- #1
Gabriel Maia
- 72
- 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
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
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