I Inner products with spherical harmonics in quantum mechanics

[Salmone]

Let $|l,m\rangle$ be a simultaneous eigenstate of operators $L^2$ and $L_z$ and we want to calculate $\langle l,m|cos(\theta)|l,m'\rangle$ where $\theta$ is the angle $[0,\pi]$. It is true that in general $\langle l,m|cos(\theta)|l,m'\rangle=0$ $(1)$ for the same $l$ even if m's are equal? Why?

For example $\langle1,0|cos(\theta)|1,0\rangle= \int_{}^{} (Y_1^0)^{*}cos(\theta)Y_1^0 \,d\Omega$ is this zero? Does the fact that equation (1) applies have anything to do with parity?

 

[PeterDonis]

we want to calculate $\langle l,m|cos(\theta)|l,m'\rangle$ where $\theta$ is the angle $[0,\pi]$.

What does this even mean? $\cos(\theta)$ isn't an operator.

 

[vanhees71]

Of course $\cos \theta$ is an operator. The trick to get the matrix elements wrt. the angular-momentum eigenstates is to use the socalled Wigner Eckart theorem to evaluate matrix elements of tensor-operator components. In your case you have $z=r \cos \theta$, i.e., you can evaluate the corresponding expectation values considering the position-vector components $\vec{x}$, using the Wigner-Eckart theorem.

https://en.wikipedia.org/wiki/Wigner–Eckart_theorem

 

[Haborix]

It isn't an operator in the form in which the OP used it, i.e. on abstract vectors in Hilbert space.

To calculate the integral in the OP I would write $\cos\theta$ in terms of spherical harmonics. Then you can look up an expansion for the product of two spherical harmonics in terms of a sum of single spherical harmonics (involves Wigner 3j -symbols which you'll need to understand the properties of). From there you can compute the integral using the orthogonality of spherical harmonics.

 

[vanhees71]

That's of course another possibility and equivalent with the use of the Wigner-Eckart theorem :-).

 

[Salmone]

Thank you for all your answers, I'd like to know how to say instantly that $\langle lm|cos(\theta)|lm'\rangle=0$ using the parity where $|lm\rangle=Y_l^m(\theta,\phi)$. Is there a way to prove it? By "using parity" I mean something like: spherical harmonics parity is equal to $(-1)^lY_l^m$ and $cos(\pi-\theta)=-cos(\theta)$

 

[vanhees71]

Since $\cos \theta=z/r$ the parity of this operator is $(-1)$, because the position vector is a polar vector and $r=|\vec{x}|$ is a scalar. Thus the parity of $\cos \theta Y_{l}^m$ is $(-1)^{l+1}$. Since the spherical harmonics are parity eigenstates, the matrix element is the same $l$ must vanish (or with different $l$ of the same parity, i.e., with both $l$ even or both odd).

 

[Salmone]

@vanhees71 I can't get the last passage, why the matrix element in the same $l$ must vanish? If possible, could you be more explicit please?

 

[vanhees71]

Obviously $\cos \theta Y_{l}^m(\theta,\phi)$ has the opposite parity than $Y_{l}^{m'}(\theta,\phi)$. That implies that the integrand in the integral of the matrix element is antisymmetric under $\theta \rightarrow \pi-\theta$, which makes the integral over $\theta \in [0,\pi]$ vanish.

Indeed if $f(\theta)=-f(\pi-\theta)$ you have on the one hand
$$I=\int_0^{\pi} \mathrm{d} \theta \sin \theta f(\theta).$$
Now substitute $\theta'=\pi-\theta$
$$I=\int_0^{\pi} \mathrm{d} \theta' \sin(\pi-\theta') f(\pi-\theta').$$
Now $\sin(\pi-\theta')=\sin \theta'$ and by assumption $f(\pi-\theta')=-f(\theta')$, i.e.,
$$I=-\int_0^{\pi} \mathrm{d} \theta' \sin \theta' f(\theta')=-I \; \Rightarrow \; I=0.$$

 

[Meir Achuz]

For example $\langle L,0|cos(\theta)L,0\rangle= \int_{}^{} (Y_L^0)^{*}cos(\theta)Y_L^0 \,d\Omega$ is this zero?

It's just simple integration. After the substitution $z=\cos\theta$, the integral becomes $\int_{-1}^{+1}dz$ of an odd function of z which vanishes.

 

[vanhees71]

The scalar product for functions on the unit sphere is defined by
$$\langle f|g \rangle=\int_0^{\pi} \mathrm{d} \theta \int_0^{2 \pi} \mathrm{d} \phi \sin \theta f^*(\theta,\phi) g(\theta,\phi).$$
In my calculation
$$f(\theta)=Y_l^{m*}(\theta,\varphi) Y_l^{m'}(\theta,\varphi) \cos \theta.$$
The integral over $\theta$ alone makes the integral vanish, no matter what the $\phi$ dependence is. The $\sin \theta$ must be in the integral because of the definition of the scalar product for $\mathrm{L}^2(\Omega)$, where $\Omega$ is the unit sphere in $\mathbb{R}^3$.

 

[Salmone]

@vanhees71 Perfect! Thank you for your time.

 

[Salmone]

One last thing, is it possible to write the integral as $\int_{-1}^{1} \mathrm{d} cos\theta \int_0^{2 \pi} \mathrm{d} \phi f^*(\theta,\phi) g(\theta,\phi)$ so that I'm integrating from $-1$ to $1$ an odd function?

 

[vanhees71]

Sure. Just substitute $u=\cos \theta$. Then $\mathrm{d} u =-\mathrm{d} \theta \sin \theta$ and thus
$$\langle f|g \rangle = \int_{-1}^1 \mathrm{d} u \int_0^{2 \pi} \mathrm{d} \phi f^* g,$$
where I wrote in a somewhat sloppy notation.