point in space?

If I was in a determinate state of position, then the wavefunction would actually look like a spike? If I was in a determinate state of momentum, the wavefunction would be a sinusoidal?

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$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no such thing as "looking like a collapsed wavefunction", even if you believe there's collapse.

Let's go to the finite dimensional case and have a simple two-level spin system, that is, our Hilbert space is spanned by, e.g., the definite spin states in the $z$-direction $\ket{\uparrow_z}$,$\ket{\downarrow_z}$.

Now, the state $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{\uparrow_z} + \ket{\downarrow_z})$ is not an eigenstate of $S_z$, and measurement of $S_z$ will collapse it with equal probability into $\ket{\uparrow_z}$ or $\ket{\downarrow_z}$. Yet, this is an eigenstate of the spin in $y$-direction, i.e. $\ket{\psi} = \ket{\uparrow_y}$ (or down, we'd have to check that by computation, but it doesn't matter for this argument). So, although you can "collapse" $\ket{\psi}$ into other states, it already looks like a collapsed state by your logic. This shows that the notion of "looking like a collapsed state" is not very useful to begin with.

Furthermore, you seem to be confused about the difference between a measurement (inducing collapse in some dictions) and the application of an operator. You say

So when the hamiltonian acts on a system in one of those eigenfunctions, that system always collapses to the same point in space?

but this is non-sensical. The action of the Hamiltonian is an infinitesimal time step, as the Schrödinger equation tells you:

$$ \mathrm{i}\hbar\partial_t \ket{\psi} = H\ket{\psi} $$

and, for an eigenstate $\ket{\psi_n}$, which is a solution to the time-independent equation with energy $E_n$, you have by definition $H\ket{\psi_n} = E_n\ket{\psi_n}$, that is, the Hamiltonian is a "do nothing" operation on eigenstates since multiplication by a number does not change the quantum state. That is, after all, why we are interested in the solutions to the time-independent equation - because these are the stationary states that do not evolve in time. This has nothing to do with collapse, or measurement.

Lastly, exactly determinate states of position are not, strictly speaking, quantum states, since the "eigenfunctions" of the position operator "multiplication by x" are Dirac deltas $\psi(x) = \delta(x-x_0)$, which are not proper square-integrable functions $L^2(\mathbb{R})$ as quantum states are usually required to be. But yes, this is "a spike", and conversely, the determinate momentum states are plane waves $\psi(x) = \mathrm{e}^{\frac{\mathrm{i}}{\hbar}px}$.

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  • $\begingroup$ Could "looking like a collapsed state" not still be useful when viewing the state after projecting onto the basis of the eigenstates of the observable? Also, this might be trivial but why is $\frac{1}{\sqrt{2}}(|\uparrow_{z} \rangle + | \downarrow_{z} \rangle) = | \uparrow_{y} \rangle$ as you stated? $\endgroup$ – Alex Sep 27 '16 at 11:12
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    $\begingroup$ @Alex I'm not sure how it would be useful - My point is that just being in an eigenstate of any observable would be "looking like a collapsed state", but that a state can be in an eigenstate for one observable but not for another. As for the second question, it follows from the commutation relations (but it might also be $\lvert \down_y\rangle$, I didn't check it for this answer as I stated, because what precisely it is didn't matter for the argument). $\endgroup$ – ACuriousMind Sep 27 '16 at 11:57
  • $\begingroup$ Yes I think we had similar ideas. You could also have used the position and momentum operators to convey your point. In the sense that if we measure the position and then project the state onto the position basis we get a spike but if we project the same state onto the momentum basis we get a large spread. But I guess you wanted to avoid getting too involved in the discussion of determinate states of the position and momentum operators which you touched on at the end of your answer. $\endgroup$ – Alex Sep 27 '16 at 12:27
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If a given state is an eigenstate of a particular observable then that observable has a standard deviation of 0, however that says nothing about the distribution of any other observables. The extreme example of this are the eigenstates of position and momentum; a momentum eigenstate is represented by delta function in momentum space but in space it is represented by an infinite plane wave and the position of the particle is completely uncertain.

An energy eigenstate in a square well is a less extreme example. Its energy is perfectly well defined but its wavefunction, which tells you about the probability distribution for position measurements, is a sin wave. In other words you can measure the energy of a system in that state as many times as you like and you will always get the same result, but if you set up a series of particles in boxes in all in the same energy eigenstate and measure there positions, you will find you get random results with a distribution given by the mod square of the wavefunction.

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