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8.3.1 Repeated measurements

If we measure E once and we find Ei as outcome we know that the system is in the ith eigenstate of the Hamiltonian. That certainty means that if we measure the energy again we must find Ei again. This is called the ``collapse of the wave function'': before the first measurement we couldn't predict the outcome of the experiment, but the first measurements prepares the wave function of the system in one particuliar state, and there is only one component left!

Now what happens if we measure two different observables? Say, at 12 o'clock we measure the position of a particle, and a little later its momentum. How do these measurements relate? Measuring $ \hat{x}$ to be x0 makes the wavefunction collapse to $ \delta$(x - x0), whatever it was before. Now mathematically it can be shown that

$\displaystyle \delta$(x - x0) = $\displaystyle {\frac{1}{2\pi}}$$\displaystyle \int_{-\infty}^{\infty}$eik(x - x0dk. (8.31)

Since eikx is an eigenstate of the momentum operator, the coordinate eigen function is a superposition of all momentum eigen functions with equal weight. Thus the spread in possible outcomes of a measurement of p is infinite!
incompatible operators
The reason is that $ \hat{x}$ and $ \hat{p}$ are so-called incompatible operators, where

$\displaystyle \hat{p}$$\displaystyle \hat{x}$ $\displaystyle \neq$ $\displaystyle \hat{x}$$\displaystyle \hat{p}$! (8.32)

The way to show this is to calculate

($\displaystyle \hat{p}$$\displaystyle \hat{x}$ - $\displaystyle \hat{x}$$\displaystyle \hat{p}$)f (x) $\displaystyle \equiv$ [$\displaystyle \hat{p}$,$\displaystyle \hat{x}$]f (x) (8.33)

for arbitrary f (x). A little algebra shows that
[$\displaystyle \hat{p}$,$\displaystyle \hat{x}$]f (x) = $\displaystyle {\frac{\hbar}{i}}$($\displaystyle {\frac{d}{dx}}$x)f (x)  
  = $\displaystyle {\frac{\hbar}{i}}$f (x). (8.34)

In operatorial notation,

[hatp,$\displaystyle \hat{x}$] = $\displaystyle {\frac{\hbar}{i}}$$\displaystyle \hat{1}$, (8.35)

where the operator $ \hat{1}$, which multiplies by 1, i.e., changes f (x) into itself, is usually not written.

The reason these are now called ``incompatible operators'' is that an eigenfunction of one operator is not one of the other: if $ \hat{x}$$ \phi$(x) = x0$ \phi$(x), then

$\displaystyle \hat{x}$$\displaystyle \hat{p}$$\displaystyle \phi$(x) = x0$\displaystyle \hat{p}$$\displaystyle \phi$(x) - $\displaystyle {\frac{\hbar}{i}}$$\displaystyle \phi$(x) (8.36)

If $ \phi$(x) was also an eigenstate of $ \hat{p}$ with eigenvalue p0 we find the contradiction x0p0 = x0p0 - $ {\frac{\hbar}{i}}$.

Now what happens if we initially measure $ \hat{x}$ = x0 with finite acuracy $ \Delta$x? This means that the wave function collapses to a Gaussian form,

$\displaystyle \phi$(x) $\displaystyle \propto$ exp(- (x - x0)2/$\displaystyle \Delta$x2) (8.37)

It can be shown that

exp(- (x - x0)2/$\displaystyle \Delta$x2) = $\displaystyle \int_{-\infty}^{\infty}$dk eikxe-ikx0exp(- 1/4k2$\displaystyle \Delta$x2), (8.38)

from which we read off that $ \Delta$p = $ \hbar$/(2$ \Delta$x), and thus we conclude that at best

$\displaystyle \Delta$x$\displaystyle \Delta$p = $\displaystyle \hbar$/2 (8.39)

which is the celeberated Heisenberg uncertainty relation.


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© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk