QM. eigenfunction help!

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Homework Statement



find eigenfunction of the the given operator. compute the vaule of px 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 +x2 /2 [after integrating]
and then solved for ѱ and i am stuck.
 
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Answers and Replies

  • #2
malawi_glenn
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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 +x2 /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?
 
  • #3
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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 px and x for the eigenfunction
 
  • #4
malawi_glenn
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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'>
 
  • #5
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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?
 
  • #6
malawi_glenn
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nono sorry, i used the notation of sakurai. This is better:

[tex]\hat{x}\psi(x) = x\psi(x)[/tex]

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

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.
 
  • #7
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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 cant then how do i find px and x for the eigenfuntion(ѱ)
 
  • #8
malawi_glenn
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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.
 
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  • #9
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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 px and x for the eigenfunction (ѱ) of O(with a hat) is the same thing as the eigenfunction of x on px and x on x?
 
  • #10
malawi_glenn
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I would keep the constants k and c and then just normalize the wave function ѱ= (exp) ((2kx+ 2c -x2)/-2ih)
 
  • #11
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nono sorry, i used the notation of sakurai. This is better:

[tex]\hat{x}\psi(x) = x\psi(x)[/tex]

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

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 px?
refering to post number 6, and when trying to find the value of px 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)?
 
  • #12
malawi_glenn
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Yes it is also an eigenfunction of the momentum operator. Remember the linearity requirement for operators.
 
  • #13
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yeas, i get that part. its just the values of px and x for the eigenfunction(ѱ) of O thats not clear to me?
 
  • #14
malawi_glenn
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why is it hard to just evaluate them by operating on the wavefunction?

Give it a try
 
  • #15
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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 px; i can solve of x2 but what about px2??
can ѱ represent any physical state, if at all?
 
  • #16
malawi_glenn
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so you don't know how to take a derivative of a general function?

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

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...
 
  • #17
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i menat px^2, px raised to the power of two? when finding px for ѱ, i first operate x on the momentum operator and then operate the result on ѱ??
 
  • #18
malawi_glenn
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It is still an eigenfunction.....
 
  • #19
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so if it were pxn 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 pxn for ѱ??
 
  • #20
malawi_glenn
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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.
 
  • #21
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yea, i wasn't sure about raising it to the nth power. thns for that. well i hope i wont have anymore questions. thnks a lot for the time and effort. its really going to help.
 
  • #22
malawi_glenn
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I can give you this nice excerice, if you want.

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

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

This wavefunction minimizes the Heisenberg Uncertainty principle:
[tex]\Delta x \Delta p = \hbar / 2[/tex]
 
  • #23
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I can give you this nice excerice, if you want.

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

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

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

yea. how do i evaluate the values of x for any given eigenfunction. like the one u've mentioned
 
  • #24
malawi_glenn
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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?
 
  • #25
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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.
 

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