How Do You Normalize and Transform a Piecewise Wavefunction?

In summary, a wavefunction is a mathematical representation of a particle's quantum state that describes its probability of being in a certain location or energy level. It is a fundamental concept in quantum mechanics and is used to calculate probabilities in experiments. When a wavefunction collapses, the quantum state becomes fixed and no longer described by a wavefunction. The Schrödinger equation is used to describe how the wavefunction evolves over time, taking into account potential energy. While the wavefunction itself cannot be directly measured, its effects can be observed through experiments.
  • #1
renegade05
52
0

Homework Statement


Ok the problem is this bad boy:
ψ(x) = \begin{cases}
A & \text{for $|x| < d$} \\
0 & \text{for $|x|>d$} \\
\end{cases}

Homework Equations



(a)Find a value of A which makes the wavefunction normalized.
(b)Find the momentum wavefunction ψ'(p).
(c)what is the relationship between d, the half width of the position probability distribution and the half-width of the momentum probability distribution. is this relationship consistent with the Heisenberg uncertainty principle?
(d)If the momentum of the particle is measured, what is the probability of it being observed moving to the right?


The Attempt at a Solution


(a) Well to normalize it I tried to break up the peicewise, such that I would get two integrals, one from neg infinity to d the other d to infinity. the one from d to infinity drops out because of the zero. so I am left with an integral that diverges? I don't know where I went wrong.

(b)no idea how to do this.. please start me off.

(c) Perhapes once I get the momentum wave function this will be become apparent? if not can someone start me off?

(d) Do I just apply the uncertainty principle? ΔxΔp>=hbar/2 ??

thanks for all the help!
 
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  • #2
renegade05 said:

Homework Statement


Ok the problem is this bad boy:
ψ(x) = \begin{cases}
A & \text{for $|x| < d$} \\
0 & \text{for $|x|>d$} \\
\end{cases}

Homework Equations



(a)Find a value of A which makes the wavefunction normalized.
(b)Find the momentum wavefunction ψ'(p).
(c)what is the relationship between d, the half width of the position probability distribution and the half-width of the momentum probability distribution. is this relationship consistent with the Heisenberg uncertainty principle?
(d)If the momentum of the particle is measured, what is the probability of it being observed moving to the right?

The Attempt at a Solution


(a) Well to normalize it I tried to break up the peicewise, such that I would get two integrals, one from neg infinity to d the other d to infinity. the one from d to infinity drops out because of the zero. so I am left with an integral that diverges? I don't know where I went wrong.

(b)no idea how to do this.. please start me off.

(c) Perhapes once I get the momentum wave function this will be become apparent? if not can someone start me off?

(d) Do I just apply the uncertainty principle? ΔxΔp>=hbar/2 ??

thanks for all the help!

Just exactly how did you get an integral that diverges for a)? ψ(x) is also zero for x<-d. I'm assuming this is a 1 dimensional problem. Show your work, ok?
 
  • #3
∫(from -∞ to d) A^2dx = 1

This integral has no answer and no way to solve for A?
 
  • #4
renegade05 said:
∫(from -∞ to d) A^2dx = 1

This integral has no answer and no way to solve for A?

The problem says ψ(x)=0 if |x|>d. That means if x<(-d) then ψ(x)=0. You only integrate A^2 from -d to +d. Do you know what |x| means?
 
  • #5
yes aboslute value.

So A = i/sqrt(2d) ?
 
  • #6
renegade05 said:
yes aboslute value.

So A = i/sqrt(2d) ?

I would have said A=1/sqrt(2d), not i/sqrt(2d), but that works as well. But why did you put the i in?
 
  • #7
dont know... haha. YA I got the answer now thank you!

so part b says Find the momentum wavefunction ψ'(p).

How can i do this?
 
  • #8
renegade05 said:
dont know... haha. YA I got the answer now thank you!

so part b says Find the momentum wavefunction ψ'(p).

How can i do this?

The momentum wavefunction is related to position wavefunction by a Fourier transform. You might have to do some research on this if you didn't know that.
 

1. What is a wavefunction?

A wavefunction is a mathematical representation of the quantum state of a particle. It describes the probability of finding a particle in a particular location or with a certain energy level.

2. How is the wavefunction related to quantum mechanics?

The wavefunction is a fundamental concept in quantum mechanics, as it is used to describe the behavior of particles on a quantum level. It is used to calculate the probabilities of different outcomes in quantum experiments.

3. What does it mean when a wavefunction collapses?

When a wavefunction collapses, it means that the quantum state of a particle has been measured or observed, causing its state to become fixed and no longer described by a wavefunction. This is known as the "measurement problem" in quantum mechanics.

4. How is the wavefunction affected by the Schrödinger equation?

The Schrödinger equation is used to describe how the wavefunction of a particle evolves over time. It takes into account the potential energy of the particle and how it changes over time. The solutions to the Schrödinger equation give the possible wavefunctions for a particle in a particular system.

5. Can the wavefunction be measured directly?

No, the wavefunction itself cannot be measured directly. It is a mathematical concept used to describe the quantum state of a particle. However, the effects of the wavefunction can be observed and measured through experiments in quantum mechanics.

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