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I know and understand why equation below holds. But i am new to operator thing in QM and would need some explaination on this.

$$\langle x \rangle = \int\limits_{-\infty}^\infty |\Psi|^2 x \, \mathrm d x $$

Q1: Is it ok if i interpret the equation above like quoted:

If i want to get expected value for position $\langle x \rangle$ i have to multiply probability $|\Psi|^2$ with a position operator $\hat x$ which in this case is equal to x.

Q2: How can i derive a momentum operator $\hat p$ and an expectated value $\langle p\rangle$ using the equation above?

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At least make an attempt to Google. I Googled "expectation value of momentum" and the first result was…, which answers your question. – John Rennie Mar 20 '13 at 9:33
Related: – Qmechanic Mar 20 '13 at 10:57

The answer for Q2 is: Let $\hat O$ be some operator, and to evaluate this operator in some basis, you have to compute $\langle m|\hat O|m\rangle$. Using the completeness relation $\hat 1=\int dx |x\rangle\langle x|$

$\langle m|\hat O|m\rangle=\int dx \langle m|x\rangle\langle x|\hat O|m\rangle=\int dx\langle x|m\rangle^{*}\langle x|\hat Om\rangle=\int dx\psi_{m}(x)^{*}O[\psi_{m}](x)$

$O$ is just the functional for $\hat O$. Substituting $\hat O$ with the momentum operator you arrive at

$\hat p[\psi_{m}](x)=-i\hbar\partial_{x}\psi_{m}(x)$

and the expectation value of $\langle p\rangle$ is

$\langle p\rangle=-i\hbar\int dx\psi_{m}(x)^{*}\partial_{x}\psi_{m}(x)$

Using this procedure you can evaluate any expectation value as long as you know how an operator acts on your wave function. And as for Q1, its not really ok to interpret the eq. you wrote as stated in the quote.

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What is $\langle x|\hat Om\rangle$ ? Is that a typo? – Magpie Apr 8 '13 at 0:37

With respect to Q1, associating $|\Psi|^2$ with a probability density for an observable (in this case $x$) is the so-called Born rule and is not a priori obvious. However, once you accept the Born rule then your equation is just a regular expectation value. As others have noted, the Born rule is formulated in terms of an eigenbasis for the observable in question.

You may find it helpful to know that the expression you ask about is a special case of the more general one:

$$ \langle \hat{O} \rangle = \sum_n O_n|\Psi_n|^2 $$ which holds when the wavefunction is given by $\sum_n \Psi_n |n\rangle$ and when the operator $\hat{O}$ is diagonal in the basis $\{|n\rangle\}$. The case wherein $n$ is a continuous index (e.g. $x$) is in some respects pathological, and I have always found fundamental issues easier to understand in terms of a discrete basis.

With respect to Q2, it is not possible to "derive a momentum operator" from the expression you refer to. More generally, in nonrelativistic quantum mechanics it is not possible to derive an expression for the momentum operator from first principles except insofar as you can deduce the consequences of possible definitions of $\hat{p}$. Some common choices are:

  • Define $\hat{p} = -i\hbar \partial_x$ with no intuitive motivation for doing so.
  • Define $\hat{p}$ as an operator satisfying the canonical commutation relation $\left[\hat{x}, \hat{p}\right] = i\hbar$. A possible motivation for this is by analogy to classical Poisson brackets.
  • Define $\hat{p}$ as the generator for the translation operator. I.e., first define the translation operator $\hat{T}_a$ by $T_a|x\rangle = |x + a\rangle$. Since $\hat{T}_a$ is unitary it follows that $\hat{T}_a = e^{-ia\hat{p}/\hbar}$ for some self-adjoint (Hermitian) operator $\hat{p}$, which is taken as the definition of $\hat{p}$.

One can show these definitions are equivalent (e.g. see this). I would argue the final definition is the most fundamental. It is the definition that makes the clearest connection between translation invariance and momentum conservation, for example.

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I think problem two is solved. Statement one is not completly correct. The position operator $\hat{x}$ is only equal to x in position space, not e.g. in momentum space. So if you have your wave function as function of x, than it is correct and $\hat{p}$ is given by the derivative with respect to x. If your wave function is in momentum space i.e. a function of p, $\hat{p}$ simply becomes p.

And as soon you operator is not as simple as x, you have to "sandwich" it between the wave function and its complex conjugate as Nijankowski explained, it does not act on the square.

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