Uncertainty Principle: Computing Δx & Δp for a Particle in a 1D Box

ChemMajor4lyf
Messages
2
Reaction score
0

Homework Statement


The uncertainty ΔB in some observable B is given by a formula ΔB = √<B^2> - <B>^2.
Use this formula to determine the uncertainty in position, Δx, and momentum Δp, for the ground state of a quantum-mechanical particle of mass m is a 1-D 'box' of length a, and show that the uncertainty principle holds.


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Welcome to PF,

ChemMajor4lyf said:

Homework Statement


The uncertainty ΔB in some observable B is given by a formula ΔB = √<B^2> - <B>^2.
Use this formula to determine the uncertainty in position, Δx, and momentum Δp, for the ground state of a quantum-mechanical particle of mass m is a 1-D 'box' of length a, and show that the uncertainty principle holds.


Homework Equations





The Attempt at a Solution


What have you done so far on this problem?
 
<x> = ∫x abs(ψ)^2 where n=1 and length = a

<x>^2 = (a/2)^2 = a^2/4

and the momentum operator = -i(h/2(pi)) ∂/∂x

I'm just really confused on where to start. I don't understand what I'm suppose to do with the ΔB equation. Plug it into the Schrodinger Equation?
 
ChemMajor4lyf said:
<x> = ∫x abs(ψ)^2 where n=1 and length = a

<x>^2 = (a/2)^2 = a^2/4

and the momentum operator = -i(h/2(pi)) ∂/∂x

I'm just really confused on where to start. I don't understand what I'm suppose to do with the ΔB equation. Plug it into the Schrodinger Equation?

The B equation is just showing you what the definition of uncertainty is for any observable. So, "B" here represents an arbitrary quantity.

This means that the uncertainty in x is given by \Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}. Similarly, \Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}. Again, the purpose of giving you the B equation was just to state that this definition holds true generically for any observable.

So it's pretty clear that in order to compute the uncertainty, you need to figure out how to compute the expectation value of an operator. The expectation value is the thing in angle brackets. From what you've posted above, you seem to know how to do that already. You compute the expectation value of a quantity by integrating the quantity in question multiplied by the modulus squared of the wavefunction. This integral takes place over all space i.e. over the entire domain over which the wavefunction is defined.

So, in order to compute the expectation value, you need to know what the wavefunction is.. For THAT (determining the wavefunction), you need to solve the Schrodinger equation for this particular 1-D potential, and then take ground state solution. However, I suspect that you've already gone over this solution in class, and therefore you have computed the wavefunctions for this "infinite square well" potential already.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top