Normalized state vector for bosons, Shankar problem

Dahaka14
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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}(&lt;\psi\phi|+&lt;\phi\psi|)(|\psi\phi&gt;+|\phi\psi&gt;)<br /> =A^{2}(&lt;\psi\phi|\psi\phi&gt;+&lt;\phi\psi|\phi\psi&gt;<br /> +&lt;\psi\phi|\phi\psi&gt;+&lt;\phi\psi|\psi\phi&gt;)<br /> =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&gt;+|\phi\psi&gt;). I am not very confident with my answer, could someone please help me?
 
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are you sure about your calculations steps not shown?

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

&lt;\psi\phi|\psi\phi&gt;<br /> = (&lt;\psi|\otimes &lt;\phi|)(|\psi&gt;\otimes|\phi&gt;)<br /> = (&lt;\psi|\phi&gt;&lt;\phi|\psi&gt;)<br /> = &lt;\phi|\psi&gt;^*&lt;\phi|\psi&gt;<br /> = |&lt;\phi|\psi&gt;|^2<br />
 
That makes sense. Thanks a lot!
 
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