Definition of a Gaussian wave packet for a Initial State

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 2K views
Gabriel Maia
Messages
70
Reaction score
1
Hi :)

I'm reading a didactic paper and the author defined the initial state ket as

|[itex]\Phi_{in}[/itex]> = [itex]{\int}dq\phi_{in}(q)[/itex]|q>

where q is a coordinate and

[itex]\phi_{in}(q)[/itex] = <q|[itex]\Phi_{in}[/itex]> = exp[itex]\left[\frac{-q^{2}}{4\Delta^{2}}\right][/itex]

I don't know if I'm missing something but isn't this definition a little flawed? I mean if you calculate the inner product of <q| with the first equation, <q|q>=1, sure, but that does not eliminate the integral, right?

I'm thinking the correct definition would be

|[itex]\Phi_{in}[/itex]> = [itex]\phi_{in}(q)[/itex]|q>

with

[itex]\phi_{in}(q)[/itex] = <q|[itex]\Phi_{in}[/itex]> = exp[itex]\left[\frac{-q^{2}}{4\Delta^{2}}\right][/itex]Thank you
 
Last edited:
Physics news on Phys.org
Just think of the unitary operator acting on [itex]|Φ_{in}>[/itex]
[itex]\hat{1}|Φ_{in}>[/itex]
but then, you also know (complete basis) that:
[itex]\hat{1}= \sum_{q} |q><q| \rightarrow \int dq |q><q|[/itex]
the arrow is for a continuous q variable...
By that you get everything straightforwardly...

In the same way, you can get:

[itex]<Φ_{in}|Φ_{in}>= \int dq' dq φ^{*}(q')φ(q) <q'|q>[/itex]
[itex]<Φ_{in}|Φ_{in}>= \int dq' dq φ^{*}(q')φ(q) δ(q'-q)[/itex]
integrating out q' due to the delta function you get:
[itex]<Φ_{in}|Φ_{in}>= \int dq φ^{*}(q)φ(q) =1[/itex]
 
  • Like
Likes   Reactions: 1 person
I see... so

<q|[itex]\Phi_{in}[/itex]> = <q|[itex]\hat{I}[/itex]|[itex]\Phi_{in}[/itex]>

right? But then

<q|[itex]\Phi_{in}[/itex]> = <q|[itex]\hat{I}[/itex]|[itex]\Phi_{in}[/itex]> = [itex]{\int}dq\phi_{in}(q)[/itex]

and the integral of [itex]\phi_{in}(q)[/itex] is not the same as = [itex]\phi_{in}(q)[/itex]

what is happening here?

Thank you.
 
you are confusing the brackets... If you have a continuous variable, then you can't generally say that <q|q>=1. This happens only for discrete ones, because the discrete version of dirac's delta function is the delta of kroenicker...
In fact you have to generalize it for continuous variables, and as I used above:
[itex]<q'|q>= δ(q'-q)[/itex]
in the same way for discrete variables you had:
[itex]<m|n>=δ_{mn}[/itex]

also avoid using the same q everywhere... since you are already taking the bra <q| in <q|Φin> when you wrote the identity operator you should have used a new symbol like |q'><q'| or generally |something else><something else|
 
  • Like
Likes   Reactions: 1 person
wait for it... wait for it...

I cannot use the very same variable in [itex]\hat{I}={\int}dq|q><q|[/itex] and in the <q| I took the inner product with, right? The identity operator must be something like

[itex]\hat{I}={\int}dq'|q'><q'|[/itex]

This will give me (when I take the inner product with <q|) a Dirac delta and save the day.

I mean... this is the rigourous precedure, right?

Thank you.