# Linear Combinations of Eigenfunctions

1. Oct 19, 2007

### Rahmuss

1. The problem statement, all variables and given/known data
Suppose that $$f(x)$$ and $$g(x)$$ are two eigenfunctions of an operator $$Q^{\wedge}$$, with the same eigenvalue $$q$$. Show that any linear combination of f and g is itself an eigenfunction of $$Q^{\wedge}$$, with eigenvalue $$q$$.

2. Relevant equations
I know that $$Q^{\wedge}f(x) = qf(x)$$ shows that $$f(x)$$ is an eigenfunction of $$Q^{\wedge}$$, with eigenvalue $$q$$. (As stated in Intro to QM 2nd Edition by Griffiths).

3. The attempt at a solution
I'm really not sure how to set this up, and that's my main problem. I starting writing:

$$h(x) = \sum a_{n} [f_{n}(x) + g_{n}(x)]$$

$$Q^{\wedge}h(x) = qh(x)$$

Here is about as far as I get because I'm not even sure I'm setting it up correctly, nor am I sure how I would begin to write out $$h(x)$$ in any kind of usable form for me to work with. I think it's mainly just a problem of setting it up. I can figure out where to lead it and how to do the basic math to manipulate it into what I want (I hope). I just can't figure out how to set it up. Any help would be appreciated.

2. Oct 19, 2007

### Dick

f and g are eigenfunctions of Q if Q(f)=q*f and Q(g)=q*g. You are now being asked to show if h=c1*f+c2*g, the Q(h)=q*h. I'm guessing you can assume Q is a linear operator.

3. Oct 19, 2007

### Coto

The key to the proof is the fact that Q is a linear operator, for then Q(f + g) = Q(f) + Q(g). Just play with that, and you'll get your proof.

4. Oct 20, 2007

### cks

I think you miss the keyword "any linear combination of f and g ",
this means (like what Dick told you)

h=c1f +c2g

and the question ask you to prove that Qh=qh, so just by substituting the h and manipulate it to get qh, then you prove it.

5. Oct 20, 2007

### Rahmuss

Dick, Coco and cks - Thank you all very much. You all seem to agree. So this problem sounds much easier than I first thought. I think they should instead have worded it 'Show that a linear combination of f and g is itself an eigenfunction of Q, with eigenvalue q.' Instead of "any" linear combination. I was thinking of a way to show that it worked with h= c1f + c2g, AND with h = c3f + c4g + c5f + c6g or any kind of random f and g with a coefficient, and I wasn't sure how to set that up.

So, do I need to prove that $$Q^{\wedge}$$ is linear? If not, then is this all I need?:

Part A
Let $$h = f + g$$. Show that $$Q^{\wedge}h(x) = qh(x)$$. $$Q^{\wedge}h(x) = qh(x)$$ --> $$Q^{\wedge}[f(x) + g(x)] = q[f(x) + g(x)]$$.

Part B
Then, as long as $$Q^{\wedge}$$ is a linear operator, then $$Q^{\wedge}f(x) + Q^{\wedge}g(x) = qf(x) + qg(x)$$.

Part C
And we know that $$Q^{\wedge}f(x) = qf(x)$$ and $$Q^{\wedge}g(x) = qg(x)$$.

Part D
So $$Q^{\wedge}[f(x) + g(x)] = g[f(x) + g(x)]$$ --> $$Q^{\wedge}h(x) = qh(x)$$.

Does that work? Or do I need to show that $$Q^{\wedge}$$ is a linear operator? And if so, then how would I do that?

6. Oct 20, 2007

### Coto

No need to show that Q is a linear operator. To show that, you would need to know what Q is. It is assumed in the quesiton.

7. Oct 20, 2007

### Rahmuss

Also, part b) (which also has to do with linear combinations of eigenfunctions) is now giving me problems.

1. The problem statement, all variables and given/known data
Check that $$f(x) = exp(x)$$ and $$g(x) = exp(-x)$$ are eigenfunctions of the operator $$\frac{d^{2}}{dx^{2}}$$, with the same eigenvalue. Construct two linear conbimations of $$f$$ and $$g$$ that are orthogonal eigenfunctions on the interval $$(-1, 1)$$.

