Normalized state vector for bosons, Shankar problem

[Dahaka14]

Homework Statement

Two identical bosons are found to be in states $$|\phi>$$ and $$|\psi>$$. Write down the normalized state vector describing the system when $$<\phi|\psi>\neq0$$.

Homework Equations

The normalized state vector for two bosons with $$<\phi|\psi>=0$$, using the fact that $$|\psi>\otimes|\phi>=|\psi\phi>$$, is:
$$\frac{1}{\sqrt{2}}(|\psi\phi>+|\phi\psi>)$$.

The Attempt at a Solution

So I thought the new normalization would be to do:
$$1=A^{2}(<\psi\phi|+<\phi\psi|)(|\psi\phi>+|\phi\psi>) =A^{2}(<\psi\phi|\psi\phi>+<\phi\psi|\phi\psi> +<\psi\phi|\phi\psi>+<\phi\psi|\psi\phi>) =A^{2}(2+C+C^{*})$$
where C is a complex number. Since in general $$C=a+bi$$ where a is real and b is imaginary, $$C+C^{*}=(a+bi)+(a-bi)=2a$$. Thus, the new, normalized state would be
$$\frac{1}{\sqrt{2(1+a)}}(|\psi\phi>+|\phi\psi>)$$. I am not very confident with my answer, could someone please help me?

 

[lanedance]

are you sure about your calculations steps not shown?

i'm not 100% sure, but would have gone something like

$$<\psi\phi|\psi\phi> = (<\psi|\otimes <\phi|)(|\psi>\otimes|\phi>) = (<\psi|\phi><\phi|\psi>) = <\phi|\psi>^*<\phi|\psi> = |<\phi|\psi>|^2$$

 

[Dahaka14]

That makes sense. Thanks a lot!