I'm reading Sakurai's Modern Quantum Mechanics. In page 99, there is a concept of Directional Eigenket. It doesn't have a definition, or any properties about it.

Because the angular dependence is common to all problems with spherical symmetry, we can isolate it and consider $$\langle \hat{n}|l,m\rangle=Y_l^m(\theta,\phi)=Y_l^m(\hat{n})$$ where we have defined a directional eigenket $|\hat{n}\rangle$. From this point of view, $Y_l^m(\theta, \phi)$ is the amplitude for the state characterized by $l,m$ to be found in the direction $\hat{n}$ specified by $\theta$ and $\phi$.

At first I thought it was the eigenket of the operator measuring the directional unit vector.(Probably it is similar or the same) But as I'm understanding it, (in position space) any function that only has a nonzero value in one direction will be such a eigenfunction. So I'm wondering,

Where is my understanding wrong, and What is this directional eigenket represented as in position space? Also, $|l,m\rangle$ refers to any eigenstate with $l,m$, right? (a linear combination of $|n,l,m\rangle$s) Degeneracy makes me a little confused here.

  • $\begingroup$ If you took $\hat n= \hat z$, can you see this is independent of r? $\endgroup$ – Cosmas Zachos Dec 29 '18 at 16:15
  • $\begingroup$ First review WP. The notation above is bad: they should have used $\hat r$ for $\hat n$. This has nothing to do with the principal n you are using in the end (which does depend on r in an indirect way...). $\endgroup$ – Cosmas Zachos Dec 29 '18 at 16:23
  • $\begingroup$ Since $|l,m\rangle$is a linear combination of $|n,l,m\rangle$, for $n=1,2,3\cdots$, and $\langle\vec{r}|n,l,m\rangle=R_{nl}(r)Y_l^m(\theta,\phi),$ $\langle\hat{r}|l,m\rangle$ would be would be $Y_l^m(\theta,\phi)$ times some linear combination of $R_{nl}(1)$, for different n.Is that correct?(Btw, I'm new at this math notation, so I'm a little slow.. sorry) $\endgroup$ – Quasi07 Dec 29 '18 at 16:32
  • $\begingroup$ Is $\langle r \hat r | n,l,m\rangle= R_{nl} (r) \langle \hat r| l,m\rangle$ helpful? $\endgroup$ – Cosmas Zachos Dec 29 '18 at 17:27

Admittedly, I never found an official reference that confirms some of the things I mention below, which therefore are just my educated guess of the topic. In particular, if someone has further mathematical insight about why it is legitimate (or illegitimate!) to write the Hilbert spaces as I'm about to do, it is welcome to correct me;

The position eigenbasis $\{|x,y,z\rangle\}$, eigenvectors of Hermitian position operators $\hat X$, $\hat Y$, $\hat Z$, is a complete basis for the Hilbert space of a 3-dimensional system, $\mathcal H_{\vec x}$, which can be considered as the tensor product of the three Hilbert spaces $\mathcal H_{x}\otimes\mathcal H_{y}\otimes\mathcal H_{z}$, each of which is a $L^2(-\infty,+\infty)$ and is spanned by one of the $\{|x\rangle\}$, $\{|y\rangle\}$, $\{|z\rangle\}$.

From a general perspective, the three Cartesian coordinates are not the unique possible choice, and we could think to translate a position eigenket in spherical coordinates as

$$|x,y,z\rangle\rightarrow |r,\theta,\phi\rangle$$

In this case, similarly, we postulate the existence of other three Hermitian operators $\hat R$, $\hat \Theta$ and $\hat \Phi$:

  • commuting with one another,

  • with spectra respectively $[0,\infty)$, $[0,\pi]$, $[0,2\pi]$,

  • with eigenbasis $\{|r\rangle\}$, $\{|\theta\rangle\}$, $\{|\phi\rangle\}$, spanning respectively a $L^2[0,\infty)$, a $L^2[0,\pi]$ and a $L^2[0,2\pi]$; the total Hilbert space now can be written as a $\mathcal H_{r}\otimes\mathcal H_{\theta}\otimes\mathcal H_{\phi}$.

$|r,\theta,\phi\rangle=|r\rangle\otimes |\theta\rangle\otimes|\phi\rangle$ is their simultaneous eigenvector with eigenvalues $r$, $\theta$, $\phi$.

With this foreword, let's now focus on the action of the angular momentum operator $\hat L=\hat X\times\hat P$:

by definition, it is the generator of infinitesimal rotations about some axis, meaning that if it acts on $|r,\theta,\phi\rangle$, it can at most modify $\theta$ and $\phi$ but will leave $r$ invariant.

As this happens for any $|r,\theta,\phi\rangle$, this means that the action of $\hat L$ on the Hilbert space $\mathcal H_{r}\otimes\mathcal H_{\theta}\otimes\mathcal H_{\phi}$ is the identity on the subspace $\mathcal H_{r}$, and some nontrivial operator on the other subspace.

In practice, we can say that $\hat L$ only acts on $\mathcal H_{\theta}\otimes\mathcal H_{\phi}$, so that's where the nontrivial part of its eigenvectors will live.

As $\hat L$ is Hermitian, its eigenvectors form a basis for the Hilbert space where it operates, so together with $\{|\theta,\phi\rangle\}\equiv\{|\theta\rangle\otimes|\phi\rangle\}$ we have another choice, which is $\{|l,m\rangle\}$.

