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 $\langle f(x)|f(x)\rangle = f^*(x)\cdot f(x)$ ?


2 Answers 2


"$| f(x) \rangle$" 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

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

  • 3
    $\begingroup$ 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
    Commented Oct 9, 2013 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
    Commented Oct 9, 2013 at 13:04

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 $f^*f=|\langle x | \alpha\rangle|^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.