Computing Eigenfunctions for a One-Dimensional Operator

[NepToon]

Homework Statement

find eigenfunction of the the given operator. compute the vaule of p_x and x for this eigenfunction

Homework Equations

O= p + x
O is made up by adding the momentum operator(p) and the position operator(x); it is one dimensional

The Attempt at a Solution

i tried doing Oѱ=ѱp+ѱx=kѱ , h'=h bar
or, -ih' *(d/dx ѱ) + x * ѱ= kѱ
or, k = -ih' *(d/dx ln ѱ) + x =k
or, kx+c=-ih' * ln ѱ +x_{2 /2 [after integrating]
and then solved for ѱ and i am stuck.}

 

[malawi_glenn]

Well this one is ok:

-ih' *(d/dx ln ѱ) + x =k

which give you:

(d/dx ln ѱ) = (i/hbar)(k-x)

the integration is also good:

kx+c=-ih' * ln kx+c=-ih' * ln ѱ +x<sub>2 /2 +x^2 /2

Why you got stucked at solving for ѱ? What did you try? How can we help you if you not tell us what you tried?</sub>

 

[NepToon]

after doing all that i get an equation for ѱ, that look slike
ѱ= (exp) ((2kx+ 2c -x²)/-2ih)
i do have an equation but its got 2 unknown k and c, how do i proceed from here to find the eigenfunciton and the values of p_x and x for the eigenfunction

 

[malawi_glenn]

k and c CAN be arbitrary constants, they are determined by normalization condition.

The eigenvalue of position operator is x|x'> = x'|x'>

The eigenvalue of momentum operator is p|x'> = -ihbar(d/dx')|x'>

 

[NepToon]

so does that mean i just take the two constants out and get an equation for the eigenfunction that looks like:
ѱ= (exp) ((2kx+ 2c -x²)/-2ih) without the k and c,
so,
ѱ= (exp) ((2x -x²)/-2ih)


and is x|x'> = x'|x'> mean x divided by x' where x' is the derivative? or is it just performing the eigenfunction(ѱ) on x?

 

[malawi_glenn]

nono sorry, i used the notation of sakurai. This is better:

\hat{x}\psi(x) = x\psi(x)

\hat{p} \psi (x) = -i\hbar\dfrac{d}{dx}\psi(x)

hat- is operator.

Should not have introduced that sloppy notation for a beginner of Qm.. sorryy

There is not ONE unique eigenfunction to this operator, the constants are arbitrary.

 

[NepToon]

yeah. i am sort of a beginner. may i bother u once more. i am asked to normalize the eigenfunction (ѱ) after finding out what it is; if there are no particular solutions for it, how do i normalize it for a given range like say (-L to L)? is it logical? and if i can't then how do i find p_x and x for the eigenfuntion(ѱ)

 

[malawi_glenn]

In general, you normalize from -infinity to +infinity if nothing else is stated.

You would have normalized to -L to L if you were given this wavefunction for instance:

Psi (X) = psi(x) if -L < x < L
Psi (X) = 0 elsewhere

A hint regarding that normalization, your obtained function is a Gaussian function.

 

[NepToon]

so normalize the equation for the eigenfunctin ѱ of O(with a hat)=p+x(both with hats) leaving the two constants k and c?? on the equation ѱ= (exp) ((2kx+ 2c -x2)/-2ih).
and p_x and x for the eigenfunction (ѱ) of O(with a hat) is the same thing as the eigenfunction of x on p_x and x on x?

 

[malawi_glenn]

I would keep the constants k and c and then just normalize the wave function ѱ= (exp) ((2kx+ 2c -x2)/-2ih)

 

[NepToon]

nono sorry, i used the notation of sakurai. This is better:

\hat{x}\psi(x) = x\psi(x)

\hat{p} \psi (x) = -i\hbar\dfrac{d}{dx}\psi(x)

hat- is operator.

Should not have introduced that sloppy notation for a beginner of Qm.. sorryy

There is not ONE unique eigenfunction to this operator, the constants are arbitrary.

okay. so after all this. does it mean that the eigen functino of ѱ is also the eigenfunctin of the momentum operator as p_x?
refering to post number 6, and when trying to find the value of p_x and x for ѱ(the eigenfunctin of O), why are we just executing the eigenvalue on the momentum and position operator, rather than plugging it on the eigenfunction(ѱ)?

 

[malawi_glenn]

Yes it is also an eigenfunction of the momentum operator. Remember the linearity requirement for operators.

 

[NepToon]

yeas, i get that part. its just the values of p_x and x for the eigenfunction(ѱ) of O that's not clear to me?

 

[malawi_glenn]

why is it hard to just evaluate them by operating on the wavefunction?

