The operators $J_1^2$, $J_2^2$, $J_{1z}$, and $J_{2z}$ are mutually commuting operators. Likewise, $J_1^2$, $J_2^2$, $J^2$, and $J_z$ are mutually commuting operators. The two groups are incompatible, and the simultaneous eigenkets along with their eigenvalues are given by:

${J_1}^2 \left|j_1,j_2;m_1,m_2\right> = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right>$ ${J_2}^2 \left|j_1,j_2;m_1,m_2\right> = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right>$ $J_{1z} \left|j_1,j_2;m_1,m_2\right> = m_1 \hbar \left|j_1,j_2;m_1,m_2\right>$
$J_{2z} \left|j_1,j_2;m_1,m_2\right> = m_2 \hbar \left|j_1,j_2;m_1,m_2\right>$


${J_1}^2 \left|j_1,j_2;j,m\right> = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;j,m\right>$ ${J_2}^2 \left|j_1,j_2;j,m\right> = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;j,m\right>$
$J^2 \left|j_1,j_2;j,m\right> = j \left(j+1\right) \hbar^2 \left|j_1,j_2;j,m\right>$
$J_z \left|j_1,j_2;j,m\right> = m \hbar \left|j_1,j_2;j,m\right>$

I read that each set of eigenkets are mutually orthogonal [1] (for eigenkets corresponding to different sets of eigenvalues). This is what I don't understand. In principle it makes sense, but when I plug in numbers I don't get zero for the inner product. For example take the first eigenket: $\left|j_1,j_2;m_1,m_2\right>$. If I choose different eigenvalues for this eigenket (e.g. let $j_1 = 0$ and then let $j_1 = 1$) I get the following:

for $j_1 = 0$ I can have:

for $j_1 = 1$ I can have any of the following, since $\left|m_1\right| \leq j_1$:

If I take the inner-product of the $j_1 = 0$ eigenket with any of the $j_1 = 1$ eigenkets I do not get zero, e.g.:

$\left<0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right> = {j_2}^2+{m_2}^2$

which is non-zero unless $j_2 = 0$.

What am I misunderstanding here? How do you show that eigenkets with different eigenvalues are orthogonal?

  • 2
    $\begingroup$ It is a general property of eigenvectors for different eigenvalues of an arbitrary Hermitian operator, see e.g., en.citizendium.org/wiki/Hermitian_operator $\endgroup$
    – Qmechanic
    May 26, 2011 at 13:44
  • $\begingroup$ The error is most likely caused by wrongly interpreting the inner product in a Cartesian fashion a la en.wikipedia.org/wiki/Inner_product_space#Examples $\endgroup$
    – Qmechanic
    May 26, 2011 at 14:22
  • $\begingroup$ As my Answer below points out, you're handling a tensor product as if it's a direct sum. It isn't. It's not an easy distinction to be completely familiar with, but it's very much worth getting there. $\endgroup$ May 26, 2011 at 15:29

3 Answers 3


The error is most likely that you are using

$$\left<j_1,j_2;m_1,m_2 \mid j'_1,j'_2;m'_1,m'_2\right> = j_1j'_1+j_2j'_2+m_1m'_1+m_2m'_2, \quad \mathrm{(Wrong!)}$$

where you should be using

$$\left<j_1,j_2;m_1,m_2 \mid j'_1,j'_2;m'_1,m'_2\right> = \delta_{j_1,j'_1}\delta_{j_2,j'_2}\delta_{m_1,m'_1}\delta_{m_2,m'_2},$$

where $\delta_{k,\ell}$ is the Kronecker delta function.

In other words, the $j$'s and $m$'s are not the coefficients $v^i$ of a vector $\vec{v}=\sum_i v^i \vec{e}_i$ in a Hilbert space, where $\vec{e}_i$ is an orthonormal basis, so that

$$ \left<\vec{v}\mid\vec{v}'\right> = \sum_i (v^i)^*v'^i, \qquad \left<\vec{e}_i\mid\vec{e}_{i'}\right> = \delta_{i,i'}. $$

Rather, the $j$'s and $m$'s correspond to the $i$-labels of the basis $\vec{e}_i$. For brevity, we often write $\left< i \mid i'\right>$ instead of $\left<\vec{e}_i\mid\vec{e}_{i'}\right>$.

Finally, to give a complete answer, let me include my comment above that it is a general property of eigenvectors for different eigenvalues of a Hermitian operator, that they are orthogonal to each other, see e.g., Lubos Motl's answer or here.

  • 1
    $\begingroup$ Brevity, beautiful brevity! Who cares about tensor products and direct sums, eigenvalues and eigenvectors, and the rest of a huge machinery? $\endgroup$ May 26, 2011 at 16:09
  • $\begingroup$ yes, I was making this mistake. Thank you for clarifying, I think part of the confusion was the idea that the indices can take non-positive values (i.e. for $j_1=1$, $m_1$ can take any of $\{-1,0,1\}$ so $\vec{e}_m$ can be any of $\{ \vec{e}_{-1}, \vec{e}_0, \vec{e}_1 \}$. All of which are mutually orthogonal. Am I understanding this correctly now? $\endgroup$
    – okj
    May 26, 2011 at 17:35
  • $\begingroup$ Dear @okj. In principle, only you yourself can say whether you understand something, but for what it's worth, yes, it seems, that you are on the right track. $\endgroup$
    – Qmechanic
    May 27, 2011 at 2:35