The so called spherical harmonics are defined as $$Y_l^m(\theta,\phi)\equiv \langle \theta,\phi|l,m\rangle$$ and can be seen either as the matrix of the change of basis between $|\theta,\phi\rangle$ and $|l,m\rangle$, or as the eigenfunctions of $\hat L$, i.e. the projection on the coordinate space of the eigenstates $|l,m\rangle$ of $\hat L$.

In some sense, as $\hat L$ generates "translations" in the angle space, we can say that the relationship between $|\theta,\phi\rangle$ and $|l,m\rangle$ is very similar to the one between $|x,y,z\rangle$ and $|p_x,p_y,p_z\rangle$, and the spherical harmonics are then the equivalent of the plane waves $\langle x|p\rangle\propto e^{ipx/\hbar}$. They also share the analogy to be a complete set of functions in their space of definition.


  • about the $\hat n$, specifying $\theta,\phi$ is equivalent to specify a direction $\hat n$, so you can use $|\hat n\rangle$ or $|\theta,\phi\rangle$ interchangeably.

  • about $|n,l,m\rangle$: as someone has already pointed out, that $n$ is not the direction $\hat n$, but the quantum number related to the energy.

If the Hamiltonian commutes with angular momentum operators $\hat{\vec L}^2$ and $\hat L_z$, then you can build a simultaneous eigenbasis $|E,l,m\rangle$ which is usually written as $|n,l,m\rangle$ if the energy is quantized and labeled by an index $n$, $E_n$.

But the operator $\hat H$, differently from angular momentum, acts on the full Hilbert space, so $|l,m\rangle$ or $|\theta,\phi\rangle$ are not enough to fully expand it and we need to resort to the full $|r\,theta,\phi\rangle$. The usual wavefunction is written as:

$$\langle r,\theta,\phi|n,l,m\rangle=\psi_{nlm}(r,\theta,\phi)$$

which thanks to the commutativity of the Hamiltonian with the angular momentum further separates as

$$\langle r,\theta,\phi|n,l,m\rangle=R_{nl}(r)Y_l^m(\theta,\phi)$$


$|\hat{n} \rangle$ is the state whose angular wavefunction is a delta function at $\hat{n}$, just like $|x \rangle$ is peaked at $x$ and is an eigenket of $\hat{x}$.

  • $\begingroup$ Does it represent any eigenstate with that property? $\endgroup$ – Quasi07 Dec 29 '18 at 16:07
  • $\begingroup$ I mean the radial part, it can be anything? $\endgroup$ – Quasi07 Dec 29 '18 at 16:10
  • 1
    $\begingroup$ Ugh... ambiguity in the use of the caret... double duty here... $\endgroup$ – Cosmas Zachos Dec 29 '18 at 16:13
  • $\begingroup$ @CosmasZachos Can you explain it in more detail?? $\endgroup$ – Quasi07 Dec 29 '18 at 16:17
  • $\begingroup$ @Quasi07 Sakurai is just talking about wavefunctions on the sphere, not wavefunctions on all of $\mathbb{R}^3$. Of course to describe a particle in $\mathbb{R}^3$, you would also have a radial part. But right now we aren't including it because it's not relevant to the stuff he's currently doing. $\endgroup$ – knzhou Dec 29 '18 at 17:16

Perhaps it will be more illuminating to write

$$ | \hat{n} \rangle = | \phi , \theta \rangle $$

Note that any direction is completely specified by any two angles $\theta, \phi$. You can think of this precisely as $| x \rangle $ except for angular coordinate. Sakurai is just trying to be general/lazy. Also note that this $\hat{n}$ has nothing to do with the radial eigenvalue $n$.

Actually more precisely, you may write, since the coordinates are independent,

$$ | \hat{n} \rangle = | \phi \rangle | \theta \rangle $$

Therefore, since we consider

$$\langle x | \Psi \rangle $$

to be the probability that the particle is within $x$ and $x+ dx$, we may identically regard

$$\langle \theta , \phi| \Psi \rangle $$

as the probability that the particle is between angles $\theta$ and $d\theta$ and $\phi$ and $d\phi$. That is, it is the angular part of the wave function, which can always be expanded in spherical harmonics due to their completeness on the surface of a sphere.

Now since in our case $|\Psi \rangle = | l, m \rangle$ we have that $$\langle \theta , \phi| l,m \rangle = \Psi_{l m}(\theta, \phi) \propto Y_{lm}(\theta, \phi) $$

All we are doing is projecting the angular momentum eigenstates $|l,m \rangle$ into position space. It turns out that the eigenfunctions corresponding to these angular momentum eigenstates are just the spherical harmonics. This is found by writing $L^2$ and $L_z$ in position space and finding the common eigenfunction (which it turns out are the $Y_{lm}$s.

  • $\begingroup$ It's my first time seeing the notation $|\alpha\rangle|\beta\rangle$. What does it mean? And I'm still having troubles understanding $|\theta,\phi\rangle$ or $|l,m\rangle$, is it in a different vector space from $|n,l,m\rangle?$ $\endgroup$ – Quasi07 Dec 30 '18 at 9:56
  • $\begingroup$ It's a direct product of kets $\endgroup$ – InertialObserver Jan 1 '19 at 2:16

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