Modern Physics: Mastering Physics: Heisenberg's Uncertainty Principle

Learning Goal: To understand, qualitatively and quantitatively, the uncertainty principle.

Understanding Heisenberg's uncertainty principle is one of the keys to understanding quantum mechanics. The principle states that you can never simultaneously know the exact location and momentum of a particle. Further, it states that the more you know about the position of the particle, the less you know about its momentum, and vice versa. The uncertainty principle is more than just a statement about the difficulty of measuring such things experimentally. Rather, it states that momentum and position are not simultaneously well defined for quantum particles. In fact, Heisenberg did not call his idea the uncertainty principle; he called it the indeterminacy principle, because position and momentum are fundamentally indeterminate, not just unknown, for the waves described by quantum mechanics.
This idea is difficult to reconcile with common experience. To understand it better, you must consider the properties of a wave.

According to the de Broglie equation, the momentum of a wave is directly related to its wavelength. For the wave in the first figure, the wavelength is clearly well defined. However, the position is not well defined at all. The question, "Where is the wave?" does not have a well-defined answer, as we expect for a particle. This is the essence of the indeterminacy principle. We could just as easily draw a single sharp point at some particular x coordinate.

This could be considered a wave with a very well determined position. However, any notion of wavelength for such a wave seems strange.
A wave like the one shown in the second figure can be built up by adding together waves with different wavelengths. Recall that if two waves with similar frequencies, f_1

and

, are added together, a wave with a beat frequency of f_1-f_2

is produced

. This gives a wave with somewhat well-defined position and wavelength. If you add contributions from all of the frequencies between f_1

and

, then you get a wave packet, which looks essentially like a single isolated beat cycle.

In this problem, you will consider such a wave packet as simply being one beat cycle of this wave. While not exactly correct, this will give a useful approxmation.
Let the distance between the two nodes of the wave be the uncertainty in position Deltax

. Since the beat frequency is given by f_1-f_2

, and the wave travels at speed

, the uncertainty in position is given by

$\Delta x = \frac{v}{f_1-f_2}$ .

Part A

The de Broglie relation $\lambda=h/p$ can be rewritten in terms of the wave number

as $p=k\hbar$ . Recall that wave number is defined by $k=2\pi/\lambda$ . Using the fact that $\lambda=v/f$ , find the wave numbers k_1

and

corresponding to frequencies f_1

and

Express your answer as two expressions separated by a comma. Use f_1

, and

ANSWER:

$\frac{2{\pi}f_{1}}{v},\frac{2{\pi}f_{2}}{v}$

Part B

Find an expression for the uncertainty $\Delta k=k_1-k_2$ in the wave number. Use your results from Part A.

Express your answer in terms of quantities given in Part A.

ANSWER:

$\frac{2{\pi}}{v}\left(f_{1}-f_{2}\right)$

Part C

What is the value of the product $\Delta x \Delta p$ ? Use $p=\hbar k$ to find the uncertainty in the momentum of the particle.

Express your answer in terms of quantities given in Part A and fundamental constants.

ANSWER:

$\Delta x \Delta p$ =

This gives you the general idea of what the uncertainty principle states mathematically. The product of the uncertainties in the momentum and position of a particle is on the order of Planck's constant. By looking more rigorously at the definition of the uncertainty, the uncertainty principle is found to state that $\Delta x \Delta p \ge \hbar$ . The greater-than-or-equal-to sign indicates that some less than ideal waveforms have greater uncertainty that the minimum value of hbar

Part D

In an atom, an electron is confined to a space of roughly $10^{-10}$ meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty Deltap

in its momentum?

Express your answer in kilogram meters per second to two significant figures.

ANSWER:

1.05×10⁻²⁴

$\rm kg\cdot m/s$

Part E

What is the kinetic energy

of an electron with momentum $1.05 \times 10^{-24}$ kilogram meters per second?

Express your answer in electron volts to two significant figures.

ANSWER:

3.8

$\rm eV$

Notice that this energy is similar to the energy scale for electrons in an atom, which typically ranges from a bit less than an electron volt up to a few dozen electron volts. A good estimate for the energy scale of a particle can often be found by calculating the energy the particle would have if you set the momentum equal to the minimum uncertainty in momentum. The justification for this sort of estimation lies in the rigorous statistical definition of the uncertainty; it is sufficient now for you to know that this will give a reasonably good order-of-magnitude estimate of the energy for a variety of quantum systems.

Part F

Suppose that you know the position of a 100-gram pebble to within the width of an atomic nucleus ( $\Delta x=10^{-15}$ meters). What is the minimum uncertainty in the momentum of the pebble?

Express your answer in kilogram meters per second to one significant figure.

ANSWER:

1.0×10⁻¹⁹

$\rm kg \cdot m /s$

For a 100-gram pebble, this corresponds to an uncertainty in the speed of about $10^{-18}$ meters per second. Such tiny values are the reason that you are unaware of the uncertainty principle in everyday situations. In practice, it would be impossible to measure the position of a pebble to such accuracy, much less its speed

13 comments:

SauleneyFebruary 16, 2012 at 7:37 AM
That was really helpful, thanks :)
JeffMay 1, 2012 at 11:03 PM
That was really helpful...if I just wanted the answers. Having an explanation would be nice too.
GymnosMay 1, 2012 at 11:07 PM
Then feel free to ask questions regarding specific parts. I'll help if I can.
DisaFearSeptember 21, 2012 at 8:52 PM
Answer to Part D is wrong?
GymnosSeptember 21, 2012 at 9:51 PM
I believe you are correct, DisaFear. If we take (dx)(dp) = h, then (dp) = h / (dx) ~= 6.626*10^-24. I can't check these answers anymore, but you could try that.
Jeffery GaoSeptember 25, 2012 at 5:49 AM
Part D should be 3.3*10^-24 kgm/s using the uncertainty formula:

dx.dp>=h/2
TravisApril 2, 2013 at 11:08 PM
I got 5.3*10^-25 for part D. I'm getting part F wrong, too.
TravisApril 2, 2013 at 11:16 PM
Part F: deltX*deltP>=hbar/2, therefore deltp>=hbar/(2*deltX)=5*10^-20 kg*m/s
BrianH321October 2, 2015 at 11:07 AM
Part D is correct. If you're using Mastering Physics then you'll in the comments above this portion of the question a different uncertainty formula; (dx)(dp)>h(bar) and not h(bar)/2. I do not understand why it is different but it is and 1.05E-24 (or 1.1E-24 for 2 sig fig's) gives a correct answer. It would be awesome if anyone could give a little more detail to Part F.
BrianH321October 2, 2015 at 11:11 AM
Disregard my issue with Part F, "... one sig fig."
HeroOctober 31, 2015 at 6:08 PM
3*10^-19 will be the answer for F

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Sunday, February 20, 2011

Mastering Physics: Heisenberg's Uncertainty Principle

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