# Finding an Operator .

1. Oct 7, 2014

### terp.asessed

Hello--I am practicing for the upcoming quiz (as part of quantum physics), but have no idea how to solve the sample problem the teacher gave (w/ solution)...

Write out the operator A2 for A = d/dx + x
(Hint: INCLUDE f(x) before carrying out the operations)

so...I tried:

A {df(x)/dx + f(x)} = d/dx{df(x)/dx + f(x)} + df(x)/dx + f(x)
= d2f(x)/dx2 + df(x)/dx + df(x)/dx + f(x)
= d2f(x)/dx2 + 2df(x)/dx + f(x)

...when the solution is A[SUP]2[/SUP] = d2/dx2 + 2xd/dx + 1 + x2

...where did I make mistake?????? Could someone please point out? Thanks!

2. Oct 8, 2014

### wle

Here, right at the beginning:
You've expanded $A f$ as $\frac{\mathrm{d}}{\mathrm{d}x} f + f$. It should be $A f = \frac{\mathrm{d}}{\mathrm{d}x} f + x f$.

3. Oct 8, 2014

### terp.asessed

Thanks! But, I still have trouble getting all the solution values:

A f(x) = df(x)/dx + xf(x)
= d (df(x)/dx + xf(x))/dx + x df(x)/dx + x2f(x)
= d2f(x)/dx2 +x df(x)/dx + x df(x)/dx + x2f(x)
= d2f(x)/dx2 +2x df(x)/dx + x2f(x).......I could NOT get (+1) part from the solution: A2 = d2/dx2 + 2xd/dx + 1 + x2

...How do I get (+1) value? Did I leave out some values while I tried to solve for A2?

4. Oct 8, 2014

### Staff: Mentor

You didn't apply the first $\frac{d}{dx}$ correctly between the last two lines.

5. Oct 8, 2014

### terp.asessed

Wait, thank you for correcting my first mistake, but I still don't see how I got the last two lines wrong.....

6. Oct 8, 2014

### terp.asessed

Thank you for correcting my first mistake, forgetting to add A2 f(x), but I still cannot see how I got last two lines wrong

.....could you pls clarify more? Thanks.

7. Oct 8, 2014

### Staff: Mentor

You should get two terms out of
$$\frac{d}{dx} \left( x f(x) \right)$$

8. Oct 8, 2014

### terp.asessed

So, I should have gotten d(xf(x))dx, NOT x df(x)/dx?

d(xf(x))dx = xdf(x)/dx + df(x)/dx?

I redid from the beginning, but I still cannot get (+1) part:

A2f(x) = d2f(x)/dx2 + df(x)/dx + 2xdf(x)/dx + x2f(x)

I got df(x)/dx INSTEAD of 1......or rather f(x) in the case of A2f(x)....

Last edited: Oct 8, 2014
9. Oct 8, 2014

### Staff: Mentor

I'm not sure what you mean here. You have written yourself the first term of $A^2 f(x)$ as d (df(x)/dx + xf(x))/dx, meaning
$$\frac{d}{dx} \left( \frac{d f(x)}{dx} + x f(x) \right)$$
which expands to
$$\frac{d}{dx} \left( \frac{d f(x)}{dx} \right) + \frac{d}{dx} \left( x f(x) \right)$$
It is the second term there that you do not calculate properly.
What is $\frac{d}{dx} \left( f(x) g(x) \right)$ ?

10. Oct 8, 2014

### terp.asessed

What is $\frac{d}{dx} \left( f(x) g(x) \right)$ ?[/QUOTE]

d(f(x)g(x)/dx = g(x)df(x)/dx + f(x) dg(x)/dx?

11. Oct 8, 2014

### Staff: Mentor

Correct, now apply it to your problem by setting $g(x) = x$.

12. Oct 8, 2014

### terp.asessed

GOTCHA! Finally got the result:

A2f(x) = d2f(x)/dx2 + f(x) + 2xdf(x)/dx + x2f(x)

Also, one more question...if A = 2x d/dx...just in case...is it:

Af(x) = 2x df(x)/dx
OR
=2xf(x) df(x)/dx?

13. Oct 8, 2014

### Staff: Mentor

You can see that A as a combination of operators, $\hat{A} = \hat{B} \hat{C}$, with $\hat{B} \equiv 2x$, meaning "multiply by $2x$," and $\hat{C} \equiv \frac{d}{dx}$, meaning "differentiate with respect to $x$." These operators are to be applied in sequence, starting from the right-most and moving to the left, meaning here "differentiate with respect to $x$, then multiply the result by $2x$." Therefore your first line above is correct.

14. Oct 8, 2014

### terp.asessed

Thank you very much!!!!!!!!!!!!!