Quantum uncertainty after passing through a slit

[md.xavier]

Homework Statement

A given particle is confined to a certain potential. At a given instant that potential is turned off and the particle is accelerated by gravity. In the initial instant t = 0, when the potential is turned off, it has an initial speed of vz = 1 m/s.

In the instant t = 0 the particle goes through a slit with width y = 1 mm = 10^-3 m in the y axis. How much time does it take for the spatial uncertainty associated with y to double? Consider the law of propagation of uncertainty for the calculation of the spatial uncertainty of y at a given instant. Consider the particle a proton.

Homework Equations

[;\sigma_{x}\sigma_{p} \geq \frac{\hbar}{2};]

[;\mathbf{p} = m\mathbf{v};]

[;v = v_0+at \,;]

[;\left |\Delta X\right |=\left |\frac{\partial f}{\partial A}\right |\cdot \left |\Delta A\right |+\left |\frac{\partial f}{\partial B}\right |\cdot \left |\Delta B\right |+\left |\frac{\partial f}{\partial C}\right |\cdot \left |\Delta C\right |+\cdots;]

The Attempt at a Solution

I understood that, as the particle passes through the slit, we have [;\Delta y\right;] equal to the slit's width.

But, from that, I don't really get how to proceed, considering the particle is moving along the z axis. I can relate the errors along the z axis, but since there's no momentum along the y axis, I am kind of lost.

How am I supposed to use the propagation of uncertainty?

Thank you in advance. I don't expect to be just given the solution, I just want a little push forward. (I am new here, so please tell me if I'm doing anything wrong!)

 

[DimReg]

I'm not sure I understand this problem entirely but...

To start, ignore the z axis. What does the uncertainty in y position tell you about the uncertainty in y momentum?

 

[md.xavier]

I'm not sure either, actually!

And well, according to the Uncertainty principle:

[;\sigma_{py} \geq \frac{\hbar}{2\sigma_{y}};]

But since they mention the propagation, and even with this info, I'm really lost.

 

[DimReg]

Ok, that's right. (I think you should check that it's really supposed to be an initial v_z and not an initial v_y)

If you have a range of potential initial y positions, and a range of potential y velocities, how would you compute the range of potential y positions in the future?

 

[md.xavier]

(Confirmed! It really is an initial v_z.)

So, taking this:

[;\sigma_{py} \geq \frac{\hbar}{2\sigma_{y}};]

[;\sigma_{y_0} = 10^{-3};]

[;\sigma_{py_0} \geq 5.275 \times 10^{-32} ;]

Which means... (dividing by the proton mass)

[;\sigma_{vy_0} \geq 3.153 \times 10^{-5};]

Do I propagate to [;\sigma_{y};], using the info I have from [;\sigma_{vy_0};] and [;\sigma_{y_0};]? (using a normal position equation, like [; x = x_0 + v_0 \times t + 0.5 \times a \times t^2 ;])

 

[DimReg]

Yeah that's the idea. Basically, what I'm thinking is: what's the furthest in one direction it could be? what's the furthest in the other direction? Presumably the particle is somewhere in between these two points.

Also I figured out why the v_z matters: if there was no v_z, then the particle would still be in the slit, and the y position uncertainty would never change.

 

[md.xavier]

That's the uncertainty, no? For example, in the y direction, at t = 0, the particle could be anywhere within the slit, from one border to the other.

The thing is, now that I propagated (using the partial derivatives of y in respect to y_0 and vy_0), I have a few questions: are the initial position / velocity null? I now have an expression for [;\sigma_{y};] with respect to t, y_0, vy_0, and a. Considering none of those apart from t really apply to the y-axis in this problem, I'm kind of confused. I got an expression similar to:

[;\sigma_{y}^{2}(t) = (1 + v_y_0 \times t + 0.5 \times a \times t^2)^{2} \times (\sigma_{y_0})^{2} + (y_0 + t + 0.5 \times a \times t^{2})^{2} \times (\sigma_{v_y_0})^{2} ;]

Am I making a complete mess of things or am I going the right way? I'm definitely understanding the problem better, but the way to get there I'm quite not grasping yet.