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The Heisenberg's uncertainty principle states the following:

$$\Delta p \cdot \Delta x \ge \frac{h}{4\pi}.$$

While studying for my high school physics exams, I fooled myself into believing that I understood the uncertainty principle (at least the implications). But suddenly the question that's nagging me is the following. If the uncertainty $\Delta x$ of an electron is 1.2 nm, does it imply that the probability that the x-coordinate lies within a 1.2 nm range, equal to 100%? Or does it mean that the probability is 95%? Or does it mean something totally different?

I wonder why no author made it clear in the high school/junior college level textbooks. I am uncertain about what uncertainty means.

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I'm sure that your textbook gave the definition somewhere: $\Delta x = \sqrt{\left<x^2\right>-\left<x\right>^2}$ –  user2963 Oct 11 '12 at 23:09
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On second thought, perhaps it didn't: it seems a lot of explanations of the uncertainty principle just gloss over the mathematical meaning. –  user2963 Oct 11 '12 at 23:15

4 Answers 4

$\Delta x$ is really the standard deviation in $x$, $\sigma_x$. So the probability that you find the particle within $\Delta x$ is about $68\%$. Same goes for momentum.

In response to the other answer, the Fourier Transform formalism in fact shows that if the probability distribution of $x$ is a normal distribution, the standard deviation = $\sigma_x$. The probability that a random variable lies within 1 sigma of the mean is 68.27%.

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The 68% only holds if the position has a normal distribution like (P(x)=Ae^-(x-x0)^2/sig^2). There is no guarantee that a given wavefunction probability distribution will have that form - see @DavidZaslavsky 's answer... –  FrankH Oct 12 '12 at 1:52
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I agree. I didn't mean to say that the probability distribution is always normal; that's why I qualified with an "if." And since so many distributions are asymptotically gaussian, 68% is a nice number to throw around. –  hwlin Oct 12 '12 at 2:00
    
I have a problem with calling it "THE standard deviation", unqualified, because in measuring x the standard deviation given by the measurement is classical and momentum can be measured simultaneously to equal accuracy, unless one is working in quantum dimensions. It is better to call it "uncertainty" in position and momentum. –  anna v Oct 12 '12 at 4:37
    
From my understanding, it is THE standard deviation --- at least the one that arises from the statistics of quantum mechanics. See aip.org/history/heisenberg/p08a.htm –  hwlin Oct 12 '12 at 22:16

For a particle which has a position-space wavefunction $\psi(x)$, the uncertainty in position, denoted $\sigma_x$ or $\Delta x$ (I prefer the former), is given by

$$\begin{align} \sigma_x^2 &= \langle x^2\rangle - \langle x\rangle^2 \\ &= \int_{-\infty}^{\infty}\psi^*(x)x^2\psi(x)\,\mathrm{d}x - \biggl[\int_{-\infty}^{\infty}\psi^*(x)\,x\,\psi(x)\,\mathrm{d}x\biggr]^2 \end{align}$$

and the uncertainty in momentum, denoted $\sigma_p$ or $\Delta p$, is given by

$$\begin{align} \sigma_p^2 &= \langle p^2\rangle - \langle p\rangle^2 \\ &= \int_{-\infty}^{\infty}\psi^*(x)\biggl(-i\hbar\frac{\partial}{\partial x}\biggr)^2\psi(x)\,\mathrm{d}x - \biggl[\int_{-\infty}^{\infty}\psi^*(x)\biggl(-i\hbar\frac{\partial}{\partial x}\biggr)\psi(x)\,\mathrm{d}x\biggr]^2 \end{align}$$

Wikipedia's page on the uncertainty principle contains a proof using these definitions.

These definitions do not imply anything about the probability of finding the particle within the range specified by the uncertainty.

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David outlined the mathematical of the uncertainty principle.

The meaning behind the uncertainty principle can be understood if you look very look at an experimental apparatus that illustrates it. If one attempts to measure the position of a particle to an accuracy, then the same experimental apparatus cannot measure momentum to an accuracy greater than that given by the momentum cut.

For an Illustration consider the Double slit experiment. The measurement of position is performed to an accuracy corresponding to that of the slit width. There is an uncontrollable exchange of momentum between the slit and the particle, that manifests itself as an uncertainty in the momentum of the photon that leaves the slit.

I would Highly recommend Bohr's exposition. http://www.marxists.org/reference/subject/philosophy/works/dk/bohr.htm

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Any textbook on quantum mechanics that I know gives the definition of $\Delta p$ and $\Delta x$.

Apart from the purely mathematical answer, consider this from a physical point of view. The only possibility to restrict an electron within a 1.2 nm wide box is by applying an infinite force at the edges. In reality the wavefunction for the electron will spread over all the space and the probability to find the electron outside the box will be non-zero.

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