/sci/ - Science & Math

[Anonymous No.4517659]

I suppose nobody here will understand that, but one can hope.

[Anonymous No.4517673]

OH SHIT TENSORS

[Anonymous No.4517677]

Anyone who has taken introductory QM will understand that.

[sage No.4517680]

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[Anonymous No.4517699]

>>4517673
that isn't tensors you fucking idiot

holy fuck OP you're an idiot you can take out the position operator god damn

[Anonymous No.4517701]

>>4517680
What the fuck are you talking about?

[Anonymous No.4517706]

>>4517699
You can; when you act it on something, it becomes a number. That's why it doesn't have a hat anymore.

[Anonymous No.4517737]

Fuck you, sci, I thought better of you.

[Anonymous No.4517780]

|x> isn't really a proper state vector. It's not surprising that you can do shit like this with it.

[Anonymous No.4517792]

>rearrange commutation relation
>get position
>zero commutator
>thinks definite position does not commute with momentum

comedy gold

[Anonymous No.4517795]

Also, generally we set <x|y> = delta(x-y), not <x|x> = 1.

[Anonymous No.4517797]

>not using Lyapunov notation for your state vectors
>showing commutation without an adjoint operator
babby's first QM. Come back when you're in grad school.

[Anonymous No.4517799]

what math is this? sorry ive only taken up to precal

[Anonymous No.4517814]

>>4517792
>definite position does not commute with momentum
It commutes, that's the point.

>>4517780
Care to elaborate?

[Anonymous No.4517815]

>>4517659
Basic problem: <x| p | x> is ill-defined. That's why you can derive an apparent contradiction.

[Anonymous No.4517824]

>>4517797
What's a Lyapunov notaithion?
And why do you need adjoint operator here?

[Anonymous No.4517827]

>>4517814
>It commutes, that's the point.
That's what I said
notice
>thinks

>>4517815
>>4517795
>>4517780
The illegal syntax works for the joke in this case, deal with it
>implying the commutator is an operator

>>4517797
>>showing commutation without an adjoint operator
>mfw

[Anonymous No.4517828]

>>4517815
Continued
For example, if you work with the position representation, | x > is delta(x), p = - i hbar d/dx. Now, p | x > = -i hbar d/dx delta(x) which is ill-defined (you can't take derivative of a dirac-delta function).

[Anonymous No.4517846]

>>4517827
>>implying the commutator is an operator
Isn't it?

[Anonymous No.4517848]

>>4517827
If you have definite position doesnt momentum reduce to mv and you get a classical limit?

[Anonymous No.4517851]

So we would have
<div class="math">i \hbar \delta(x-y) </div><div class="math">
= \langle x | i \hbar | y \rangle </div><div class="math">
= \langle x | [x, p] | y \rangle </div><div class="math">
= \langle x | (xp - px) | y \rangle </div><div class="math">
= \langle x | xp | y \rangle - \langle x | px | y \rangle </div><div class="math">
= x \langle x | p | y \rangle - \langle x | p | y \rangle y </div><div class="math">
= (x-y) \langle x | p | y \rangle </div><div class="math">
= (x-y) \langle x | p \int dz ~ \delta(z-y) | z \rangle </div><div class="math">
= (x-y) \langle x | \int dz ~ (-i \hbar) \delta'(z-y) | z \rangle </div><div class="math">
= (-i \hbar) (x-y) \delta'(x-y)
</div> which some mathematician will probably complain about, but doesn't seem problematic to me.

[Anonymous No.4517857]

>>4517851
So? Isn't the last line zero when x=y?

[Anonymous No.4517862]

>>4517857
Look closer faggot

[Anonymous No.4517892]

>>4517857
What's the left-hand side's value when x=y?

You can't take notation that treats dirac deltas as functions from reals to reals too literally.

If we think of <span class="math">\delta(x)[/spoiler] as an infinitesimal peak with an integral of 1, we've manipulated it into two infinitesimally separated infinitesimal peaks. It's not the value at zero that matters, but the integral, so it's reasonable to consider the two equivalent.

[Anonymous No.4517920]

>>4517892
OK, I could beleive that.
Thank you.