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How does the 3d position operator look like in position representation? I know that in 1d the position operator $\hat{x}$ is just multiplication by $x$.

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up vote 4 down vote accepted

The position operator in three dimensions is a vector operator, which, as in mpv's answer, acts as $$ \hat{\mathbf r}\psi(\mathbf r)=\mathbf r \psi(\mathbf r). $$ Note that the hat denotes an operator and not a unit vector.

The definition of a vector operator is somewhat tricky, and it is indeed startling that an operator can have vector eigenvalues. The way to make this rigorous, surprisingly enough, is to do it component by component. What that means is that a vector operator like the position (but also momentum, angular momentum, and a bunch of others) is actually a trio of operators, $$ \hat x_1, \hat x_2, \text{ and }\hat x_3, $$ which are forced to play well with rotations. Thus, if you rotate to a reference system $(x_1',x_2',x_3')=(x_2,-x_1,x_3)$ by rotating 90° about the $z$ axis, say, then the same will happen to the operators.

Some vector operators have components which all commute with each other, like position. In this case, you can have simultaneous eigenstates, $|\mathbf r⟩=|x_1,x_2,x_3⟩$, for which each component acts as a scalar, so $$ \hat x_j|\mathbf r⟩=\hat x_j|x_1,x_2,x_3⟩ = x_j|x_1,x_2,x_3⟩= x_j|\mathbf r⟩\text{ for each }j. $$ Other vector operators, like angular momentum or the Runge-Lenz vector, don't have this property, so they're a bit harder to handle.

The reason this is done component-by-component is to keep things simple, particularly when you generalize past vector operators towards tensor operators like quadrupole couplings and so on. If you want to read more about this, Edmonds is the standard reference.

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The position operator in 3D is a vector in 3D:

$$ \hat{\bf r} \psi({\bf r}) = {\bf r} \psi({\bf r}) $$

See here.

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here the eigen value r is a vector.Can eigen value be a vector? – user43641 Jul 16 '14 at 10:01
the usual space of $d$-dimensional QM is $L^2(\mathbb{R}^d)$ endowed with the usual $d$-dimensional Lebesgue measure. So in calculations the operators are usually scalars, like multiplication by the coordinate $x_i$ or the modulus operator $(\sum_i x_i^2)^{1/2}$. – yuggib Jul 16 '14 at 10:27
@user43641 See the answer by Emilio Pisanty. You can treat the vecor operator as a collection of 3 operators, each having a scalar eigenvalue. – mpv Jul 16 '14 at 12:39

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