From Sakurai eq. 6.4.21a we have that $$\langle {\bf k} \vert E,l,m \rangle=\frac{\hbar}{\sqrt{M k}}\delta\left(E-\frac{\hbar^2 k^2 }{2M}\right) Y_l^m({\bf\hat k}),$$ where $M$ is the mass of the particle in question. Now I wonder what the generalization of this formula is. What if we have a two particle system where the particles have 3-momenta ${\bf p_1}$ and ${\bf p_2}$ respectively. What is then $$\langle {\bf p_1}, {\bf p_2} \vert E,l,m \rangle=~???$$

The simplest thing is to copy: $$\langle {\bf p_1}, {\bf p_2} \vert E,l,m \rangle=\frac{\hbar}{\sqrt{M_1 k_1}}\delta\left(E_1-\frac{\hbar^2 k_1^2 }{2M_1}\right) Y_l^m({\bf\hat k_1}) \frac{\hbar}{\sqrt{M_2 k_2}}\delta\left(E_2-\frac{\hbar^2 k_2^2 }{2M_2}\right) Y_l^m({\bf\hat k_2})$$ but I really don't know if this makes any sense?


1 Answer 1


Recall $|E,l,m \rangle$ is the joint eigenvector of the Hamiltonian $H$, the total spin $L^2$, and a spin component (typically z) $L_z$, and the $E$, $l$ and $m$ label their respective eigenvalues. Notice all three of these operators act on the single particle that we're considering.

Also recall $\langle k |$ is an element in the momentum basis, also in the Hilbert space of the single particle,$\mathscr{H}_1$.

Now consider $\langle p_1,p_2|$: this is not in the Hilbert space of the single particle, namely this state lives in the tensor product space of two particles $\mathscr{H}_1 \bigotimes \mathscr{H}_2$.

This is why strictly speaking $\langle p_1,p_2|E,l,m \rangle$ is not well defined: the bra and ket are not in the same space. Equivalently if you think of the bra as a row vector and the ket as a column, the dimensions of the vectors don't match and you don't produce a scalar inner product.

We can however, try to "fix" this by doing the following. When considering the 2 particle problem, we can extend the operators as $H \bigotimes I$, $L^2 \bigotimes I$ and $L_z \bigotimes I$, and define the state $|E,l,m \rangle$ as a joint eigenvector of these operators. Now the inner product in question is indeed well defined, however by doing this we have restricted ourselves to a rather uninteresting case, namely we can only do this when we know the two particles are not interacting: otherwise, for example, you will not be able to write down the Hamiltonian for just the first particle, as we did here.

To see how to treat multiple particle angular momentum properly, flip forward a few pages in Sakurai for a treatment of angular momenta addition.

  • $\begingroup$ It doesn't look well-defined to me either, hence the question. What if one uses the relative momentum of the two particles in the formula. Then we have a single momentum but I don't know if this approach is correct or how to interpret it. $\endgroup$ Commented Apr 11, 2014 at 8:21
  • $\begingroup$ If you use a relative momentum approach, your Hamiltonian will also involve the center of mass momentum, therefore we will still have to deal with a tensor product Hilbert space (namely between the CM and relative spaces) and the inner product in question is still ill-defined. $\endgroup$
    – zzz
    Commented Apr 11, 2014 at 17:24

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