Linear Operators and Eigen Values

[Nezva]

I'm looking for a good website for understanding Quantum Mechanics (i.e. Time Independent Schrodinger Eq'n, Harmonic Oscillators, Rigid Rotors, etc)

The operator is linear if the following is satisfied:
A[c*f(x)+d*g(x)]=c*A[f(x)]+d*A[fg(x)], where A = an operator of any kind

I'm having trouble coming up with cases where they are NOT equal to each other... Maybe I am doing the math incorrectly. As examples we were told to determine if the following operators were linear and if so then what their eigenvalues are:

a) the first derivative operator (d/dx)
b) the natural log operator (ln)
c) the doubling operator (2*)

Maybe my math is wrong but they all come out linear. I'm struggling with the "A[c*f(x)+d*g(x)]" portion of the equation. I've been plugging in simple equations to test their linearity and I'm simply lost. Any guidance would be greatly appreciated.

 

[radou]

Are you sure the natural logarithm operator is linear? Take, for example, f(x) = g(x) = 1, for all positive x, and c = d = 1 (for simplicity).

 

[jdwood983]

I'm looking for a good website for understanding Quantum Mechanics (i.e. Time Independent Schrodinger Eq'n, Harmonic Oscillators, Rigid Rotors, etc)

I don't think websites are the best resource, but here is one of the better ones I have found: Lecture notes from University of Virginia. You can view them in HTML format or download them in PDF format--the latter requires Adobe Reader or similar PDF viewers.

I recommend JJ Sakurai's Modern Quantum Mechanics textbook in lieu of a website. There are other texts out there (Griffiths, Townsend, Shankar, etc) that are good, but I'd recommend Sakurai's first.

 

[Nezva]

Wow, I did a lot of work for it to be disproven by $$ln(1+1)\neq ln(1)+ln(1)$$
*sigh* Thank you very much.

Also thank you for the book recommendation I will try to find a copy of it.

 

[Nezva]

How do I get eigenvalues out of d/dx operator? It seems like the eigenvalues are specific to the function being used...

 

[Dick]

f(x) is an eigenvector if df(x)/dx=c*f(x) with c a constant. Sure, how you find eigenvalues and eigenvectors depends on the operator. In this case you solve a differential equation.

 

[Nezva]

Part of the question is to identify eigenvalues and eigenfunctions of the example.

In d/dx operator, can d/dx be the eigenvalue... (The constant in front of the function after being operated on, $$\lambda$$ some books call it.) Depending on the style of function this lambda can be very different (i.e. ln[x], x^2, cos(x), etc.). However if you use d/dx itself as the eigenvalue is magically fits... Is this an incorrect form of an eigenvalue? Is my thinking off for this answer?

I assumed that the d/dx operator is linear due to the fact that the sum of the separated derivatives is equal to the total sum of all derivatives.

 

[Anti-Meson]

Nezva, any eigenvalue problem where the operator operating on a function does not yield a single value and the function then the function is said not to be an eigenfunction of that operator.

 

[Dick]

Part of the question is to identify eigenvalues and eigenfunctions of the example.

In d/dx operator, can d/dx be the eigenvalue... (The constant in front of the function after being operated on, $$\lambda$$ some books call it.) Depending on the style of function this lambda can be very different (i.e. ln[x], x^2, cos(x), etc.). However if you use d/dx itself as the eigenvalue is magically fits... Is this an incorrect form of an eigenvalue? Is my thinking off for this answer?

I assumed that the d/dx operator is linear due to the fact that the sum of the separated derivatives is equal to the total sum of all derivatives.

The eigenvalue is a NUMBER. d/dx is a OPERATOR. I've already told you you need to solve the differential equation df(x)/dx=c*f(x). Why don't you do that?

 

[Nezva]

Because there is no f(x), these are more like 'theory' problems. They address the general problem "Is this operator linear" In 2's case, obviously. In ln's case no. However in d/dx's case the eigenvalue can vary from function to function, implying it is not linear (c = not constant)? Since no functions are given, how can I solve "df(x)/dx=c*f(x)" in general terms. c=[tex]df(x)/dx[f(x)] ?

 

[Dick]

d/dx is linear. To know what the eigenvalues are you need to solve for them. That means solve df/dx=c*f. It's a separable ordinary differential equation. You must have solved something like that before.

 

[Nezva]

if $$f(x)=sin(x)$$
the deriv is $$cos(x)$$

So, how do you find a constant that satisfies the statement:
$$dsin(x)/dx = c*sin(x) = cos(x)$$

Even if you can manipulate that to get a constant it will not be the same constant as if the function was $$f(x)=ln(x)$$

If I knew exactly what the problem I'm having was, I wouldn't be asking...

 

[Dick]

df/dx=c*f. Separate the variables. df/f=c*dx. Integrate both sides. ln(f)=c*x+C. Exponentiate both sides. f(x)=e^(c*x+C), or f(x)=D*e^(c*x) where D=e^C is an arbitrary constant. Those are your eigenfunctions. They are exponential functions. What are the possible eigenvalues?

 

[vela]

if $$f(x)=sin(x)$$
the deriv is $$cos(x)$$

So, how do you find a constant that satisfies the statement:
$$dsin(x)/dx = c*sin(x) = cos(x)$$

You can't because sin x isn't an eigenfunction of d/dx. Only specific functions are eigenfunctions of an operator, just as only certain vectors are eigenvectors of a given matrix. To find these functions, you do as others have told you to do: you solve the differential equation.

If you haven't learned how to solve differential equations yet, you need to do so before you can have even a hope of understanding quantum mechanics.

 

[Nezva]

Every derivative of e^(cx) spits out a c, so the eigenvalue is c?

 

[Nezva]

You can't because sin x isn't an eigenfunction of d/dx. Only specific functions are eigenfunctions of an operator, just as only certain vectors are eigenvectors of a given matrix. To find these functions, you do as others have told you to do: you solve the differential equation.

If you haven't learned how to solve differential equations yet, you need to do so before you can have even a hope of understanding quantum mechanics.

That makes sense! The eigenfunctions being specific to the operator part. I'm taking Physical Chemistry, we're barely touching the surface of QM and this is as far as my math needs to go. Though I am thinking of taking Cal 2 over again, maybe Cal 1...