Quantum Mechanics and Eigenfunctions Checks

In summary, we are checking whether given wave functions are eigenfunctions of the momentum operator by applying i \hbar \partial_x to the function and checking if the result is proportional to the original function. If the result can be written as a constant times the original function, then it is an eigenfunction and the constant is the eigenvalue. However, if the result cannot be written in this form, then the function is not an eigenfunction. We also need to calculate the expectation value 〈 \hat{p} 〉 and the standard deviation Δ\hat{p} ̂ for each eigenfunction.
  • #1
TFM
1,026
0

Homework Statement



For each of the following wave functions check whether they are eigenfunctions of the momentum operator, ie whether they satisfy the eigenvalue equation:

[tex] \hat{p} \psi(x) = p\psi(x) with \hat{p} = i \hbar \frac{\partial}{\partial x} [/tex] and p is a real number.

For those that are eigenfunctions, calculate the eigenvalue p, the expectation value〈[tex] \hat{p} [/tex]〉, and the standard deviation Δ[tex]\hat{p}[/tex] ̂.

(a)

[tex] \psi(x) = \sqrt{\frac{2}{L}} cos(\frac{x\pi}{L}) for -L/2 \leq x \leqL/2[/tex]

0 for x > L/2, x < -L/2

(b)

[tex] \phi(x)= \frac{1}{\sqrt{L}}e^{ikx} for 0 \leq x \leq L with k is real [\tex]

0 for x > L, x < 0


(c)

[tex] \chi (x) = 2xe^{-x} for 0 \leq x [/tex]
0 for x < 0


Homework Equations



N/A

The Attempt at a Solution



Okay, I have gone some way through the first part.

[tex] \hat{p} = -i \hbar \frac{\partial}{\partial} [/tex]

[tex] -i\hbar \frac{\partial}{\partial x}(\sqrt{\frac{2}{L}}cos\frac{x\pi}{L}) [/tex]

[tex] -i\hbar \sqrt{\frac{2}{L}} \frac{\partial}{\partial x}(cos\frac{x\pi}{L}) [/tex]

With Limits between L/2 and -L/2

[tex] -i\hbar \sqrt{\frac{2}{L}} (-\frac{\pi}{L}sin\frac{x\pi}{L})^{L/2}_{-L/2} [/tex]

[tex] -i\hbar \sqrt{\frac{2}{L}} ([-\frac{\pi}{L}sin\frac{L\pi}{2L}]-[-\frac{\pi}{L}sin\frac{-L\pi}{2L}]) [/tex]

[tex] -i\hbar \sqrt{\frac{2}{L}} ([-\frac{\pi}{L}sin\frac{\pi}{2}]-[-\frac{\pi}{L}sin\frac{-\pi}{2}]) [/tex]

But I am unsure where I am supoosed to go from here...

To show it is a Eigenfunction, I need to get rid of the i at the very begining, but I am unsure how to do this?

Any suggesstions where to rpoceed?

Thanks in advance,

TFM
 
Physics news on Phys.org
  • #2
You do not need the boundary conditions to check if a function is an eigenstate so no need to replace x by L/2 or -L/2. These limits only enter when you calculate the expectation values.


You must ask yourself if the function you get after applying [itex] i \hbar \partial_x [/itex] to [itex] \psi [/itex] is proportional to the wavefunction itself. If yes, then the wavefunction is an eigenstate and the constant of proportionality is the eigenvalue. If not, the function is not an eigenstate.
 
Last edited:
  • #3
So do I need to differnitiate it then?
 
  • #4
TFM said:
So do I need to differnitiate it then?

Yes. Sorry, there was a typo in my post and the partial derivative did not show up. Yes, you must apply [itex] i \hbar \partial_x [/itex] to [itex] \psi [/itex] and check if the result is proportional to the wavefunction itself. Is it possible to write [itex] \sin(\pi x/l) [/itex] as a constant times [itex] \cos (\pi x/L) [/itex] in such a way that the result will be valid for any value of x ? If the answer is no, then this wavefunction (the cos) is not an eigenstate of the momentum operator
 
  • #5
Okay so the original wavefunction is:

[tex] \psi(x) = \sqrt{\frac{2}{L}} cos(\frac{x\pi}{L}) [/tex] --- (1)

The [tex] \hat{p} [/tex] function is:

[tex] -i\hbar \sqrt{\frac{2}{L}} (-\frac{\pi}{L}sin\frac{x\pi}{L}) [/tex] Which can be simplified to:

[tex] i\hbar \sqrt{\frac{2}{L}} * \frac{\pi}{L}sin\frac{x\pi}{L} [/tex]

So now I have to see if this result is proportional to (1).

So I have to find how to write [tex] sin(\frac{\pi x}{L}) [/tex] as [tex] C cos(\frac{\pi x}{L}) [/tex]

Can this actually be done, though, since the only relation I know that will link sin and cos is [tex] sin^2(\theta) + cos^2(\theta) = 1 [/tex]?

Also, do I need to get rid of that i at the beginning, since p needs to be a real number?
 

What is Quantum Mechanics?

Quantum Mechanics is a branch of physics that deals with the behavior and interactions of particles on a microscopic scale, such as atoms and subatomic particles. It describes the fundamental laws that govern the behavior of these particles and their interactions with each other.

What are Eigenfunctions in Quantum Mechanics?

An Eigenfunction is a special type of function that represents a state or property of a quantum system. It is a solution to the Schrödinger equation, which describes the time evolution of a quantum system, and it represents the state of a system with a particular energy level.

What is the importance of Eigenfunctions in Quantum Mechanics?

Eigenfunctions play a crucial role in the mathematical formulation of quantum mechanics. They represent the possible states of a system and help us calculate the probabilities of a system being in a particular state. They also provide a basis for understanding the behavior of quantum systems and predicting their future states.

How are Eigenfunctions checked in Quantum Mechanics?

To check if a function is an Eigenfunction, we use the Eigenfunction Equation, which states that when a function is operated on by a linear operator, the result is equal to a constant times the original function. This constant is known as the Eigenvalue, and the function is an Eigenfunction if it satisfies this equation.

What are some real-world applications of Quantum Mechanics and Eigenfunctions?

Quantum Mechanics and Eigenfunctions have numerous applications in various fields, such as electronics, chemistry, and engineering. Some examples include the development of quantum computers, the design of new materials, and the understanding of chemical reactions at the molecular level.

Similar threads

  • Introductory Physics Homework Help
Replies
28
Views
352
  • Introductory Physics Homework Help
Replies
6
Views
272
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
555
Replies
8
Views
221
  • Introductory Physics Homework Help
Replies
11
Views
215
  • Introductory Physics Homework Help
Replies
5
Views
456
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
690
  • Introductory Physics Homework Help
Replies
6
Views
952
Back
Top