Finding eigenfunctions of the linear momentum operator

In summary, the conversation discusses the concept of eigenfunctions and eigenvectors in quantum mechanics. Specifically, the conversation focuses on determining whether a given function is an eigenfunction of the linear momentum operator. The process involves applying the operator to the function and seeing if it returns the same function multiplied by a constant. If it does, the function is an eigenfunction. The conversation also clarifies the meaning of the imaginary number, i, and highlights the difference between discrete and continuous eigenvalues. Ultimately, the conclusion is reached that the given function is not an eigenfunction of the linear momentum operator.
  • #1
t_n_p
595
0

Homework Statement



http://img508.imageshack.us/img508/7199/46168034nt3.jpg

The Attempt at a Solution



I'm just totally lost with this question. The theory just eludes me totally. Just how do you determine whether it is/isn't an eighenfunction of the linear momentum operator?
 
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  • #2
t_n_p said:

Homework Statement



http://img508.imageshack.us/img508/7199/46168034nt3.jpg

The Attempt at a Solution



I'm just totally lost with this question. The theory just eludes me totally. Just how do you determine whether it is/isn't an eighenfunction of the linear momentum operator?

A function psi is an eigenfunction of an operator A if applying A on psi gives the function back times some constant (and the constant is then the correspodning eigenvalue of that eigenfunction).

So apply the operator [tex] p_x [/tex] on that wavefunction and see if the result is a constant times the initial wavefunction
 
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  • #3
In other words, one of the postulates of QM relates observables by eigenfunctions and eigenvectors. For example

[tex]\hat{H}\psi = E \psi[/tex]

would be the hamiltonian operator giving energy eigenvalues. You could also have

[tex]\hat{x_o}\psi = x_n \psi[/tex]

or more relevantly

[tex]\hat{p} \psi = p_n \psi[/tex]

In the end, do exactly what kdv told you, see if when you apply the operator you get the function back with a constant.
 
  • #4
pardon my lack of knowledge, but what is the linear momentum operator?
 
  • #5
t_n_p said:
pardon my lack of knowledge, but what is the linear momentum operator?
The linear momentum operator is a partial differential operator and in three dimensions has the form,

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

And in one dimension,

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

So for example, in 3D

[tex]\hat{p}\left(xyz\right) = -i\hbar\left(yz,xz,xy\right)[/tex]

So as both Mindscrape & kdv have said, all you need to do is apply the operator to the wave function and see if you get the wave function (eigenfunction) multiplied by a constant (eigenvalue).
 
  • #6
Ok, so I want to be using the one dimensional form. Pardon my lack of knowledge again, but what do you mean by apply?
 
  • #7
t_n_p said:
Ok, so I want to be using the one dimensional form. Pardon my lack of knowledge again, but what do you mean by apply?

it means take the derivative with respect to x of the wavefunction then multiply by [itex] -i
hbar [/itex]
 
  • #8
oh no, I havn't done calculus for a while...

Do I just treat A and i as constants?

If yes, by chain rule I get

http://img299.imageshack.us/img299/4323/15423623yp4.jpg

Then I multiple by i*h/2pi (although it's probably wrong, worth a try though)...?
 
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  • #9
Yes, you have taken the derivative correctly, now just multiply it by [itex]-i\hbar[/itex] and see if you can arrange it into the form,

[tex]\hat{p}\psi = p\psi[/tex]

Where p is constant. If you can, then [itex]\psi[/itex] is a linear eigenfunction of [itex]\hat{p}[/itex]; if not, then [itex]\psi[/itex] is not a linear eigenfunction of [itex]\hat{p}[/itex].
 
  • #10
Just one question, is i the imaginary number? i.e. i^2=-1?
 
  • #11
t_n_p said:
Just one question, is i the imaginary number? i.e. i^2=-1?

that is correct! :-)
 
  • #12
Sorry, I was looking at my help and I shouldn't have put [itex]p_n[/itex] as eigenvalues because that would imply they are discrete, but really the momentum eigenvalues are continuous, as well as the position eigenvalues. I should have distinguished with maybe capitals as operators and lower case as eigenvalues.

[tex]
\hat{X}\psi = x \psi
[/tex]

[tex]
\hat{P}\psi = p \psi
[/tex]

We should really be talking about an eigenket, or state vector, that lives in Hilbert space (or its suburbs), but it would be bad to burden you with technicalities, so don't worry about it unless you want to.
 
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  • #13
Ok, so after all that I end up with...

http://img258.imageshack.us/img258/6994/20815713op7.jpg

So my constant is i*(hbar)*x + (hbar) ?
 
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  • #14
Hold on there, is x a constant?
 
  • #15
In that case, it isn't an eigenfunction of the linear momentum operator?
 
  • #16
Does it satisfy the postulate?
 
  • #17
"A function psi is an eigenfunction of an operator A if applying A on psi gives the function back times some constant (and the constant is then the correspodning eigenvalue of that eigenfunction)."

x is not a constant, therefore applying the linear momentum operator on psi does not give the wavelength function back times some constant.

Therefore, psi is not an eigenfuction of the linear momentum operator.

(Well, that's what I think anyway...)
 
  • #18
Sounds good to me :approve:
 
  • #19
Excellent! Thanks
 

1. What is the linear momentum operator?

The linear momentum operator is a mathematical operator used in quantum mechanics to represent the momentum of a particle. It is denoted by the symbol p and it is defined as the derivative of the position operator with respect to time.

2. Why is it important to find eigenfunctions of the linear momentum operator?

Finding eigenfunctions of the linear momentum operator is important because they represent the possible states or energy levels of a particle's momentum. These eigenfunctions can then be used to calculate the probability of a particle having a certain momentum value.

3. How do you find eigenfunctions of the linear momentum operator?

To find eigenfunctions of the linear momentum operator, you need to solve the Schrödinger equation, which is a differential equation that describes the behavior of quantum systems. The solution to this equation will give you the eigenfunctions of the linear momentum operator.

4. What are the properties of eigenfunctions of the linear momentum operator?

The properties of eigenfunctions of the linear momentum operator include being orthogonal to each other, meaning they are perpendicular in function space, and they form a complete set, meaning any other function can be expressed as a linear combination of these eigenfunctions.

5. How are eigenfunctions of the linear momentum operator used in quantum mechanics?

Eigenfunctions of the linear momentum operator are used in quantum mechanics to calculate the probability of a particle having a certain momentum value, as well as to describe the energy levels and states of a particle's momentum. They are also used in various mathematical operations, such as calculating expectation values and solving the time-dependent Schrödinger equation.

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