$$ \newcommand{\op}[1]{\hat{#1}} \renewcommand{\vec}[1]{\mathbf{#1}} \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\bra}[1]{\langle {#1}|} $$ Let's say we have a spherically symmetric system, like a hydrogen atom, in which we label the eigenstates of the three (commuting) observables $\op H$, $\op{\vec L}^2$ and $\op L_3$, the hamiltonian, the squared angular momentum and its third component, such that $$ \op H\ket{n,l,m}=E_n\ket{n,l,m}\\ \op{\vec L}^2\ket{n,l,m}=l(l+1)\ket{n,l,m}\\ \op L_3\ket{n,l,m}=m\ket{n,l,m} $$ A state with $l=0$, which is of the type $$ \ket{\psi}=\sum_{n\in\mathbb N}a_n\ket{n,0,0} $$ has a radially symmetric wave function $\psi(\vec x)$ so the expected value $\bra{\psi}\op x_i\ket{\psi}$ has to be $0$ for each $i=1,2,3$. This tells us that $\ket{n,0,0}$ and $\op x_i\ket{n,0,0}$ are orthogonal states, so $\op x_i\ket{n,0,0}$ is in a subspace with $l\ge 1$.

Now, the spherical harmonics with $l=1$ are three linear combinations of the spatial coordinates $x_1,x_2,x_3$ and are linearly independent, so we could write $x,y,z$ as linear combinations of $Y_{1,-1}$, $Y_{1,0}$ and $Y_{1,1}$, e.g. $x=\frac12(Y_{1,1}+Y_{1,-1})$. Can this be seen as a "proof" that $\op x_i\ket{n,0,0}$ is a state with $l=1$? I get the broader picture, but I'm stuck mathematically: are there operators associated to the spherical harmonics which we can use in this case? Could they be just, for example, $\op Y_{1,1}=\op x+i\op y$? That seems just circular reasoning to me. And, having "defined" these spherical harmonic operators, how does applying the operator $\frac12(\op Y_{1,1}+\op Y_{1,-1})$ to $\ket{n,0,0}$ bring it to a state with $l=1$?

  • $\begingroup$ It looks like you've basically got it. I'm not sure what you think is circular here, exactly. $\endgroup$ – march Nov 20 '15 at 18:53
  • $\begingroup$ It's just that I don't get anything new from this. I was expecting to find $\hat{x}_i$ expressed as combination of some other operators, but here it is just $\hat{x}_i=\hat{x}_i$. $\endgroup$ – yellowquark Nov 20 '15 at 19:19

I think you are looking for the Wigner-Eckart theorem. The operators $x,y,z$ themselves form a representation of the rotation group with $l=1$. That is if we take the operator $T^0$ to be $z$ and $T^{\pm 1}$ to be $\mp (x\pm i y)$, and have $R$ be a member of the rotation group, and $U(R)$ the unitary rotation operator

$$U(R) T^i U(R)^{-1} = {R^{(1)}}^{i}_{\,\,\,j} T^j$$

where $R^{(1)}$ is the irreducible representation with $l=1$.

The Wigner-Eckart theorem pretty much says you can add the angular momentum quantum numbers of the operator and the state in the usual way, so an operator $x,y,z$ acting on a $l=0$ state has $l=1$.

It's not so difficult to see why:

$$U(R) (T^i |l,m\rangle) =U(R) T^i U(R)^{-1} (U(R)|l,m\rangle) = {R^{(1)}}^{i}_{\,\,\,j}{R^{(l)}}^{m^\prime}_{\,\,\,m} T^j |l,m^\prime\rangle$$

This looks just like the tensor product of two representations, which we know leads to the addition of angular momenta.

  • $\begingroup$ Is $R^{(1)}=R(1)$ or are these different things? Anyway, what is $T^i$? Is it shorthand for $T^0$ and $T^{\pm}$? $\endgroup$ – yellowquark Nov 20 '15 at 20:02
  • $\begingroup$ They're the same thing, just a notation error I fixed. But they would be different for instance from $R^{(1/2)}$ which would be the 2x2 rotation matrices on the 1/2 representation. You can think of the 0 and +-1 to be like the m value of the operators. $\endgroup$ – octonion Nov 20 '15 at 20:05

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