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I am reading some introduction on quantum mechanics. I don't understand all but I get the point that the wavefunction tells some probability aspects. In one book, they show one example of the wavefunction $f(x)$ in position space as a complex function, so they said the probability of finding the particle is $f^*(x) f(x) = |f(x)|^2$. In other book, the same example shown but in so-called bra and ket vector form, I know if I calculate the absolute square, I should get the same answer. But I am still learning the bra, ket notation, so I wonder if $\langle f(x)|f(x)\rangle$ or $|\langle f(x)|f(x)\rangle|^2$ gives the probability? If the last one gives the probability, what is $\langle f(x)|f(x)\rangle$? Is $\langle f(x)|f(x)\rangle = f^*(x)\cdot f(x)$ ?

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"$| f(x) \rangle$" does not mean anything and is not proper bra-ket notation. For translating back and forth beteween wavefunction and bra-ket notation, here is the #1 thing to keep in mind:

$$ f(x) = \langle x \mid f \rangle $$

So, the probability density to find the particle at $x$ is

$$ \left|f(x)\right|^2 = \left| \langle x \mid f \rangle \right|^2 $$

Since $\langle a \mid b \rangle = \langle b \mid a \rangle^*$, this can also be written

$$ |f(x)|^2 = \langle f \mid x \rangle \langle x \mid f \rangle $$

Remember, this represents a probability density in $x$. What this means is that

$$ \int dx\, A(x) \left| f(x) \right|^2 = \left< f \right| \left\{ \int dx \, A(x) \left| x \rangle \langle x \right| \right\} \left| f \right> $$

should be the expected value of the function A(x). The quantity in the brackets is an operator:

$$ \hat A = \int dx \, A(x) \left| x \rangle \langle x \right| $$

(Edit: As pointed out by Trimok, the above is not true for most operators. It is only true for any operator that is diagonal in the x basis, or equivalently that can be written as a function of the operator $\hat x$. These are the only kind of operator for which expectation value and higher moments can be computed using $|f(x)|^2$ as a probability density function.)

The expectation value of this operator is

$$ \langle A \rangle = \left< f \right| \hat A \left| f \right> $$

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    $\begingroup$ Be careful. For an operator $\hat A$, the notation $A(x)$, as a function, does not make sense, except if $|x\rangle$ is an eigenvector of $\hat A$ : $\hat A|x\rangle = A(x)|x\rangle$. In this particular case, $A(x)$ is the eigenvalue of the operator $\hat A$ corresponding to the eigenvector $|x\rangle$. But generally, this does not make sense. For instance, with the momentum operator $\hat P$, a notation "$P(x)$" would be a nonsense, because $|x\rangle$ is not a eingenvector for $\hat P$. $\endgroup$ – Trimok Oct 9 '13 at 8:29
  • $\begingroup$ I wanted to generalize from the specific case of thinking about $|f(x)|^2$ as a probability density, in which only operators diagonal in $x$ make sense, to the general case of finding the expectation value of any operator. I didn't mean to imply that any operator can be diagonalized in the x basis. It is a bit misleading though. $\endgroup$ – jwimberley Oct 9 '13 at 13:04
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In Bra-Ket language: Wave function is defined to be the coefficient of expansion of arbitary state in space base ket. $|\alpha \rangle=\int dx | x\rangle \langle x | \alpha \rangle=\int dx | x\rangle f_{\alpha}(x)$ . For a general state $|\alpha \rangle$, $f_{\alpha}$ is the wave function. The probability is $f^*f=|\langle x | \alpha\rangle|^2$.

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