In quote box below, there is an inner product of an angular momentum eigenvector. Why can you use this inner product as a new eigenvector for the next part of the work?

And why do they "of course" satisfy the same relations?

In coordinate representation, the eigenvectors of angular momentum are known as spherical harmonics $$ \langle \theta, \phi | lm_l \rangle = Y_{lm_l}(\theta, \phi)$$ They of course satisfy the same relations: $$\hat{\vec{L}}^2 Y_{l m_l}(\theta, \phi) = \hbar^2 l (l + 1)Y_{l m_l}(\theta, \phi) \qquad \hat{L}_z Y_{l m_l} (\theta, \phi) = \hbar m_l Y_{l m_l}(\theta, \phi)$$

enter image description here

  • 2
    $\begingroup$ Are you familiar in general with how the Dirac notation relates to the coordinate representation of wavefunctions and are just confused about this specific example, or is this more general? $\endgroup$ – jacob1729 Apr 22 at 20:35
  • 1
    $\begingroup$ I'm very good at remembering ways to solve equations and other things with algebra but when it comes to linear algebra I'm sometimes stumped. I have a brief understand of Dirac notation but I don't understand how a complex number can be used as an eigenstate. $\endgroup$ – hi hi Apr 22 at 20:37
  • $\begingroup$ WP. You are really, really fussing about a definition, of course! $\endgroup$ – Cosmas Zachos Apr 24 at 1:48

Recall that in linear algebra there are two types of objects you usually deal with: vectors $|\psi\rangle$ and linear maps $\hat{L}:|\psi\rangle \mapsto |\phi\rangle$. The vectors geometrically represent arrows or quantum states whilst the linear operators represent rotations, scalings, shears or various transformations on quantum states. However the abstract approach is hard to actually do calculations with, so we put everything in terms of concrete numbers by projecting onto a basis $\{|x\rangle\}$.

$$\psi(x) = \langle x | \psi \rangle$$

Likewise the matrix elements of an operator become:

$$L(x,x')=\langle x' | \hat{L} | x \rangle$$

So there are two types of thing: the abstract geometric objects, and their components. The rules for multiplying the components (ie matrix multiplication) are chosen to match up with the rules for manipulating the geometric objects. So if you have an operator $\hat{L}$ with coordinate description $L(x,x')$ and an eigenvector $|\Lambda\rangle$ with coordinate description $\Lambda(x)$ then the fact that

$$\hat{L}|\Lambda\rangle = \lambda |\Lambda \rangle$$

must mean that when you 'multiply' the coordinate representations of the things on the left, you get the coordinate representation of the things on the right, ie:

$$L(x,x')[\Lambda(x')] = \lambda \Lambda(x)$$

if you didn't, then the whole scheme of converting things to coordinates, multiplying and then converting back to abstract quantities wouldn't work.

  • $\begingroup$ What exactly are these bra and ket vectors ⟨θ,ϕ| , |Lm⟩? I just can't visualise what components and what dimensions and just on a whole what they are to become another eigenvector. I don't know if my brain works with linear algebra or if it's just that I've never been taught it well. $\endgroup$ – hi hi Apr 22 at 22:35
  • $\begingroup$ @hihi the $|\theta,\phi\rangle$ vectors correspond the the basis vectors $|x\rangle$ in my answer and the $|L,M\rangle$ vectors to the $|\Lambda\rangle$ eigenvectors. The $L,M$ are simply convenient labels (they're integers or half-integers) to label the eigenstates of angular momentum. $\endgroup$ – jacob1729 Apr 23 at 13:53
  • 1
    $\begingroup$ @hihi I'd originally written out what all the corresponding notation for finite dimensional vector spaces looks like in index notation - I don't know if that would be more familiar to you? In that notation you extract the components of a vector via $v^i = \vec{x}^i \cdot \vec{v}$ and abstract linear operators end up represented as matrices with respect to a basis. $\endgroup$ – jacob1729 Apr 23 at 13:55

For angular momentum you have the equations

\begin{align} \hat{L}_z|l,m\rangle &= m\hbar|l,m\rangle,\\ \hat{L}^2|l,m\rangle &= l(l+1)\hbar^2|l,m\rangle. \end{align}

What you have in your equations is simply the representation of the states $|l,m\rangle$ in the real space, i.e.

\begin{align} \langle \theta,\phi|\hat{L}_z|l,m\rangle &= m\hbar\langle \theta,\phi|l,m\rangle,\\ \langle \theta,\phi|\hat{L}^2|l,m\rangle &= l(l+1)\hbar^2\langle \theta,\phi|l,m\rangle. \end{align}

If you want to know how $\hat{L}_z$ and $\hat{L}^2$ are represented in real space, you can check this, Eqs. (3.76.26) and (3.6.28).

  • $\begingroup$ How come ⟨θ,ϕ|L^2|l,m⟩ = L⃗^2Y(θ,ϕ). Thought you couldn't swap bras with operators. $\endgroup$ – hi hi Apr 23 at 17:03
  • $\begingroup$ You can't in this case and that's why I didn't :). I'm not sure that the equality that you wrote (implied by the book), is correct. $\endgroup$ – user2820579 Apr 23 at 17:36
  • $\begingroup$ If you check the reference, the representation of e.g. $\hat{L}^2$ is a differential operator... Where did you got your expressions? $\endgroup$ – user2820579 Apr 23 at 17:37
  • $\begingroup$ Its in my university QM notes. He knows what hes talking about but I think he might be simplifying somewhere because ive been stuck on this part of the notes for a couple days trying to find the maths involved that can make me able to move on. $\endgroup$ – hi hi Apr 23 at 17:52
  • 1
    $\begingroup$ I just saw this in Cohen's book, Quantum Mechanics Vol 1 Ch 6. Take it as an abuse of notation. As I said, $\hat{L}^2$ represented in real space is a differential operator. Beware that if you project with another abstract bra $\langle v|$, the differential representation will change. $\endgroup$ – user2820579 Apr 24 at 0:19

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.