How to Use Induction to Prove Quantum Commutator Relations?

[belleamie]

PLEASE help asap with quantum physics!

Hi there, i need help in a couple of questions that I'm just stumped
one of them :
A) use induction to show that
[ x (hat)^n, p(hat) sub "x" ] = i (hbar)n x(hat)^(n-1)

- so far I've figured out this equation is in relation to solve the above eq, but I'm not entirely sure how to connect the two
[ f (x (hat)), p(hat) sub "x"] = i h(bar) (partial F/ partial x) * (x (hat))

B) I'm not sure how to show the symbol "pitch fork" but i will refer to it as "tsi"

Show for infinitesimal translation for
|tsi> --> |tsi'> =T(hat) (dirac delta x)|tsi>
that <x> ---> <x> + dirac delta x and
< P (as in momentum) sub x > ----> < P subx >

SO far I have gotten {T is for hte translator)
<tsi| T(hat with dagger) (diarc delta x) x T hat (diarc delta x)| tsi>
= <tsi| (1+(idelta sub x P hat sub x/ hbar) x hat (1- i diarc delta x P sub x /hbar) |tsi>

I don't know where to go from there tho...

 

[bigplanet401]

Part A:
You can either prove the second equation you have listed by induction (harder), or use this convenient commutator identity,
<br /> [AB, C] = A[B,C] +[A,C]B \, ,<br />
to get what you want. Here is a presentation of the first few steps (but only the 1st, nth, and (n+1)th are really important!)
<br /> \begin{align*}<br /> [\hat{x}^2, p_x] \equiv [x^2, p] &amp;= [xx, p] = x[x,p] + [x,p] x = 2i\hbar x \, ,\\<br /> [x^3,p] = [xx^2, p] &amp;= x[x^2,p] + [x,p]x^2 = x(2i \hbar x) + i \hbar x^2 = 3i \hbar x^2 \,\\ <br /> &amp;\vdots <br /> \end{align*}<br />
Do you see the pattern? Induce the final result.
Note: Your more general statement about position-dependent functions is not correct.
Get rid of the "x(hat)" at the end.
Part B:
You're on the right track, but your last line isn't self-consistent. It should read
\langle \psi \mid \left( 1 + \frac{i \delta(x) p}{\hbar} \right) x \left( 1 - \frac{i \delta(x) p}{\hbar} \right) \mid \psi \rangle \, .<br />
The Dirac delta function is a distribution that goes to infinity at x = 0. I'm not sure why they call it "infinitesimal." But if you want to get the right expressions for the expectation value, you'll need to pretend that the (\delta(x))^2 you get after carrying out the multiplication above, is zero.

 

[Physics Monkey]

I don't mean to intrude, bigplanet, but don't you find it a bit odd that a dirac delta function would be inside the translation operator. The exponential of a Dirac delta function is pretty highly singlular object and it certainly can't be Taylor expanded. As written, the expression is also not dimensionally correct since the Dirac delta function has units of one over length. It seems to me that the OP probably confused \delta, which was supposed to be a small length, with the Dirac delta function. Do you agree?

 

[bigplanet401]

Hi Physics_Monkey,
Thanks! I didn't know what to do with "Dirac Delta." It is not well-behaved at the origin. But Delta (a small constant) seems OK (see below).
To the OP: The line above the last paragraph does not involve Delta functions. Replace \delta(x) with some constant \Delta. Then you need not worry about singularities. You can also neglect the term in \Delta^2, which is a second-order contributon.