Give it a try

 

[NepToon]

yea, i was about to get to that; to find the value of x in ѱ, i just operate for x; but what do i do for p_x; i can solve of x² but what about p_x²??
can ѱ represent any physical state, if at all?

 

[malawi_glenn]

so you don't know how to take a derivative of a general function?

<br /> \hat{p} \psi (x) = -i\hbar\dfrac{d}{dx}\psi(x) <br />

You will find out that the momentum is the generator of translations..

now just do this, don't think so much, be axiomatic.

Yes this is a Gaussian wavefunction and have many nice properties.

"but what do i do for px; i can solve of x2 but what about px2"
I can't understand what you are thinking about here.. just evaluate the derivative...

 

[NepToon]

i menat px^2, px raised to the power of two? when finding p_x for ѱ, i first operate x on the momentum operator and then operate the result on ѱ??

 

[malawi_glenn]

It is still an eigenfunction...

 

[NepToon]

so if it were p_xⁿ then i would just apply the momentum function on any given function n times, and plug it in for on ѱ to find the value of p_xⁿ for ѱ??

 

[malawi_glenn]

what is the "momentum function"? and "plug it in"?

But yes, if you want to evaluate the eigenvaule of the momentum operator to the n-th power for some wavefunction in position space, you take the n-th derivative with respect to x and then multiply it with (-i\hbar) to the n-th power.

Now that has nothing to do with the normalization.

 

[NepToon]

yea, i wasn't sure about raising it to the nth power. thns for that. well i hope i won't have anymore questions. thnks a lot for the time and effort. its really going to help.

 

[malawi_glenn]

I can give you this nice excerice, if you want.

Given the wavefuction \psi(x) = (1/(\pi ^{1/4}\sqrt{d}))\exp (ikx - x^2/(2d^2))

Evaluate the expectation values of x, x^2, p and p^2

This wavefunction minimizes the Heisenberg Uncertainty principle:
\Delta x \Delta p = \hbar / 2

 

[NepToon]

I can give you this nice excerice, if you want.

Given the wavefuction \psi(x) = (1/(\pi ^{1/4}\sqrt{d}))\exp (ikx - x^2/(2d^2))

Evaluate the expectation values of x, x^2, p and p^2

This wavefunction minimizes the Heisenberg Uncertainty principle:
\Delta x \Delta p = \hbar / 2

yea. how do i evaluate the values of x for any given eigenfunction. like the one u've mentioned

 

[malawi_glenn]

you mean the eigenvaule of x-hat on psi(x)?

It is simple!

x-hat Psi(x) = x Psi(x)

You did that thing when you did the operator:
O = p + x

Don't you remember?

What are you asking?

 

[NepToon]

for the exercise you gave me in post 22, how do i evaluate the expected value of x, x^2 and so on for any given wave function.

 

[malawi_glenn]

click on that hyperlink: expected value

Or wait until it is introduced in class, I thought you have done that.

 

[NepToon]

what hyperlink. is it hard? i might understand if u give me the instruction on here.

 

[malawi_glenn]

In your post #25, there was a hyperlink created for the library item

https://www.physicsforums.com/library.php?do=view_item&itemid=206

It is very easy, we learned operators, eigenfunction, expectation values etc, in one lecture I recall..

This should be introduced to you next in your class I think.

The example I gave you will learn you to master the operators and taking integrals with Gaussian functions.

 

[NepToon]

I can give you this nice excerice, if you want.

Given the wavefuction \psi(x) = (1/(\pi ^{1/4}\sqrt{d}))\exp (ikx - x^2/(2d^2))

Evaluate the expectation values of x, x^2, p and p^2

This wavefunction minimizes the Heisenberg Uncertainty principle:
\Delta x \Delta p = \hbar / 2

hey glenn, do u mind evaluating for x and p^2 for me, showing the steps. so i can learn how to do it. i went through the library item and didnt quite get the end.
to find x do i integrate x|ѱ(x)²| ?? what does ѱ(x)² mean??

 

[malawi_glenn]

you mean a physical interpretation or what it is mathematically?

Since ѱ is a complex valued function, in order to get something with is positive and real, one has to take the MODULUS square. Now for a complex number a, a* means that you take the complex conjugate of a.

a = Re(a) + Im(a)
a* = Re(a) - Im(a)

So, |a|^2 means a* a

I thought complex number algebra was a pre requirement for Quantum Mechanics.. what kind of Quantum Class do you attend? :P I mean, no book :P

That was the mathematical explanation what |ѱ(x)|^2 is. The physical is that |ѱ(x)| ^2 is the probability DENSITY. Just as you have in mathematical statistics, which is also an requirement for quantum mechanics.

<x> = int psi* x psi dx

and

<p^2> = (-i hbar)^" int psi* (d/dx)^2 psi dx