Proving the Commutator Relationship with Power Series Expansion | Homework Help

  • Thread starter Thread starter cahill8
  • Start date Start date
  • Tags Tags
    Commutator Proof
cahill8
Messages
29
Reaction score
0

Homework Statement


Show \left[x,f(p)\right)] = i\hbar\frac{d}{dp}(f(p))\right.


Homework Equations



I can use \left[x,p^{n}\right)] = i\hbar\\n\right.p^{n}\right.
f(p) = \Sigma f_{n}p^{n} (power series expansion)


The Attempt at a Solution


I started by expanding f(p) to the power series which makes

\left[x,\Sigma\\f_{n}\\p^{n}\right)]

and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C]
but the power series cannot be split up into two products(BC) ? So I'm not sure how to go on
 
Physics news on Phys.org
cahill8 said:
and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C]
How do you know that?
 
In a textbook it says it can be shown using that equation

Trying a different method:

[x, f(p)] = [x,\sum_{n}\\f_{n}p^{n}] = [x,fnpn + \sum_{n-1}\\f_{n}p^{n}]

using [A, B+C] = [A,B] + [A,C]

= [x, fnpn] + [x, \sum_{n-1}\\f_{n}p^{n}]

using [A, BC] = C[A,B] + B[A,C]

= fn[x, pn] + pn[x, fn] + [x, \sum_{n-1}\\f_{n}p^{n}]

using [x, pn] = i\hbarnpn-1

= fni\hbarnpn-1 + pn[x, fn] + [x, \sum_{n-1}\\f_{n}p^{n}]

[x, fn] = 0 as fn is a const.

= fni\hbarnpn-1 + [x, \sum_{n-1}\\f_{n}p^{n}]

am I on the right track?
 
Last edited:
I'm curious why you used
[A, rC] = [A,r]C + r[A,C]​
to pull out a scalar, rather than just using
[A, rC] = r [A,C]​

I'm also curious why you stopped using
[A, B + C] = [A,B] + [A,C]​
after a single addition.

But that aside, everything you wrote looks correct. We won't know if you're on the right track until we see where this path leads, though!
 
I see what you mean. [x, fnpn] = fn[x, pn] is fine.

I kept going with the addition and noticed a pattern and managed to solve it. Thanks for the hints :)
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top