How to Find the Operator A2 for A = d/dx + x

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In summary: You've been a great help! In summary, the conversation discusses a sample problem on quantum physics and the solution provided by the teacher. The student makes a mistake in expanding the operator A squared and asks for clarification. Through the conversation, the student is able to correct their mistake and solve for the correct solution. The conversation also touches on the application of operators in quantum physics.
  • #1
terp.asessed
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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!
 
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  • #2
terp.asessed said:
...where did I make mistake?

Here, right at the beginning:
A {df(x)/dx + f(x)} = d/dx{df(x)/dx + f(x)} + df(x)/dx + f(x)

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##.
 
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  • #3
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
terp.asessed said:
A f(x) = df(x)/dx + xf(x)
A2f(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)
You didn't apply the first ##\frac{d}{dx}## correctly between the last two lines.
 
  • #5
Wait, thank you for correcting my first mistake, but I still don't see how I got the last two lines wrong...
 
  • #6
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

DrClaude said:
You didn't apply the first ##\frac{d}{dx}## correctly between the last two lines.

...could you pls clarify more? Thanks.
 
  • #7
You should get two terms out of
$$
\frac{d}{dx} \left( x f(x) \right)
$$
 
  • #8
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:
  • #9
terp.asessed said:
So, I should have gotten d(xf(x))dx, NOT x df(x)/dx?
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.
terp.asessed said:
d(xf(x))dx = xdf(x)/dx + df(x)/dx?
What is ##
\frac{d}{dx} \left( f(x) g(x) \right)
## ?
 
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  • #10
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
terp.asessed said:
d(f(x)g(x)/dx = g(x)df(x)/dx + f(x) dg(x)/dx?
Correct, now apply it to your problem by setting ##g(x) = x##.
 
  • #12
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
terp.asessed said:
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?
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.
 
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  • #14
Thank you very much!
 

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