Quantum Mechanics

1. Apr 11, 2010

latentcorpse

Ok so i need to prove that $[ \hat{x} , \hat{k} ] = i$ in this operator notation where
$\hat{x}=\int dx | x \rangle \langle x | x$
$\hat{k}=\int dk | k \rangle \langle k | k$
and $\langle x | k \rangle = \frac{1}{\sqrt{2 \pi}}e^{ikx}$

so i have worked out
$\hat{x} \hat{k} = \int dx \hat{k} | x \rangle \langle x | x$
but $\hat{k} | x \rangle = i \frac{\partial}{\partial x} | x \rangle$
so $\hat{x} \hat{k} = \int dx i \frac{\partial x}{\partial x} | x \rangle \langle x|$
$=i \int dx | x \rangle \langle x | = i \hat{1}=i$ where $\hat{1}$ is the unit/identity operator

similarly i find that $\hat{k} \hat{x}=-i$
and so $[\hat{x},\hat{k}]=i-(-i)=2i$
i can't see where i was supposed to get rid of that factor of 2 though????

thanks.

2. Apr 12, 2010

thebigstar25

i just have one comment (in line 8), why did you assume that the partial operator will just be applied on x? im curious since im taking the same course now ..

3. Apr 12, 2010

latentcorpse

i think it's because the bra and the ket are basis states so their derivative would be zero. It says on this article
http://en.wikipedia.org/wiki/Matrix_mechanics
that
$D \int \psi(x) | x \rangle dx = \int \psi'(x) | x \rangle dx$
where D is a derivative operator.

4. Apr 12, 2010

thebigstar25

one other question .. What is the effect if we are applying x operator on k state? .. I want to try working out this problem maybe i would be helpful ..

5. Apr 12, 2010

latentcorpse

it's as follows:

$\hat{x}|k \rangle = \int x | x \rangle \langle x | k \rangle dx = \int \frac{x}{\sqrt{2 \pi}} e^{-ikx} | x \rangle dx = \frac{dx}{\sqrt{2 \pi}} i \frac{\partial}{\partial k} e^{-ikx} | x \rangle = \int dx i \frac{\partial}{\partial k} |x \rangle \langle x | k \rangle = i \frac{\partial}{\partial k} | k \rangle$

6. Apr 12, 2010

vela

Staff Emeritus
You seem to have mistakenly commuted the operators. The order of x-hat and k-hat are different on the LHS and RHS.

7. Apr 12, 2010

latentcorpse

so the one i've actually shown the working for is $\hat{k} \hat{x}$. This gives me the i that i want so i just have to show $\hat{x} \hat{k}=0$

i have $\hat{x} \hat{k} = \int dx x | x \rangle \langle x | \hat{k}$

i don't know what to do now. should i sub for the k operator? then i'll have an integral over both x and k and i'll still have something operating to the right when there's nothing on the right to operate on.

8. Apr 12, 2010

vela

Staff Emeritus
This isn't quite right either (even after correcting the order of x and k). It might be clearer to see what's going on if you consider

$$\hat{k} \hat{x}|\psi(x)\rangle = \int dx \, i \frac{\partial}{\partial x} | x \rangle \langle x|x|\psi(x)\rangle$$

9. Apr 12, 2010

latentcorpse

not following this. how did you know to change $\hat{k}$ to $i \frac{\partial}{\partial x}$ without doing the lines i did above to prove it?

anyway i get following it through
$\int dx i \partial_x | x \rangle \langle x | \hat{x} | \psi \rangle$
$=i \int dx \int dx' \partial_x |x \rangle \langle x | x' | x' \rangle \langle x' | \psi \rangle$
$=i \int dx \partial_x | x \rangle \langle x | x \psi(x)$
$=i \partial_x (x \psi(x)) = i ( \psi(x) + x \psi'(x))$
which looks a bit....erm...off!

10. Apr 12, 2010

vela

Staff Emeritus
I was just picking up from the middle of what you did. You'd still want to show the steps you did earlier to get to this point.
Actually, this is correct. So you can see that $\hat{k}\hat{x}$ is the operator represented by $i(1+x\partial_x)$, and if you look at that last term, you should see it's the representation of $\hat{x}\hat{k}$. Hence, the commutator is...?

11. Apr 12, 2010

latentcorpse

well it will be i.
this however, relies on me being able to show that $\hat{x} \hat{k} = x \partial_x$

$\hat{x} \hat{k} | \psi \rangle = \int dx x | x \rangle \langle x| \hat{k} \psi \rangle$
$= \int dx \int dk x | x \rangle \langle | k \rangle \langle k | \psi \rangle k$
$=\int dx \int dk x | x \rangle (-i \partial_x) \langle x | k \rangle \langle k | \psi \rangle$
$=-i \int dx x | x \rangle \partial_x \langle x | \psi \rangle$
$=-i x \partial_x | \psi \rangle$

which cancels the other term leaving just the i

also the term in the previous post should have been in terms of the ket $|psi \rangle$ instead of $\psi(x)$ if we're being pedantic about the algebra, shouldn't it?

12. Apr 12, 2010

vela

Staff Emeritus
Yes, you're right. I was just looking at the derivative part. You got rid of one of the integrals when you shouldn't have. You had:

$$\hat{k}\hat{x}|\psi\rangle = i \int dx \int dx'\, \partial_x |x\rangle\langle x| x' | x'\rangle \langle x'|\psi\rangle = i \int dx \int dx' \, \partial_x |x\rangle \langle x|x' \rangle x'\psi(x')$$

You can then do one integral because of the <x|x'> delta function to get:

$$\hat{k}\hat{x}|\psi\rangle = i \int dx \, |x\rangle \partial_x (x\psi(x))$$

P.S. It looks like there might be a sign error somewhere. I'm not sure what the proper signs are, but I figure you can straighten all that out.

Last edited: Apr 12, 2010
13. Apr 13, 2010

thebigstar25

im just wondering, in the last step where you carried on the integral, is it really fine to ignore the fact you have an operator on the state x, and take out the identity operator?? .. I just have a feeling that there is something missing here? ..

14. Apr 13, 2010

vela

Staff Emeritus
No, it's not. You should end up with

$$-i \int dx\,|x\rangle x\partial_x\psi(x)$$

Last edited: Apr 13, 2010
15. Apr 14, 2010

latentcorpse

surely the final answer in post 9 should have an integral in it as well then?

16. Apr 14, 2010

vela

Staff Emeritus
Yes, that's what I said in post 12.