- #1

- 32

- 0

^{2}π

^{2})/2ma

^{2}. So the momentum is given by (2mE)

^{1/2}= ħπ/a. Since this is a precise value, doesn't that mean that we know momentum with 100% certainty? And if that is the case shouldn't the uncertainty in position be infinite?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Ananthan9470
- Start date

- #1

- 32

- 0

- #2

Simon Bridge

Science Advisor

Homework Helper

- 17,873

- 1,655

Your relation for the momentum is incorrect.

You can find the momentum in QM by applying the momentum operator - just like you find the energy by applying the energy operator. So what happes when you apply the momentum operator to the ground-state energy eigenfunction for a particle in a box?

Note: The momentum wavefunction is the fourier transform of the position wavefunction.

You can find the momentum in QM by applying the momentum operator - just like you find the energy by applying the energy operator. So what happes when you apply the momentum operator to the ground-state energy eigenfunction for a particle in a box?

Note: The momentum wavefunction is the fourier transform of the position wavefunction.

Last edited:

- #3

- 8,919

- 2,897

^{2}π^{2})/2ma^{2}. So the momentum is given by (2mE)^{1/2}= ħπ/a. Since this is a precise value, doesn't that mean that we know momentum with 100% certainty? And if that is the case shouldn't the uncertainty in position be infinite?

But if all you know is [itex]E[/itex], then the momentum is [itex]\pm \sqrt{2mE}[/itex]. So the expectation value of [itex]p[/itex] is 0 (it's just as likely to be pointing in one direction as another). A rough estimate of the uncertainty in a quantity is the standard deviation, which would tell us:

[itex]\delta p = \sqrt{\langle p^2\rangle - \langle p \rangle^2}[/itex] where [itex]\langle X \rangle[/itex] means the expectation value, or average value. So for this problem:

[itex]\langle p \rangle = 0[/itex]

[itex]\langle p^2 \rangle = 2mE[/itex]

So

[itex]\delta p = \sqrt{2mE} = \frac{\hbar \pi}{a} [/itex]

So the uncertainty of [itex]p[/itex] is not zero.

Last edited by a moderator:

- #4

jtbell

Mentor

- 15,803

- 4,034

Note particularly the graph of the momentum probability distribution for the ground state, ##|\Phi_1|^2##.

Share: