Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
3  
Have your studies progressed to the point of talking about "commutators" yet? This is very easy to answer with them, a bit more involved without. –  dmckee Jul 7 '13 at 21:27
1  
All operators are constructed from position, momentum and spin operators by adding or multiplying them together. Addition/multiplication of operators shouldn't be confused with simple addition/multiplication of their eigenvalues. In particular position + momentum would be a completely different operator. Eigenfunctions of "position + momentum" are neither eigenfunctions of position nor of momentum; hence its eigenvalues can not be interpreted as simple sum of eigenvalues of position and momentum operators. –  user10001 Jul 7 '13 at 22:21
2  
@user10001 that should be an answer. (In fact much the same answer I was about to post, until I noticed your comment) –  David Z Jul 7 '13 at 22:31
    
@David You should post your answer. May be you can explain better. –  user10001 Jul 7 '13 at 22:43

3 Answers 3

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.

share|improve this answer

Go look up "Fractionally Iterated Fourier Transforms". This is a rich area of applied mathematics for physics and engineering.

Position and momentum operators are Fourier Transforms of each other. When you form linear combinations, generally like

$$\cos(\theta)x + i\sin(\theta){{d}\over{dx}},$$

you find an interesting relation with the plain position operator, a sort of partial Fourier transform. This linear combination has its evil partner with which it's incompatible, just like position is with momentum:

$$\sin(\theta)x - i\cos(\theta){{d}\over{dx}},$$

So no, you can't cheat Heisenberg.

The Fractionally Iterated Fourier Transform is an operator $\mathscr{F}_\alpha$ such that

$$\begin{align} \mathscr{F}_0(f) &= f \\ \mathscr{F}_1(f) &= \mathscr{F}f \\ \mathscr{F}_\alpha \mathscr{F}_\beta &= \mathscr{F}_{\alpha+\beta} \end{align}$$

Note that $\mathscr{F}_4(f) = f$. (If you didn't know that four Fourier transform give back the original function, it is good to know now.)

The coefficients for the mixed position+momentum operator use $\theta$, related to $\alpha$ by $\theta = {\pi\over{2}}\alpha$.

The eigenfunctions of the FIFT are the energy state wavefunctions $\phi_n$ of the harmonic oscillator. An arbitrary function can be analyzed in terms of these eigenfunctions, which are orthogonal and complete.

$$ f(x) = \sum A_n\phi_n(x) $$

To find the ordinary Fourier transform, multiply the coefficients $A_n$ by phase factors:

$$\begin{gather} \mathscr{F}(f(x)) = g(y) = \sum B_n \phi_n(y) \\ B_n = i^n A_n \end{gather}$$

This easily generalizes to arbitrary orders of iteration:

$$\begin{gather} \mathscr{F}_\alpha(f(x)) = g^{(\alpha)}(y) = \sum B^{(\alpha)}_n \phi_n(y)\\ B^{(\alpha)}_n = i^{\alpha n} A_n \end{gather}$$

If $\alpha$ is $2\pi$, then $B_n=A_n$.

If you examine Schrodinger's equation for the harmonic oscillator starting with an arbitrary initial wavefunction, the time evolution of the wavefunction can be described using FIFT with $\alpha$ varying linearly with time.

Fun things to ponder:

  • $\alpha$ being close to zero. Make approximations.
  • Differentiate w.r.t. $\alpha$
  • $\alpha$ being pure imaginary.

In real-world engineering, FIFT is used in optics to describe near-field diffraction patterns in some situations.

share|improve this answer
3  
go home Fourier transform, you're drunk –  lurscher Jul 8 '13 at 0:38

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".

Try it.

Define

$$\psi_0 = \exp \left( \frac{x^2}{2i}−cx \right)$$

and act on it with $x + p$ to get

$$ \begin{array} \\ \psi_1 & = (x+p)\psi_0 \\ & = x\exp\left(\frac{x^2}{2i}−cx\right) - i\frac{d}{dx}\exp\left(\frac{x^2}{2i}−cx\right) \\ & = x\exp\left(\frac{x^2}{2i}−cx\right) - i\exp\left(\frac{x^2}{2i}−cx\right) \frac{d}{dx}\left(\frac{x^2}{2i}−cx\right) \\ & = x\exp\left(\frac{x^2}{2i}−cx\right) - i\exp\left(\frac{x^2}{2i}−cx\right) \left(\frac{2x}{2i} - c\right) \\ & = c\exp\left(\frac{x^2}{2i}−cx\right) = c \psi_0 \end{array}$$ which shows that $\psi_0$ is an eigenfunction of $(x+p)$ (which you already knew, but I exhibit it to show the difference with the position operator below).

So we try the same thing with just $x$, but

$$\psi_2 = x\psi_0 = x \exp\left(\frac{x^2}{2i}−cx\right) $$

is not a constant times $\psi_0$ so your your wave function is not an eigenfunction of the position operator. That result turns out to be general, $\hat{O} = \hat{x} + \hat{p}$ does not share eigenfunctions with either the position or the momentum operator.


I get the feeling that you are groping toward a method to understand these issues without the (potential) complexity of working the problem long hand every time. That's a good idea; the usual framework is to define the "commutator" between to operators $\hat{A}$ and $\hat{B}$ as

$$ \left[\hat{A},\hat{B}\right] = \hat{A}\hat{B} - \hat{B}\hat{A} \,.$$

With a little effort you can show that when the commutator is zero the operators share eigenstates, and when it is non-zero they do not.

With this in mind we could have written

$$\begin{array} \\ \left[ x,(x+p) \right] & = x\left(x-i\frac{d}{dx}\right) - \left(x-i\frac{d}{dx}\right)x \\ & = \left( x^2 - ix\frac{d}{dx} \right) - \left(x^2 - i\frac{d}{dx}x \right) \\ & = \left( x^2 - ix\frac{d}{dx} \right) - \left(x^2 - i \right) \\ & = i \left( \frac{d}{dx}x - x\frac{d}{dx} \right) \\ & \ne 0 \end{array} $$

and drawn the same conclusions.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.