Eigenvalues of a Hermitian operator or their set - such as components of $\vec J$ and/or $J^2$ - corresponding to different eigenvalues are always orthogonal to each other because $$\langle \psi | M | \phi \rangle = m_\psi \langle \psi | \phi \rangle = m_\phi \langle \psi | \phi \rangle $$ I could get any eigenvalue $m_\phi$ or $m_\psi$ by acting with $M$ on the two sides. Because those two things are equal, we have $$ (m_\psi-m_\phi) \langle \psi | \phi \rangle = 0 $$ which implies - because the eigenvalues differ $$\langle \psi | \phi \rangle = 0$$ Your derivation of a nonzero inner product is incorrect and you couldn't have derived it from the formulae above your final result because none of them contains any information about the inner product - in fact, they contain no bra vector whatsoever, so your reversal of a ket vector and its interpretation of an inner product was clearly some beginner's misunderstanding of what the symbols mean.

  • $\begingroup$ I admit I'm a beginner and probably am making A LOT of beginner's mistakes. I know the derivation is incorrect, I want to know what I did wrong so that I can correctly understand. $\endgroup$
    – okj
    May 26, 2011 at 14:18
  • $\begingroup$ Dear @okj, I would like to help you to locate the error more precisely but I just honestly don't understand how you came with the $j_2^2+m_2^2$ inner product. At any rate, I don't think you should be trying to refine your mistakes - it's more sensible to try to learn the right way to calculate the inner products. ;-) $\endgroup$ May 26, 2011 at 15:52

your line $\left<0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right> = {j_2}^2+{m_2}^2$ is making an assumption about what an inner product must look like. A technical way to express your problem is to note that the sequence of eigenvalues identify an element in a tensor product of vector spaces, not a direct sum. If it was a direct sum, your sum of squares would be right, but it isn't.

Luboš's Answer is completely right, enough so that I upvoted it because he needs all the rep he can get, but it looks like it needs you to know what you're doing already for you to understand it. When you use objects like $\left|j_1,j_2;m_1,m_2\right>$ to represent a state, you implicitly claim that the operators that have the eigenvalues $j_1,j_2;m_1,m_2$ are self-adjoint and mutually commutative. In elementary terms, we can take this to define the inner product on the Hilbert space. We don't know what the inner product is until we've defined it. If the eigenvalues are different, the inner product is defined to be $0$, if the eigenvalues are the same, the inner product is defined to be $1$.

As Luboš says, all the lines above your first introduction of a bra are only about operators acting on a vector space, you haven't got a Hilbert space until you've defined an inner product (and more than that, closure in the norm). Once you've defined an inner product, $(\left|U\right>,\left|V\right>)$, you can define a bra as the object that acts on a vector to get this value, $\left<U \mid V\right>=(\left|U\right>,\left|V\right>)$. There's a theorem that says we can do this if we're properly careful, http://en.wikipedia.org/wiki/Riesz_representation_theorem.

  • $\begingroup$ @Peter: thank you for your explanation that was very helpful. So if I'm understanding you correctly the dual bra to a given eigenket (in my case a simultaneous eigenket) is defined to give 0 as their inner-product. If this is correct, then in practical terms, given some numerical values for $j_1, j_2, m_1, m_2, j, m$, how do I compute the inner product? (i.e. the definition we've imposed tells me what the inner-product of two orthogonal kets is, but it tells me nothing about what the inner-product of two general kets is so that I can check if they are orthogonal). $\endgroup$
    – okj
    May 26, 2011 at 15:38
  • $\begingroup$ Dear @Peter, OK comments but I actually don't need to know that the operators are commutative to prove that the inner product is zero (it's needed for the simultaneous eigenstates to exist in the first place, however). The two vectors have different eigenvalues of $J^2$, so they're orthogonal. $\endgroup$ May 26, 2011 at 15:48
  • $\begingroup$ Dear @okj, the inner product of two different vectors belonging to the same basis - either the $j_1m_1 j_2 m_2$ basis or the $jm j_1 j_2$ basis - are always exactly zero. That's why we say that the bases are orthogonal bases. The inner products of basis vector from the first basis with basis vectors with the second basis are given by the Clebsch-Gordan coefficients. $\endgroup$ May 26, 2011 at 15:50
  • $\begingroup$ If the definition of the inner product is given in terms of a basis of eigenvectors of some complete set of mutually commutative self-adjoint operators, then to find the inner product of two arbitrary vectors you have to express them as linear superpositions of the basis vectors. You must know enough about the vectors relative to the intrinsic structure of the Hilbert space to be able to do that. If you know why you introduced a particular Hilbert space in the first place, you hopefully also know why you want to work with specific vectors in it. $\endgroup$ May 26, 2011 at 15:58
  • 1
    $\begingroup$ OK, but keep careful track of what is complex linear and what is complex anti-linear. Your $\left<j_1j_2m_1m_2\mid j_1j_2jm\right>$ should be $\left<j_1j_2jm\mid j_1j_2m_1m_2\right>$. You're effectively using a resolution of the identity here, $\sum_\alpha\left|\alpha\right>\left<\alpha\right|=1$, with the $j_1j_2$ eigenvalues unchanged. $\endgroup$ May 26, 2011 at 16:48

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