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

[ChemMajor4lyf]

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

 

[cepheid]

Welcome to PF,

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?

 

[ChemMajor4lyf]

<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?

 

[cepheid]

<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.

 

[paranormal]

The uncertainty principle, first proposed by Werner Heisenberg, states that it is impossible to know with certainty both the position and momentum of a particle at the same time. This is mathematically expressed as ΔxΔp ≥ ħ/2, where Δx is the uncertainty in position and Δp is the uncertainty in momentum, and ħ is the reduced Planck's constant.

To determine the uncertainty in position, we can use the formula given in the problem, Δx = √<x^2> - <x>^2, where <x^2> is the expectation value of x^2 and <x> is the expectation value of x. For a particle in a 1-D box of length a, the wave function can be expressed as Ψ(x) = √(2/a)sin(nπx/a), where n is the quantum number for the energy level.

Using this wave function, we can calculate <x^2> and <x> as follows:

<x^2> = ∫Ψ*(x)x^2Ψ(x)dx = (2/a)∫sin^2(nπx/a)x^2dx = a^2/3

<x> = ∫Ψ*(x)xΨ(x)dx = (2/a)∫sin^2(nπx/a)xdx = a/2

Substituting these values into the formula for uncertainty in position, we get:

Δx = √(a^2/3 - (a/2)^2) = a/√12

Similarly, we can calculate the uncertainty in momentum using the formula Δp = √<p^2> - <p>^2, where <p^2> is the expectation value of p^2 and <p> is the expectation value of p. The momentum operator in 1-D is given by p = -iħd/dx.

Using the same wave function, we can calculate <p^2> and <p> as follows:

<p^2> = ∫Ψ*(x)(-iħd/dx)^2Ψ(x)dx = (2/a)∫sin^2(nπx/a)(-iħd/dx)^2dx = (nπħ)^2

<p> = ∫Ψ*(x)(-i