Linear combinations of non-eigenfunctions to create eigenfunctions

[mark.laidlaw19]

Homework Statement

Consider the Parity Operator, P', of a single variable function, defined as P'ψ(x)=P'(-x).

Let ψ1=(1+x)/(1+x^2) and ψ2=(1+x)/(1+x^2). I have already shown that these are not eigenfunctions of P'.

The question asks me to find what linear combinations, Θ=aψ1+bψ2 are eigenfunctions of P', and what are their eigenvalues.

Homework Equations

I have already done most of the calculations through this method: Taken P'Θ=P*Θ=P*(aψ1+bψ2) [P here is the eigenvalue], expanded and equated coefficients of the resultant polynomial.

This has given me: (a+b)(P-1)=0 and (P+1)(b-a)=0

So the solutions are a=-b, P=1 (from the first equation) and P=-1 and a=b (from the second).

The Attempt at a Solution

Now, I understand that when a=b, the linear combination will have even parity, and this should have an eigenvalue of P=1. What confuses me is that the equation that gives the a=b solution also gives P=-1, which corresponds to odd parity.

I feel like there is something simple I am missing, as I seem to have the right relations between a and b, and eigenvalues, but I've somehow mixed them up.

!Further more!, the following questions asks "what can we add to a given wave function, which is not an eigenfunction of P', to make parity eigenfunctions?". I feel this might be more obvious when I understand the above question, but my reasoning would be that we should add the complex conjugate of the wave function. This is because this would cancel out the imaginary part. Am I on the right track here?

I would greatly appreciate any guidance in making sense of my results.

Many thanks.

 

[vanhees71]

You should have two different wave functions. Otherwise you cannot find a linear combination that is a eigenfunction of the parity operator if the given function isn'd one already!

 

[mark.laidlaw19]

The two functions, Psi1 and Psi2, are not parity eigenfunctions, but a linear combination of them (for example, a=b=1, or a=1, b=-1) creates a function that has even and odd parities respectively. So, I'm not sure if I understand what you are saying, because in this way, you do get an eigenfunction of the operator as a result of a linear combination of two functions that are not eigenfunctions. I didn't mean to imply, for example, that Psi1 and Psi2 might be the basis vectors of the set of parity eigenfunctions. All that confuses me is why the a,b relations and the eigenvalues seem to come from opposite equations.

Thanks

 

[vela]

Homework Statement

Consider the Parity Operator, P', of a single variable function, defined as P'ψ(x)=P'(-x).

Let ψ1=(1+x)/(1+x^2) and ψ2=(1+x)/(1+x^2). I have already shown that these are not eigenfunctions of P'.

The question asks me to find what linear combinations, Θ=aψ1+bψ2 are eigenfunctions of P', and what are their eigenvalues.

Homework Equations

I have already done most of the calculations through this method: Taken P'Θ=P*Θ=P*(aψ1+bψ2) [P here is the eigenvalue], expanded and equated coefficients of the resultant polynomial.

This has given me: (a+b)(P-1)=0 and (P+1)(b-a)=0

So the solutions are a=-b, P=1 (from the first equation) and P=-1 and a=b (from the second).

The Attempt at a Solution

Now, I understand that when a=b, the linear combination will have even parity, and this should have an eigenvalue of P=1. What confuses me is that the equation that gives the a=b solution also gives P=-1, which corresponds to odd parity.

I feel like there is something simple I am missing, as I seem to have the right relations between a and b, and eigenvalues, but I've somehow mixed them up.

Both relationships need to be satisfied, but you can only choose $a$ and $b$ such that one will be satisfied at a time. To satisfy the other relationship, P has to take on the right value. For example, when $a=b$, the second relationship holds, but then you can only make the first equation hold if P=+1. In other words, $a=b$ yields an even parity solution.

!Further more!, the following questions asks "what can we add to a given wave function, which is not an eigenfunction of P', to make parity eigenfunctions?". I feel this might be more obvious when I understand the above question, but my reasoning would be that we should add the complex conjugate of the wave function. This is because this would cancel out the imaginary part. Am I on the right track here?

Kind of. It's not the complex conjugate that you want. What if your solutions are real? For example, suppose you have $\psi_1(x) = 1+x$ and $\psi_2(x) = 1-x$? How can you construct even and odd solutions as a linear combinations of those two functions?

I would greatly appreciate any guidance in making sense of my results.

Many thanks.

 

[mark.laidlaw19]

Thank you very much vela! I now understand the my first problem. I knew I just needed someone to explain it to me clearly.

For the second part, yes, I forgot to account for real wave functions. For Psi1 and Psi2, any linear combination that satisfies a=b or a=-b will give parity solutions. But what about any other non - eigenfunctions? I feel this is an important question because the wording of the questions seems to speak generally about any non-eigenfunction, not just Psi1 and Psi2.

If so, then my guess would be, add anything that has equal but negative coefficients of odd powers of x, or add anything that has equal but negative coefficients of even powers of x (and include x^0 as an even power);
e.g. for a non eigenfunction of 1(x^0)+4x^2-3x^3, this does not have any parity, and I've included x^0 to make my question clearer
But if we add -1(x^0)-4x^2-3x^3, then we get -6x^3, which has odd parity, and so it is an eigenfunction of the parity operator.
Same for adding 1(x^0)+4x^2+3x^3, as this gives 2(x^0)+8x^2, which has even parity.

If this is correct, how can I succinctly describe it? It is a similar idea to the conjugate, but I'm not sure what it is. And I'm not sure if this is a complete answer either.

Thank you again for your previous help, vela :)

 

[vela]

In the original post, $\psi_1$ and $\psi_2$ are identical. One is supposed to have 1-x in the numerator, right?

 

[mark.laidlaw19]

In the original post, $\psi_1$ and $\psi_2$ are identical. One is supposed to have 1-x in the numerator, right?

Yes, that was my mistake sorry.

 

[vela]

Did you note that $\psi_1(x) = \psi_2(-x)$?

 

[mark.laidlaw19]

Ahh, no I didn't, but thank you! I clearly wasn't looking at it hard enough.
So my conclusion from that observation would be just add the same function, but with an inversion of the argument sign.
I.e. If a general function f(x) is not an eigenfunction of the parity operator, adding f(-x) will form an eigenfunction.
I can see this is a much better and more complete and clever way of describing what i was mentioning before (with inverting the sign of the coefficients of the odd powers of x etc.) but makes much more sense geometrically and algebraicly!

Thank you for your help vela!