# Evaluating [C, D] commutator

1. Oct 15, 2014

### terp.asessed

1. The problem statement, all variables and given/known data
I am practicing problems from the textbook, but have no idea how to get to some of the solutions available in the back of the textbook....

6-16. Evaluate the commutator [C,D] where C and D are given below:

(e) C = d2/dx2, D = x
(g) C = integral (x = 0 to infinite) dx, D = d/dx

2. Relevant equations
[C,D] = CD - DC

3. The attempt at a solution
e)
CDf(x) = C(xf(x)) = d2xf(x)/dx2 = 0 + xd2f(x)/dx2
DCf(x) = D(d2f(x)/dx2) =xd2f(x)/dx2

[C,D] = xd2f(x)/dx2 - xd2f(x)/dx2 is what I got, but the solution, according to the textbook is:

[C,D] = 2df(x)/dx...how do I get to here?

g)
CDf(x) = C(df(x)/dx) = integral (x=0 to infinite) df(x)/dx * dx = f(x) (integration from x = 0 to infinite)
DCf(x) = D(integral f(x)) = d/dx integral f(x)*dx...which is what? I am lost here....

Again, for this part, the solution is:
[C,D]f(x) = -f(0)...could someone explain how this value was hinted?

2. Oct 15, 2014

### Simon Bridge

(e) CDf = (xf)'' ... you made an error in the next two steps.
(g) p1. cancel the dx's inside the integral: $\int_a^b df = ?$
p2. how does the definite integral result depend on x?

3. Oct 15, 2014

### RUber

For (e),
$(CD-DC)f(x)= \left[\frac{d^2}{dx^2}x - x\frac{d^2}{dx^2}\right] f(x)$
$= \frac{d^2}{dx^2}xf(x) - x\frac{d^2}{dx^2}f(x)$
By product rule $=\frac{d}{dx}(xf'(x)+f(x)) -xf''(x)$

4. Oct 15, 2014

### terp.asessed

(e) d2xf(x)/dx2 =d/dx(d/dx xf(x)) = d/dx (f(x) + xdf(x)/dx) = df(x)/dx +df(x)/dx + xd2f(x)/dx2
...so:
[C,D] = df(x)/dx +df(x)/dx + xd2f(x)/dx2 - xd2f(x)/dx2 = 2df(x)/dx---got it, thanks!

5. Oct 15, 2014

### terp.asessed

By the way, for (g),

As for (g)...so I guess I get integral (x = a to b) df = f (x= a to b) = b-a?

Also, by definite integral....could you clarify as to what does "definite integral" mean? An integral without any infinity?

6. Oct 15, 2014

### RUber

For (g), where you have $\frac{d}{dx} \int_0^\infty f(x) dx$, you can think of a definite integral as a constant. Every f has an associated $k_f=\int_0^\infty f(x) dx$. The fact that you are evaluating the interval at fixed points makes it definite. If the limit of integration were a function of x, you would have a different derivative of the integral.

7. Oct 15, 2014

### Simon Bridge

Well done - note: Leibnitx notation is somewhat eyewatering in text, it is usually clearer to write, say,
CDf=(xf)'' = (f+xf')' = f' + (xf')' = 2f' + xf''.

... but this is where LaTeX comes into it's own.
$$CDF = \frac{d^2}{dx^2}xf = \frac{d}{dx}\left(f + x\frac{df}{dx}\right)$$

8. Oct 15, 2014

### terp.asessed

Okay! I have yet to use LaTex...which I have not yet figured out yet.

Aside back to (g):

is ABf(x) = f(x) - f(0) , for as you have so said, "integral df(x)/dx * dx" cancels out dx, making integral df(x)...which I have trouble thinking about.
Also, for BAf(x)...d/dx integral f(x)*dx, dx is not cancelled out, so, I guess it would become something like f2(x)/2 over x=0 and x?

I am sorry, but I am really confused when there are no equations within the integral, which I am used to.

9. Oct 15, 2014

### RUber

$\int_0 ^\infty \frac {d}{dx} f(x) dx = \int_0 ^\infty f'(x) dx = \left. f(x) \right|_0^\infty$
Also you could look at
$\frac {d}{dx} \int_0 ^\infty f(x) dx = \lim_{x \to \infty } \frac {d}{dx} \int_0 ^x f(x) dx$

10. Oct 15, 2014

### Simon Bridge

Right ... but in your case the limits in the integral are for x, not f ... so how do you change variables in an integral?

Of course you could treat the integral as an antiderivative ....

11. Oct 15, 2014

### RUber

What do you mean by?
There are general rules for integrals.
For example:
$\int_a^b f'(x) dx = f(a)-f(b)$
$\frac{d}{dx} \int_a^x f(x) dx = f(x)$
You are not looking to evaluate f, you just need to know how the derivative and integral act upon eachother.

12. Oct 15, 2014

### Simon Bridge

You don't have to memorize more rules - you can just follow the definitions...

if $g(x)=f'(x)$, then $\int g(x)=f(x)+c$ is the indefinite integral;
then $\int_a^b g(x)\;dx = f(b)-f(a)$ is the definite integral.

putting b=x in the definite integral would give f(x)-f(a) so I can do:
$\frac{d}{dx}\int_a^x g(x')\;dx' = \frac{d}{dx}\big(f(x)-f(a)\big) = f'(x) = g(x)$​

Note: $\int_a^x g(x)\;dx$ is sometimes considered an abuse of notation.

If you just treat the definite integral as a constant, you'd end up with $$\int_0^\infty \frac{d}{dx}f(x)\;dx - \frac{d}{dx}\int_0^\infty f(x)\;dx = [f(\infty)-f(0)] - 0$$... so unless you know that $\lim_{x\to\infty}f(x)=$ some specific value, you should consider writing the integral this way:
$$\int_0^\infty f(x)\;dx = \lim_{x\to\infty}\int_0^xf(x')\;dx'$$ ... the limit notation is sort-of implied by the integration notation.

You should already have done a lot of work on limits.

13. Oct 15, 2014

### terp.asessed

GOTCHA--I finally remember--thank you very much. I figured out everything!