2. Relevant equations
$$Q^{\wedge}f(x) = qf(x)$$ (not given; but I use it when maybe I shouldn't).

3. The attempt at a solution
For the first part of the problem I've done:

$$\frac{d^{2}}{dx^{2}}exp(x) = \lambda exp(x)$$ and $$\frac{d^{2}}{dx^{2}}exp(-x) = \lambda exp(-x)$$

--> $$exp(x) = \lambda exp(x)$$ and $$exp(-x) = \lambda exp(-x)$$

Which only happens when $$\lambda = 1$$.

For the second part of the problem I'm thinking of something like $$h_1 = f_{n} + g_{m}$$ and $$h_2 = f_{x} + g_{y}$$ where $$n\neq m$$ and $$x\neq y$$; but then I'm not sure what they mean about the interval $$(-1, 1)$$.

Any thoughts?

8. Oct 20, 2007

### Coto

The first part is fine. Let Q = d/dx^2 , and find that indeed, Q is a linear operator. The second part needs you to find two orthogonal eigenfunctions:

Orthogonality =>

$$\int_{-1}^{+1} [h_1(x)*h_2(x)] dx = 0$$

where h_1 and h_2 are two functions that are linear combinations of f(x) and g(x).

The most general way to do this (IMO) is to start by defining two new functions, h_1 and h_2, as general linear combinations of f and g (i.e. use a, b, c, and d as coefficients.). Then proceed in the integration, and determine what coefficients you require such that the integral is zero.

Last edited: Oct 20, 2007
9. Oct 20, 2007

### Rahmuss

Coto - Oh, ok, so again I mis-understood what was meant. The math usually isn't too big of a deal (although I sometimes have to look up info), it's really just understanding what in the world they're talking about. So here is my attempt:

Part A
Let $$h_{n} = Af + Bg$$ and $$h_{m} = Cf + Dg$$ where $$n \neq m$$ for orthogonality.

Part B
$$\int^{1}_{-1}h_{n}^{*} h_{m} dx = 0$$ --> $$\int^{1}_{-1}(Af + Bg)^{*} (Cf + Dg) dx = 0$$

Part C
$$\int^{1}_{-1}Af^{*} Cf dx + \int^{1}_{-1}Bg^{*} Dg dx = 0$$ --> $$AC \int^{1}_{-1}f^{*} f dx + BD \int^{1}_{-1}g^{*} g dx = 0$$

Part D
$$\int^{1}_{-1}f^{*} f dx = 1$$ and $$\int^{1}_{-1}g^{*} g dx = 1$$

Part E
$$AC + BD = 0$$ Let $$AC = E$$ and $$BD = F$$ --> $$E = -F$$.
Which shows at least that the (new) coefficients are equal and opposite. Is that kind of what we're looking for?

Last edited: Oct 20, 2007
10. Oct 21, 2007

### cks

You have proved it right. but $$\int_{-1}^1 f*f dx$$ isn't 1 because it's not normalized yet.

11. Oct 21, 2007

### cks

You can do it in a general way.

$$\frac{d^2\phi}{dx^2} = -k^2\phi$$

with -k^2 as the eigenvalues.

$$\phi = Aexp(ikx) +Bexp(-ikx)$$

by substituting k=i , you get the two eigenfunctions.

and -(i)^2 = 1 show that they have the same eigenvalues.

12. Oct 21, 2007

### Rahmuss

cks - Oh, ok, right. So it should actually be:

$$AC \int ^{1}_{-1}f^{*} f dx = 1$$ and $$BD \int ^{1}_{-1}g^{*} g dx = 1$$.

Is that right? Wait, if so, then we end up with $$1 + 1 = 0$$??? I guess I'm really confused on this topic then. I'm not sure what's going on.

13. Oct 21, 2007

### cks

From your first part, you have already shown that f=exp(x) and g=exp(-x)

$$\int_{-1}^1 f*f dx = \int_{-1}^1 exp(2x) dx = \frac{1}{2}[exp(2)-exp(-2)]$$

Similarly for

$$\int_{-1}^1 g*g dx=-\frac{1}{2}[exp(-2)-exp(2)]$$

$$AC \int^{1}_{-1}f^{*} f dx + BD \int^{1}_{-1}g^{*} g dx +AD\int fg dx +BC\int gf dx= 0$$

$$\frac{1}{2}(AC+BD)exp(2)-\frac{1}{2}(AC+BD)exp(-2) + 2(AD+BC)=0$$

$$(AC+BD)\frac{exp(2)-exp(-2)}{2} +2(AD+BC)=0$$

EDIT: I MADE A MISTAKE AND IT'S CORRECTED NOW. $$\int fg dx$$ is not zero.

