the probability of decay of a particle into two due to cubic in $\hat{x}$ interaction is given by $\langle f \mid \hat{x}^3\mid i\rangle$. The $\hat{x}^3$ term is written in the basis of ladder operators and takes the form: $$\sum_{k,k',q}\langle f \mid(a_k +a_k^\dagger) (a_{k'} + a_{k'}^\dagger) (a_q + a_q^\dagger)\mid i\rangle \delta (E_f-E_i)$$ where the states are number states of the form $\mid n_k \rangle \otimes \mid n_{k'} \rangle \otimes\mid n_q \rangle$. The are other factors of matrix elements and c-numbers. But let's get to the point. How do I simplify this expression and minimize the effort of considering all 8 terms? Properties like symmetry, cases where $k=k'=q$, etc., and other conditions might simplify the expression and make the calculation easier. Any words of wisdom or reference to sources like books, papers, and review articles will be a great help. Thank you.

Edit: @doublefelix pointed out that no commutator was prescribed. So let the commutator be $$[a_k,a_q^\dagger]=\delta_{k,q}$$.

  • $\begingroup$ What do you know about $f$ and $i$? $\endgroup$ Commented Aug 21, 2022 at 19:57
  • $\begingroup$ Also, what is the commutation relation between the $a,a^{\dagger}$? There can be different conventions here. $\endgroup$ Commented Aug 21, 2022 at 19:58
  • $\begingroup$ The form of $\mid i \rangle$ is mentioned below the expression. Let us consider the states of a truly harmonic Hamiltonian. The states are then equally spaced in energy and excitations are bosons. That implies that each state can have any number of particles. $\endgroup$ Commented Aug 21, 2022 at 19:59
  • $\begingroup$ @doublefelix I made edits and mentioned the commutator. It's a bosonic system. $\endgroup$ Commented Aug 21, 2022 at 20:03

1 Answer 1


You can reach some nice simplifications by expanding out the bra and kets as

$$|i \rangle =|n_k \rangle \otimes |n_{k'} \rangle \otimes |n_q \rangle $$

$$|f \rangle =|m_k \rangle \otimes |m_{k'} \rangle \otimes |m_q \rangle $$

Since each $a$ acts only on bras or kets for its own particle, the summand then reads

$$\langle m_k |(a_k +a_k^\dagger)| n_k \rangle \langle m_{k'} |(a_{k'} + a_{k'}^\dagger)| n_{k'} \rangle \langle m_q | (a_q + a_q^\dagger)|m_q \rangle \delta (E_f-E_i)$$

Consider just the first factor:

$$\langle m_k |(a_k +a_k^\dagger)| n_k \rangle$$

If you have $a |n_k \rangle = c_1 |n_{k-1} \rangle$ and $a^{\dagger} |n_k \rangle = c_2 |n_{k+1} \rangle$ (what $c_1$ and $c_2$ are can again depend on conventions of how $a, a^{\dagger}$ are defined, so it's best if I leave those for you to fill), then this becomes

$$ = c_1 \delta_{m_k, n_{k-1}} + c_2 \delta_{m_k, n_{k+1}}$$

and similar can be done for each of the three factors.

Lastly, notice that depending on what the dependence between $E_f$ and the particle numbers are, you can potentially express that in terms of a delta of the particle numbers as well. I would need the function for $E_f$ to do that step, though depending on your application you may or may not need to.

  • $\begingroup$ That's a good simplification. Another point is the range of values of ${k,k',q}$. The summation is over all possible values $[1,N]$ but we already have an intuition about the process. For example, a particle in higher state $q$ can decay into two particles $k,k'$. which means that $q>k,k'$. Further, one of the final states will have a higher energy (assuming linear dispersion), i.e. $k>k'$. In other words, there might be a double counting? Or will there be a half counting? Then there is the special case of $k=k'$. May be if there is a worked out problem that might help a lot. $\endgroup$ Commented Aug 21, 2022 at 20:45
  • $\begingroup$ So, is the sum over $k,k' \in [1,N )$ and $q \in [1,N]$ with the constraint $k,k'<q$? This isn't apparent from the previous information. Or are you just saying that some of the terms in the sum correspond to this possibility? If there are any constraints on $k, k', q$ it would be good to add them to the question text. $\endgroup$ Commented Aug 21, 2022 at 21:49
  • $\begingroup$ Also, what is the form of the function $E(k, k', q)$? $\endgroup$ Commented Aug 21, 2022 at 21:51
  • $\begingroup$ There are no constraints on $(k,k',q)$. The energy delta function and the $k$ conserving kronecker delta that comes about due to inner product of states take care of everything. The dispersion is linear $E(k) = h w k$. The reason I am asking for a shortcut is that there can be a huge number of terms in the summation depending on the power of $\hat{x}$. I know the usual way to solve this by considering each term but wondering if something clever can be done to make this less tedious. Like swapping the lables or moving around operators. $\endgroup$ Commented Aug 22, 2022 at 0:45
  • 1
    $\begingroup$ Got it; thanks for checking back on your question then. Cheers $\endgroup$ Commented Aug 25, 2022 at 0:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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