If I have states $\psi_{a,b,c,d}$ ,then is the following relation true: $$\langle \psi_a \otimes \psi_b| \psi_c \otimes \psi_d\rangle = \langle \psi_a |\psi_c \rangle \cdot \langle \psi_a |\psi_d \rangle \cdot \langle \psi_b |\psi_c \rangle \cdot \langle \psi_b |\psi_d \rangle?$$

I believe that it is true when the states are orthonormal; e.g. with momentum states we have: $$\langle p_a p_b| p_c p_d \rangle = \langle p_a |p_c \rangle \cdot \langle p_a |p_d \rangle \cdot \langle p_b |p_c \rangle \cdot \langle p_b |p_d \rangle = \delta(p_a - p_c)\delta(p_a - p_d)\delta(p_b - p_c)\delta(p_b - p_d) $$

The context for this question is evaluating a scattering amplitude. The probability for an initial state $|\phi_A \phi_B \rangle$ to scatter and become a final state of $n$ particles whose momenta lie in a small region $d^3 p_n ... d^3 p_n$ is given by:

$$P(AB \to 1, 2, ... n) = \left( \Pi_f \frac{d^3 p_f}{(2\pi)^3} \frac{1}{2E_f}\right) |\langle p_1 ... p_n | \phi_A \phi_B \rangle |^2$$

How are we to treat the term $$\langle p_1 ... p_n | \phi_A \phi_B \rangle ?$$

  • $\begingroup$ Are $\psi_{a,b,c,d}$ elements of same Hilbert space? $\endgroup$
    – paul230_x
    Commented Nov 3, 2021 at 15:15
  • $\begingroup$ @KP99 I'm not sure. If I have two free particles, then Hilbert spaces are certainly isomorphic, however they describe different particles, so I wouldn't say they are the same $\endgroup$
    – Jbag1212
    Commented Nov 3, 2021 at 15:20

1 Answer 1


That would not be the correct way to take the inner product.

One way to realize this is that if we have a states $\psi_1 \otimes \psi_2$ that live in the Hilbert space $H_1 \otimes H_2$, there is no well defined notion of inner product between states in $H_1$ and $H_2$. For example, we could have an electron wavefunction with both orbital and spin parts: $\psi_{radial} \otimes \chi_{spin}$, where in a particular basis $\psi_{radial}$ is a continuous function $\psi(r, \theta, \phi$), and $\chi$ is a two component vector. Then, clearly, we cannot take $\langle \psi | \chi \rangle$.

The correct way to take such an inner product is the most intuitive:

$$\langle \psi_a \otimes \psi_b | \psi_c \otimes \psi_d \rangle = \langle \psi_a | \psi_c \rangle \cdot \langle \psi_b | \psi_d \rangle$$

For the scattering states in question, the same interpretation holds: $\langle p_1 ... p_n | \phi_A \phi_B \rangle$ gives the product of the momentum wavefunctions (to be more precise, the bras and kets are out and in states respectively, so the inner product is the momentum information of the incoming wavepackets). As I believe you are using Peskin and Schroeder, you can follow the tedious algebra from equation 4.74 to 4.79 for more detail, and eventually find that the particular momentum dependence of the incoming waves does not matter for finite resolution detectors.

  • $\begingroup$ Suppose we have 3 particles. From equation 4.74 to 4.76 it looks like the following is used: $\langle p_1 p_2 p_3 | \phi_A \phi_B \rangle = \langle p_1 | \phi_A \rangle \cdot \langle p_1 | \phi_B \rangle \cdot \langle p_2 | \phi_A \rangle \cdot \langle p_2| \phi_B \rangle \cdot \langle p_3| \phi_A \rangle \cdot \langle p_3 | \phi_B \rangle .$ This is what I have written in OP though, and that is incorrect. $\endgroup$
    – Jbag1212
    Commented Nov 3, 2021 at 17:05

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