Last edited: Oct 22, 2007
14. Oct 21, 2007

### cks

I'm afraid I might make some mistakes. if other users see my mistakes, please correct me also.

I'm here to share what I know, I'm not here to teach.

15. Oct 21, 2007

### cks

Why you say $$\int^{1}_{-1}f^{*} f dx = 1$$ and $$BD \int ^{1}_{-1}g^{*} g dx = 1$$

are equal to one.?

16. Oct 21, 2007

### cks

The normalized eigenfunction of f is not exp(x) but

$$\frac{exp(x)}{[\frac{1}{2}[exp(2)-exp(-2)]]^{1/2}}$$

so if you integrate the square of the above function, then it's equal to one.

Last edited: Oct 21, 2007
17. Oct 21, 2007

### Coto

The problem lies with your initial equations.

$$\int_{-1}^{+1} [(Af(x) + Bg(x))*(Cf(x) + Dg(x))] dx$$

is what you're starting with. Now separating this out, you've assumed that f(x) and g(x) are orthogonal, so you've automatically set these integrals to zero, but this is not the case at all, in fact f*g = 1, and the integral of that between -1 and 1 is 2, not zero.

Start by solving the entire equation, plug in your f(x) and g(x), solve the integrals, and come up with a system of equations for your constants. Essentially you'll end up with 2 equations for 4 constants, however you'll notice that you only require one solution of the problem, i.e. any set of 4 constants that make the above integral zero will do. In the end you'll find that A = -B, and C=D are one such solution, and so using this, we find that:
h1(x) = A(f - g) and h2(x) = C(f + g)
are orthogonal on the interval -1 and 1.

EDIT: Also with what cks is saying. f*f, g*g are not normalized, nor are you seeking to normalize them.

Last edited: Oct 21, 2007
18. Oct 22, 2007

### Rahmuss

cks - I guess I was treating the first part (defing f and g) as separate from the second part (constructing two functions). The reason I set the integrals to 1 was because it seemed as if you were saying I needed to normalize them and to normalize them I set the integral WITH the coefficients this time, equal to 1. I guess that's not what you were trying to convey.

Coto - You're right. I was setting $$f g = 0 = g f$$. My mistake. I take back what I said about knowing how to do math. :D Here is what I have so far:

$$\int^{1}_{-1} (Af + Bg)^{*} (Cf + Dg) dx = 0$$

$$\int^{1}_{-1} AC(f^{*}f) + BC(g^{*}f) + AD(f^{*}g) + BD(g^{*}g) dx = 0$$

$$AC\int^{1}_{-1} (f^{*}f) + BC\int^{1}_{-1} (g^{*}f) + AD\int^{1}_{-1} (f^{*}g) + BD\int^{1}_{-1} (g^{*}g) dx = 0$$

$$BC\int^{1}_{-1} (g^{*}f) dx = 2BC$$ and $$AD\int^{1}_{-1} (f^{*}g) dx = 2AD$$

$$\frac{AC}{2}\exp ^{2x}|^{1}_{-1} + \frac{-BD}{2}\exp ^{-2x}|^{1}_{-1} + 2BC + 2AD = 0$$

$$(AC + BD)(\frac{e^{2} - e^{-2}}{2}) + 2BC + 2AD = 0$$

So now, which two equations should I have? Do you mean with the $$h_{m} = (Af + Bg)$$ and $$h_{n} = (Cf + Dg)$$?

19. Oct 22, 2007

### Coto

Rahmuss, you have to do some of this problem alone, so you're on your own from here on out. You pretty much have the solution, just think about the problem. It is quite literally trivial from here to get your 2 equations that govern the coefficents (Grade 9 math?).

20. Oct 22, 2007

### Rahmuss

Coto - I failed grade 9 math. :( JK. Ok, I'll try and get it from here. My homework is due tomorrow, so I'll be done then one way or another. :D

Thanks again for the help to you and everyone else.