/sci/ - Science & Math

at least give us the answer before some dickwad mod deletes the thread again..

>>4094310
You can't normalize |x> states by 1, you have to normalize them by Kronecker <span class="math">\delta[/spoiler].

>>4094330
fu

so |x> is an eigenstate of both position and momentum at the same time?

>>4094328
ah yeah, it seems obvious now .. position eigenstates are treated differently than some objects |n> in Fock space. It all comes back to me.

so xp = px? hur dur it doesnt

Yep, basically it all goes down to the kind of counterintuitive fact that
<div class="math">(x_2-x_1)\delta '(x_1-x_2)=\delta(x_1-x_2)</div>, rather then <div class="math">(x_2-x_1)\delta '(x_1-x_2)=0</div>.
You can put this expression into the integral and check it yourself.

>>4094328 kronecker delta
Did you mean delta distribution?

<div class="math">\mathrm i\hbar
= \int_{\mathbb R^n}\mathrm d^nx\;\delta(x)\mathrm i\hbar
= \int_{\mathbb R^n} \mathrm d^nx\;
\langle x|\mathrm i\hbar|x\rangle
= \int_{\mathbb R^n} \mathrm d^nx\;\langle x|[\hat x,\hat p]|x\rangle
= \int_{\mathbb R^n} \mathrm d^nx\;\langle x|(\hat x\hat p-\hat p\hat x)|x\rangle
= \int_{\mathbb R^n} \mathrm d^nx\;x\big(\langle x|\hat p|x\rangle-\langle x|\hat p|x\rangle x\big)
= \int_{\mathbb R^n} \mathrm d^nx\; 0
= 0</div>

>>4094399
That's a counterintuitive fact?
The converse doesn't seem intuitive to me.

>>4094426
Josef here kindly illustrated why this fact is counterintuitive.
One generally would expect that <span class="math">(x_1-x_2)\cdot f(x)[/spoiler] would be zero if <span class="math">x_1=x_2[/spoiler].

>>4094400
<span class="math">\delta(x)=<x|0>[/spoiler], not <span class="math"><x|x>[/spoiler] (which is just infinity).

>>4094434
Oh, sorry, I meant <span class="math">f(x_1,x_2)[/spoiler], not <span class="math">f(x)[/spoiler].

>>4094441
But <span class="math">\langle a|0\rangle=0~\forall \langle a|\in V^*[/spoiler]!

>>4094451
I'm sorry, I'm not familiar with terminology. What's <span class="math">V^*[/spoiler]? I only rememver that <span class="math"><x_1|x_2>=\delta(x_1-x_2)[/spoiler]

>>4094469
The zero vector.

>>4094472
What.
That's not a zero vector, that's eigenstate of a particle at x=0.

OK, to make long story short
<div class="math">i\hbar\delta(x_1-x_2)=
i\hbar\langle x_1|x_2\rangle=
\langle x_1|[\hat{x},\hat{p}]|x_2\rangle=
\langle x_1|\hat{x}\hat{p}-\hat{p}\hat{x}|x_2\rangle=
(x_2-x_1)\langle x_1|\hat{p}|x_2\rangle=
(x_2-x_1)\delta '(x_1-x_2)
</div>
The last expression looks like 0, but in fact if you actually use it you get
<div class="math">
\int (x_2-x_1)\delta'(x_1-x_2)f(x_1) dx_1=-\int \delta(x_1-x_2) [(\x_2-x_1)f(x_1)]' dx_1=\int \delta(x_1-x_2)f(x_1) dx_1-\int \delta(x_1-x_2)(x_2-x_1)f'(x_1) dx_1=f(x_2)-0=f(x_2)
</div>
Which means that <span class="math">(x_2-x_1)\delta'(x_1-x_2)=\delta(x_1-x_2)[/spoiler], that is math is not broken after all.

Nice mindfuck though.

OK, to make long story short
<div class="math">i\hbar\delta(x_1-x_2)= i\hbar\langle x_1|x_2\rangle= \langle x_1|[\hat{x},\hat{p}]|x_2\rangle= \langle x_1|\hat{x}\hat{p}-\hat{p}\hat{x}|x_2\rangle= (x_2-x_1)\langle x_1|\hat{p}|x_2\rangle= i\hbar (x_2-x_1)\delta '(x_1-x_2)
</div>

The last expression looks like 0, but in fact if you actually use it you get
<div class="math">\int (x_2-x_1)\delta'(x_1-x_2)f(x_1) dx_1=-\int \delta(x_1-x_2) [(x_2-x_1)f(x_1)]' dx_1=\int \delta(x_1-x_2)f(x_1) dx_1-\int \delta(x_1-x_2)(x_2-x_1)f'(x_1) dx_1=f(x_2)-0=f(x_2)</div>

Which means that <span class="math">(x_2-x_1)\delta'(x_1-x_2)=\delta(x_1-x_2)[/spoiler], so math is not broken after all.

Nice mindfuck though.

Sorry for the typos in the previous post.

are you retarded?

>>4094310
How the fuck do you go from the 2nd last step to zero? The momentum operator takes an x derivative of everything to the right, product rule causes shit to fuck up.

>>4095041
x on the outside is just a constant, p doesn't act on it.

>>4095045
x was initially inside the bracket, so it should be actable upon by p. Since p takes an x derivative, it does NOT leave x alone. Have you ever taken a quantum mechanics course?

>>4095061
x acts on the right state for <span class="math">\hat{p}\hat{x}[/spoiler] and on the left for <span class="math">\hat{x}\hat{p}[/spoiler]. In either case it doesn't have to go over p.

>>4095061 Have you ever taken a quantum mechanics course?
Funny you say that.
<div class="math">\langle x|\hat x = x\langle x|</div><div class="math">\hat x|x\rangle = x|x\rangle</div>

>>4094310
This is the type of trolling that /sci/ needs. I approve!