QM. eigenfunction help!

1. The problem statement, all variables and given/known data

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

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


3. 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.
 
Last edited:

malawi_glenn

Science Advisor
Homework Helper
4,783
22
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?
 
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
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
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'>
 
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

Science Advisor
Homework Helper
4,783
22
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.
 
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)
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
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.
 
Last edited:
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?
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
I would keep the constants k and c and then just normalize the wave function ѱ= (exp) ((2kx+ 2c -x2)/-2ih)
 
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)?
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
Yes it is also an eigenfunction of the momentum operator. Remember the linearity requirement for operators.
 
yeas, i get that part. its just the values of px and x for the eigenfunction(ѱ) of O thats not clear to me?
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
why is it hard to just evaluate them by operating on the wavefunction?

Give it a try
 
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?
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
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...
 
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??
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
It is still an eigenfunction.....
 
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??
 

malawi_glenn

Science Advisor
Homework Helper
4,783
22
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.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top