/sci/ - Science & Math

[Anonymous No.3868926]

homwork tredd

Suppose I have a wavefunction

ψ(r1, r2)= (∅1s(r1) ∅1p(r2) - ∅1s(r2) ∅1p(r1))

And I know that ∅1s(r1) and ∅1p(r1) are normalized. How would I go about finding the normalization constant for ψ(r1, r2)?

OBS: ∅1s(r1) denotes the 1s orbital, ∅1p the 1p. These are not just random names.


s and p orbital wavefunctions aren't orthogonal to each other are they? Since they occupy the same space at some points and all.

[Anonymous No.3868963]

ψ(r,R)=N (s(r)·p(R) - s(R)·p(r))
with
∫ |s(x)|^2 dx = 1
∫ |p(x)|^2 dx = 1

we want
∫ |ψ(r,R)|^2 dr dR = 2

|ψ(r,R)|^2=N^2 ( |s(r)·p(R)|^2 + |s(R)·p(r)|^2 - 2 *combinations of s(r)p(r) and so on* )

now if you integrate you'll use

∫∫ f(x)g(y) dx dy = ∫f(x)dx · ∫g(y)dy = ... = 1 in your case for the two sides. the mixed terms will probably be zero due to orthogonality and you'll solve for N. I guess it'll be 1/sqrt(2)

[Anonymous No.3868993]

>>3868963
Tell me your secrets, wizard!

[Anonymous No.3869000]

>>3868963

Alright, so I got to

∫ |s(r)p(R)|² + |s(R)p(r)|² - 2*derp dr dR = 1/N²

Using the relation you gave, I got

∫ |s(r)|² dr + ∫ |p(R)|² dR + ∫ |s(R)|² dR + ∫ |p(r)|² dr = 1/N²

2 + ∫ |p(R)|² dR + ∫ |s(R)|² dR = 1/N²

Now I know the answer is 1/sqrt(2), but for that,
∫ |p(R)|² dR = ∫ |s(R)|² dR = 0

why the hell?

[Anonymous No.3869015]

>>3869000

wait, I summed the terms instead of multiply (although fixing this didn't really solve everything).

I got to

∫ |p(R)|² dR + ∫ |s(R)|² dR = 1/N²

now. Now those things have to equal 1 for some reason. But that involves me claiming they are automatically normalized. Can I say that if s(r) is normalized, then s(R) also is? because that would solve everything

[Anonymous No.3869066]

Bump

[Anonymous No.3869067]

∫ |s(r)p(R)|² + |s(R)p(r)|² - 2*derp dr dR = 1/N²
∫ |s(r)|² dr · ∫ |p(R)|² dR + ∫ |s(R)|² dR · ∫ |p(r)|² dr = 1/N²
1 · 1 + 1 · 1 = 1/N²
2 = 1/N²
????
profit

[Anonymous No.3869072]

>>3869067

So is that a yes then?
Can I say that if s(r) is normalized, then s(R) also is?

[mtp !EKFQOBEsKo No.3869080]

and when i post simple geometry problem and ask if ive done it right, no replies. never change /sci/.

[Anonymous No.3869083]

>>3869072
Are you studying physics or what?
you're asking here if
∫ |s(x)|^2 dx = ∫ |s(R)|^2 dR = ∫ |s(OPisfag )|^2 dOPisfag
?

[Anonymous No.3869086]

>sage hw threads
>unless it gives me a chance to pretend i'm a genius
>/sci/

[Anonymous No.3869095]

>>3869080
>>3869086
It might look harder than it is. There is nothing going on but using basic integral relations and facts about wavefunctions, like ∫ |f(x)|^2 dx = 1

[Anonymous No.3869097]

>>3869083
Oh derp, fuck me. I guess you're right there

I guess it's pretty obvious when you look at it mathematically, I was just getting confused at what that would mean for the electron and the wavefunction, but I guess they're indistinguishable and both share the same orbitals then it would be just as obvious

Thanks dood, that really helped

[Anonymous No.3869105]

>>3869080
>>3869086
It might look harder than it is. There is nothing going on but using basic integral relations and facts about wavefunctions, like ∫ |f(x)|^2 dx = 1