/sci/ - Science & Math

Are you talking about the quantum harmonic oscillator?

>>9292683
ψ(x, t=0)=(mω/πℏ)^1/4 e^(-mωx^2/2ℏ)
Break it down to the free particle bias ψ_k(x,t) = Ae^i(kx-ωt) by Fourier transforming it at t=0
Bring back the time term in the series (with ℏ^2 k^2/2m = ℏω)
∆x=√E[x^2]=√<ψ|x^2|ψ> ∆p=√E[p^2]=√<ψ|p^2|ψ>

>>9292727
>bias
*basis

>>9292683
Let [math]U(x,t) = \begin{cases}x^2 &; t < 0 \\
0 &; t \geq 0[/math] and use the interaction picture.
>>9292727
Completely wrong. And not in a funny way too.

>>9292683
Let [math]U(x,t) = \begin{cases}x^2 &; t < 0 \\ 0 &; t \geq 0\end{cases}[/math] and use the interaction picture.
>>9292727
Completely wrong. And not in a funny way either.

>>9292817
I think >>9292727 is correct, I remember classmates discussing that your answer requires a Forrier transform.

>>9292818
Nothing can "require" a FT, it's a fucking change of basis.

>>9293034
>the fourier transform is just a change of basis XD

>>9293041
It is though. It's a change in coordinate basis to momentum basis. The Fourier kernel is literally [math]\langle x |p \rangle[/math].
Or are you saying you have no idea how infinite dimensional vector spaces work?

To see what the uncertainty is for such a system, all that is needed is the Heisenberg Picture of the observablessence. For the sake of brevity, I'll spare most of the details.

At time t=0, we have:

<x>=0
<p>=0
<x^2>=h/2mw
<P^2>=hmw/2

At time t>=0, H=p^2/2m. Hence,

dp/dt=1/ih [p,H]=0 and dp^2/dt=0.

This implies that p (t)=p (0) and p^2 (t)=p^2 (0). Hence var (p)=hmw/2.

Along the same lines, it is possible to show that var (x)=h/2mw (1+w^2t^2). So finally we have that:

uncertainty: var(x)var (y)>=h^2/4 (1+w^2t^2)

*obsevables. What the actual Fucking autocorrect.

>>9294449
*fuck. I might as well just quiet

>>9294443
>>9294449
>>9294477
Fucking Christ...

>>9293679
I'm saying you don't. Hint, what you learned in your QM course isn't proper math.

>>9294652
Lol@bigboi here
In this context, there is always a natural Pontryagin duality isomorphism which implies that is indeed a change of basis
The context has been clearly set in the OP so nobody cares about the general case, and assuming nobody here knows actual math is a 2nd year student arrogant
Stop embarrassing yourself, anon