Which momenta can never be measured?

[hb1547]

Homework Statement

"A particle is described by:
$$\psi(x) = \left\{ \begin{array}{lr} C & : \left|x\right| \leq +\frac{1}{2w}\\ 0 & : \left|x\right| > \frac{1}{2w} \end{array} \right.$$

What momenta can never be measured?"

Homework Equations

$$\Delta p \Delta x \geq \frac{\hbar}{2}$$
- Others?

The Attempt at a Solution

Not entirely sure what to do. I tried solving for Delta p, but that answer didn't really make sense to me. From what I gathered, Delta x must be equal to W, so that'd mean:

$$\Delta p = \frac{\hbar}{2w}$$

But I can tell this isn't right, but I'm not sure if I'm going in the right direction.

 

[Bhumble]

I'm not sure about the 1/2 coefficient on the right hand side but just looking at the wave function being equal to zero raises flags for me. I'd say that you can't observe the momenta when -1/2w < x < 1/2w. The wave function is giving you a probability density then telling you that it is zero outside of this interval.

Just to clarify $$\Delta$$x is not x. Typically you are given a range.

 

[gabbagabbahey]

When you measure Momentum, the state will collapse into an eigenstate of the momentum operator. The probability of measuring a given value of momentum is therefore going to be the inner product of the corresponding momentum eigenstate with the original state. Which momenta give a probability of zero?

 

[Bhumble]

Correct me if I'm wrong but my understanding is that the wave function can be interpreted as a probability density function thus the probability is the integral over all space. With this problem the probability density is zero except from -1/2w to 1/2w and is constant within that range.
So the momentum is obervable where a value of x is possible or within the specified range. Is my reasoning flawed somewhere?

 

[fzero]

Correct me if I'm wrong but my understanding is that the wave function can be interpreted as a probability density function thus the probability is the integral over all space. With this problem the probability density is zero except from -1/2w to 1/2w and is constant within that range.
So the momentum is obervable where a value of x is possible or within the specified range. Is my reasoning flawed somewhere?

The wavefunction here is given in position space, so the corresponding probability density is the one that gives you the probability of finding the particle in a position between x and x + dx. If you want the probability density for momentum values, you need to compute the Fourier transform to find the wavefunction in momentum space.

 

[hb1547]

The wavefunction here is given in position space, so the corresponding probability density is the one that gives you the probability of finding the particle in a position between x and x + dx. If you want the probability density for momentum values, you need to compute the Fourier transform to find the wavefunction in momentum space.

The class I'm taking doesn't expect us to use Fourier transformations yet, is there another way of looking at this?

 

[fzero]

The class I'm taking doesn't expect us to use Fourier transformations yet, is there another way of looking at this?

It may not have been called a Fourier transform, but the momentum space wavefunction is related to that in position space by the integral

$$\psi(p) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-ipx/\hbar} \psi(x)~dx.$$

You will have to normalize $$\psi(p)$$ to finish the problem.

 

[Dickfore]

Find the wave function in momentum representation:

$$a(p) = \frac{1}{(2\pi \, \hbar)^{\frac{1}{2}}} \, \int_{-\infty}^{\infty}{\exp\left[-\frac{i}{\hbar} \, p \, x\right] \, \psi(x) \, dx}$$

The probability density function for different values of momentum is then:

$$|a(p)|^{2}$$

Therefore, whenever $a(p) = 0$, those values of momentum have zero probability of being measured, i.e. they are never measured.