# Marginalisation of a joint probability distribution in bra-ket notation

Given a wave function $$\Psi(\vec r_1, \vec r_2)$$, where $$\vec r_1$$ and $$\vec r_2$$ are the positions of particle 1 and 2, respectively, the probability of finding particle 1 at position $$\vec r$$ (irrespective of particle 2) is the marginalised probability distribution $$P_1(\vec r) = \int |\Psi(\vec r,\vec r_2)|^2 d^3\vec r_2. \tag 1$$

Similarly, the probability of finding particle 2 at position $$\vec r$$ is the marginalised distribution $$P_1(\vec r) = \int |\Psi(\vec r_1,\vec r)|^2 d^3\vec r_1. \tag 2$$

Question: How can I do this marginalisation in bra-ket notation? Given a ket $$|\Psi\rangle \in \mathcal{H}_1 \otimes \mathcal{H}_2$$, where $$\mathcal{H}_1$$ and $$\mathcal{H}_2$$ are the Hilbert spaces of particle 1 and 2, respectively, how do I define the probability of finding particle 1 at position $$\vec r$$ (irrespective of particle 2)?

Attempt: My guess to translate $$(1)$$ into bra-ket notation would be something like $$P_1(\vec r) = |\langle \vec r_1 | \Psi \rangle |^2,$$

where $$\langle \vec r_1|$$ only acts on the part of $$|\Psi\rangle$$ which is in $$\mathcal{H}_1$$. However, this is a notational mess because $$\langle \vec r_1|$$ (which is actually $$\langle \vec r|$$, an eigenstate of the position operator) does not have the same symbol as the argument of the function $$P_1$$.

Furthermore, how do I denote that nothing happens to the part of $$|\Psi\rangle$$ which is in $$\mathcal{H}_2$$? Shouldn't one write something like $$P_1(\vec r)=|\left(\langle \vec r_1 | \otimes I \right) |\Psi \rangle|^2,$$

where $$I$$ is the identity operator on $$\mathcal{H}_2$$. But this is wrong because $$\langle \vec r_1 | \otimes I$$ is meaningless, as it is the tensor product of a vector and an operator.

• Here are a few hints/ comments: 1.Read about the partial trace. 2. Your equations cannot be correct because the LHS is independent of $r_1$, whereas the RHS is not. 3. Regarding the meaning of the tensor product of the operator with the bra / ket, see for example here. You will see exactly this type of object a lot in the field of quantum information theory, when defining the partial trace. 4. You're almost there :) (again, compare to the notion of partial trace, e.g. in the linked post). Commented Feb 8, 2022 at 0:27
• @JasonFunderberker Thanks! I'll have a look at the partial trace. Regarding point 2.: To fix this problem, I was thinking of writing $P_1(\vec r) = |\langle \vec r | \Psi \rangle |^2$. But this has the problem that it isn't clear that $\langle r|$ only acts on the $\mathcal{H}_1$ part of $|\Psi\rangle$. Commented Feb 8, 2022 at 0:47
• Btw, I assumed that you meant two distinguishable particles, or, more generally, a Hilbert space structure like $H = H_1 \otimes H_2$. Commented Feb 8, 2022 at 0:53
• Actually, I initially had two indistinguishable particles in my mind... (e.g. two electrons in the state $\Psi(\vec r_1, \vec r_2) =\frac{1}{\sqrt{2}}[ \Psi_A(\vec r_1)\Psi_B(\vec r_2) ± \Psi_B(\vec r_1) \Psi_A(\vec r_2)]$, where $\Psi_A$ and $\Psi_B$ are single-particle wave functions.) Commented Feb 8, 2022 at 1:05
• @JasonFunderberker Is the answer perhaps $P_1(\vec r)=|\left(\langle \vec r | \otimes I \right) |\Psi \rangle|^2$, where $\langle \vec r | \otimes I$ is actually a well-defined object because $\langle \vec r|$ and $I$ are operators. Commented Feb 8, 2022 at 1:18