- #1

Gabriel Maia

- 72

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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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