[Physics] Eigenstate of position+momentum

commutatorheisenberg-uncertainty-principlemeasurement-problemquantum mechanics

I'm studying Quantum Mechanics on my own, so I'm bound to have alot of wrong ideas – please be forgiving!

Recently, I was thinking about the quantum mechanical assertion (postulate?) that states with a definite value for an observable are eigenvectors of that observable's operator. This might sound kind of silly, but what if, for a one dimensional particle, one decided to measure position+momentum? It's easy to construct the corresponding operator (in atomic units $\hbar=1$): $x-i\frac{d}{dx}$

It's pretty easy to find an eigenfunction of this operator: $\exp(\frac{x^2}{2i}-cx)$.

So, why can't the uncertainty relations be violated in such a case, if I could, say, measure the position of the object with this wave function, and immediately know the momentum (c-position)?

I understand that if the wave function collapses into a delta function (as it would in an idealized position measurement) then the momentum wave function would necessarily be a constant (or at least have constant mod) over all space, meaning that the momentum would be completely unknown.

UPDATE:

  1. I understand that the momentum wave function is the Fourier transform of the position wave function, and that the uncertainty relations can be shown with purely mathematical arguments about the std deviaion of fourier transform pairs – I have no problem with the uncertainty principle in general.

  2. I'm very appreciative for the answers, but I don't think the question was answered completely:

I think there was some confusion (my fault and mostly on my part) about exactly what was meant about "measuring the quantity x+p". I was imagining that if a system's vector is an eigenvector, with eigenvalue k, of some observable O, then any measurement of O will come out to k and will leave the system in that same eigenvector state – even if O is some kind of "compound observable".

So if I have a quantum system in some state that is an eigenvector of an operator that involves two quantities, I thought one could measure only one of the quantities and then infer the second one using the known eigenvalue (in much the same way that the spin of an entangled electron can be measured and then immediately the other spin will be known because the entangled state was an eigenvector of the "spin 1 + spin 2 operator").

If I had instead asked if x1 + p2 were measured for a two particle system, then the first position could be measured and immediately p2 would be known (if the system started out in the appropriate eigenvector state), right?

I think the problem in my original reasoning was the lack of freedom in the momentum wave function once the position wave function collapsed – i.e. one determines the other. In a several part system one part's wave function can be manipulated independently of the others.

Best Answer

So, why can't the uncertainty relations be violated in such a case, if I could, say, measure the position of the object with this wave function

That's the catch. You can't.

Or rather, you can measure the position, but the result you get will vary from one measurement to the next, because the wavefunction $\exp(x^2/2i - cx)$ is not an eigenstate of position. It's not an eigenstate of momentum, either, so if you tried to measure the momentum, you'd get similar variation between measurements. You could compute the standard deviations of the position and momentum measurements, $\sigma_x$ and $\sigma_p$ respectively, and you would find that they satisfy the uncertainty principle.

It's only if you measure the specific combination of position plus momentum that you would get the same result every time.

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