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In quantum mechanics we describe a state not yet measured by $|\psi\rangle=c_1|\phi_1\rangle+c_2|\phi_2\rangle+c_3|\phi_3\rangle+\dots $ where $\langle\phi_m|\phi_n\rangle=\delta_{mn}$ and $|c_n|^2$ is probability that we get the state $|\phi_n\rangle$ when measured. But at the same time the quantities like $\langle\phi_n|x\rangle\langle x|\phi_n\rangle$ i.e. $|\phi_n^*(x)\phi_n(x)|$ represented in some '$x$' representation are defined as probability density functions. Isn't this quantity called probability density function always equal to unity as $\langle \phi_m|\phi_n\rangle=\delta_{mn}$ ?

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The states $|\phi_m\rangle$ can be expressed as wave functions, as follows $$ \langle x|\phi_m\rangle = \phi_m(x) . $$ The probability density function for this state would then be $$ |\phi_m(x)|^2 = |\langle x|\phi_m\rangle|^2 = \langle\phi_m |x\rangle \langle x|\phi_m\rangle . $$ The probability density function is not equal to one. It is a function of $x$. However, it integrates to one (as all probability density functions should): $$ \int |\phi_m(x)|^2 dx = \int \langle\phi_m |x\rangle \langle x|\phi_m\rangle dx = \langle\phi_m |\phi_m\rangle = 1 . $$

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  • $\begingroup$ ,Thanks for pointing me a silly mistake. I have edited my question. $\endgroup$ – NewStudent Nov 4 '16 at 5:26
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You need to integrate over some region of space. If you integrate from -$\infty$ to $\infty$ then yes the probability density is unity: the particle is indeed somewhere. You'll get a finite probability if your integral has finite bounds.

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For a orthonormal basis $e$ (in your case, $e$ is position $x$), we have the outer product: $$ \int \lvert e\rangle\langle e\rvert d e = \hat{I}$$ Thus we have: $$ \langle\psi\lvert\phi\rangle = \langle\psi\rvert\hat{I}\lvert\phi\rangle =\int\langle\psi|e\rangle\langle e\rvert\phi\rangle\, \mathrm{d}e$$ The probability distribution function $\rho$ has the form:$$ \int\rho(x)\,\mathrm{d}x $$ And we call $\langle\psi\lvert e\rangle\langle e\rvert\phi\rangle$ as a probability distribution function.

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