5
$\begingroup$

Generally we have that $$|\psi\rangle=\int_{all space} \psi(\mathbf x)|\mathbf x\rangle d^3\mathbf x$$ and therefore $\psi(\mathbf x)=\langle\mathbf x|\psi\rangle$.

When discussing the mutual eigenfunctions of the orbital angular momentum operators $L^2$ and $L_z$, we find that $\psi_{lm}(\mathbf x)=\langle r,\theta,\phi|l,m\rangle=R(r)Y_l^m(\theta,\phi)$ where the function $Y_l^m(\theta,\phi)$ is a spherical harmonic and $R(r)$ is some undetermined function of radial position which can only be found when we have more information about the system (so being in an eigenstate of one of these operators fixes only the angular part of the wavefunction).

Due to the fact we generally don't know $R(r)$, I have seen sources write that the angular part of the wavefunction of an eigenstate of these two operators is given by $\langle\theta,\phi|l,m\rangle=Y_l^m(\theta,\phi)$. However I am struggling to understand this notation. My thinking is that we could write $|r,\theta,\phi\rangle=|r\rangle|\theta,\phi\rangle$ because a definite position is just a combination of a definite radius with a definite solid angle. Then we would have $$|l,m\rangle=\int_{all space}\psi_{lm}(r,\theta,\phi)|r,\theta,\phi\rangle r^2drd\Omega$$$$=\int_{all space}R(r)Y_l^m(\theta,\phi)|r\rangle|\theta,\phi\rangle r^2drd\Omega$$$$=\int^\infty_{0}R(r)|r\rangle r^2dr\int_{allangles}Y_l^m(\theta,\phi)|\theta,\phi\rangle d\Omega$$ However, when acting with $\langle\theta,\phi|$ we do not get just $Y_l^m(\theta,\phi)$ as I would like: $$\langle\theta,\phi|l,m\rangle=Y_l^m(\theta,\phi)\int^\infty_{0}R(r)|r\rangle r^2dr$$ So what is the correct way to interpret this notation?

$\endgroup$
9
$\begingroup$

The equation you phrase as $$|l,m\rangle=\int_\text{all space}\psi_{lm}(r,\theta,\phi)\,\left|r,\theta,\phi\right\rangle r^2\,\mathrm dr\,\mathrm d\Omega$$ is, and must be, wrong. The reason is that $|l,m⟩$ inhabits the orbital part of Hilbert space, $\mathcal H_\Omega$, and the right-hand side is a vector in the full Hilbert space $\mathcal H$, which is the tensor product between $\mathcal H_\Omega$ and the radial Hilbert space $\mathcal H_R$.

When we deal with polar coordinates, we are factorizing $\mathcal H=L_2(\mathbb R^3)$ as a tensor product of two Hilbert spaces, $\mathcal H_R=L_2(\mathbb R^+)$ for the radial part and $\mathcal H_\Omega=L_2(\mathbb S^2)$ for the angular dependence. Thus we write $$ \mathcal H=\mathcal H_R\otimes\mathcal H_\Omega,\quad\text{i.e.}\quad L_2(\mathbb R^3) \cong L_2(\mathbb R^+)\otimes L_2(\mathbb S^2). $$

In your example, $|\psi⟩\in\mathcal H$ factorizes as a direct tensor product of a radial state vector $|R⟩\in\mathcal H_R$ and an angular state vector $|l,m⟩\in\mathcal H_\Omega$, so you can write it as $|\psi⟩=|R⟩\otimes|l,m⟩$. Similarly, the position ket $|r,\theta,\phi⟩$ also factorizes into a radial ket and an angular ket, $|r⟩$ and $|\theta, \phi⟩$.

The key point, then, is that when you take the inner product between your position ket $|r⟩\otimes|\theta,\phi⟩$ and your state $|\psi⟩=|R⟩\otimes|l,m⟩$, the inner product factorizes over the tensor product: $$ \left(⟨r|\otimes⟨\theta,\phi|\right)\left(|R⟩\otimes|l,m⟩\right)=⟨r|R⟩\,⟨\theta,\phi|l,m⟩. $$ The first factor, $$⟨r|R⟩=R(r),$$ is the radial wavefuntion, over the restricted coordinate space $\mathbb R^+=(0,\infty)$. Similarly, the second factor is $$⟨\theta,\phi|l,m⟩=Y_{lm}(\theta,\phi),$$ and it is the wavefunction of the state $|l,m⟩$ on the coordinate space $\mathbb S^2$, i.e. the unit sphere.

Finally, if you have some vector $|\psi⟩=|R⟩\otimes|l,m⟩\in\mathcal H=\mathcal H_R\otimes\mathcal H_\Omega$ in a tensor-product Hilbert space, you can use different bras from the different factor spaces to partially cancel out the different ket factors. Thus, you can multiply on the left by $⟨\theta,\phi|$ to obtain $$⟨\theta,\phi|\psi⟩=(⟨\theta,\phi|)(|R⟩\otimes|l,m⟩)=Y_{lm}(\theta,\phi)|R⟩,$$ which is a scalar factor times a state vector in the radial Hilbert space $\mathcal H_R$. Alternatively, if what you want is $|R⟩$, you can be a bit cleverer and multiply on the left with $⟨l',m'|$ to get $$⟨l',m'|\psi⟩=(⟨l',m'|)(|R⟩\otimes|l,m⟩)=\delta_{ll'}\delta_{mm'}|R⟩.$$ Similarly, you could multiply on the left by some ket $|R'⟩$ in the radial space to isolate the angular state vector $|l,m⟩$ (modulo a scalar factor): $$ ⟨R'|\psi⟩=(⟨R'|)(|R⟩\otimes|l,m⟩)=⟨R'|R⟩|l,m⟩=|l,m⟩\int_0^\infty R'(r)^\ast R(r)r^2\mathrm dr. $$ And so on.

I hope this clarifies things - let me know if it doesn't.

$\endgroup$
  • $\begingroup$ @pisanty Can you please give a book/reference where the decomposition of 3D Hilbert space as a tensor product of radial and angular parts is discussed in somewhat more detail? $\endgroup$ – Sashwat Tanay Sep 16 '17 at 18:39
  • $\begingroup$ If i remember correctly, the quantum mechanics book by cohen-tannoudji goes into this in some detail. It is well worth reading in any case. $\endgroup$ – Anton Quelle Sep 16 '17 at 21:11
  • $\begingroup$ I could not find much detail in either Sakurai or Cohen Tannoudji. @pisanty Any other references? $\endgroup$ – Sashwat Tanay Sep 16 '17 at 22:06
  • $\begingroup$ @Bingo I can't think of any at the moment. $\endgroup$ – Emilio Pisanty Sep 16 '17 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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