# How do we translate a two particle system in bra-ket notation into a wavefunction as a function of the two particle positions?

Consider the two particle system given by the following bra-ket notation

$$| \psi _1 , \psi _2 \rangle$$

where $$\psi_1, \psi_2$$ each describe a particle. I then want to apply the projector $$\langle x \rvert$$ - or some other projector, to find $$\psi (x_1, x_2 )$$.

Is the following true:

$$\langle x |\psi_1 , \psi _2 \rangle = \psi (x_1, x_2 ) \, ,$$

or do I need two projectors $$\langle x_1 \rvert$$ and $$\langle x_2 \rvert$$, or am I horribly off base with any of this?

• The equation $\langle x | \psi_1 \psi_2 \rangle = \psi(x_1, x_2)$ doesn't make sense. How can there be symbols $x_1$ and $x_2$ on the right hand side if they're not on the left hand side? Yes, you need two projectors :-) Commented Apr 13, 2020 at 17:54

The vector you have present is a direct product of vectors from two closed Hilbert spaces. It is of the form: $$| {\psi_1,\psi_2}\rangle= |{\psi_1}\rangle\otimes |{\psi_2}\rangle$$ Thus naturally any basis you’d want to express them in must also necessarily be a product of basis of two closed Hilbert spaces. As such:$$|{x_1,x_2}\rangle= |{x_1}\rangle\otimes |{x_2}\rangle$$

Given some single-particle Hilbert space $$\mathcal H$$ (e.g. $$L^2(\mathbb R)$$), the generalized position eigenvectors $$|x\rangle$$ form a continuous basis of the space. Therefore, the identity operator takes the form $$\mathbb I = \int dx |x\rangle\langle x|$$, and any state $$|\psi\rangle\in\mathcal H$$ can be expanded as

$$|\psi\rangle = \mathbb I |\psi\rangle = \int dx |x\rangle\underbrace{\langle x|\psi\rangle}_{\equiv \psi(x)} = \int dx\ \psi(x) |x\rangle$$

We can construct a two-particle Hilbert space by stitching two copies of $$\mathcal H$$ together to form the tensor product space $$\mathcal H^2 = \mathcal H \otimes \mathcal H$$. Given any choice of basis $$\{\hat e_i\}$$ for $$\mathcal H$$, the set $$\{\hat e_i \otimes \hat e_j\}$$ forms a basis for $$\mathcal H^2$$.

Therefore, the $$|x\rangle$$'s form a basis for $$\mathcal H$$, but not for $$\mathcal H^2$$. If you want a basis for the latter, you need objects of the form $$|x\rangle \otimes |y\rangle \equiv |x,y\rangle$$. The identity operator on $$H^2$$ then takes the form

$$\mathbb I = \int dx dy |x,y\rangle\langle x,y|$$

and a generic state $$|\Psi\rangle \in \mathcal H^2$$ can be expanded $$|\Psi\rangle = \mathbb I |\Psi\rangle = \int dx dy |x,y\rangle\underbrace{\langle x,y|\Psi\rangle}_{\equiv \Psi(x,y)} = \int dx dy \Psi(x,y) |x,y\rangle$$

So it's not quite that you need two projectors, but rather that you need one projector which is taken from a complete basis for the space - which takes the form of a tensor product of two single-particle states.

You didn't say whether you had identical particles, for which you also have to take into account whether they are fermions or bosons. In this case you must act with $$\langle x;y| = \frac{1}{\sqrt 2}(\langle x| \langle y| \pm \langle y| \langle x|)$$ and your state has the form $$|f;g\rangle = \frac{1}{\sqrt 2}(|f\rangle |g \rangle \pm |g\rangle |f )\rangle$$ Then $$\langle x;y|f;g\rangle = \langle x|f\rangle \langle y|g\rangle \pm \langle x|g\rangle \langle y|f\rangle = f(x)g(y) \pm g(x)f(y)$$

More generally you may have an entangled state in which the wave function cannot be factorised.

• in the case its an entangled state, what projection would we use to go from the Dirac notation to the function of position(s)? Does it make a difference that the particles are entangled only by conservation of angular momentum - i.e. an atom and a photon it just emitted? Thank you in advance
– John
Commented Apr 13, 2020 at 19:45
• You just have something like $\Psi(x,y) = \langle x;y|\Psi\rangle$ (with the obvious simplification for distinguishable particles, as in other answers). The principle is the same for spin states. You will need to sum over the possible combinations. Commented Apr 13, 2020 at 19:59

I hope I can clear up your conceptual confusion a little bit.

As base vectors we choose $$|x_1, x_2\rangle$$, meaning that one particle is at position $$x_1$$, the other at position $$x_2$$.

Using these base vectors we can compose any arbitrary two-particle state $$|\psi\rangle$$
(I prefer to call it just $$|\psi\rangle$$ instead of $$|\psi_1,\psi_2\rangle$$): $$|\psi\rangle=\iint\psi(x_1,x_2)|x_1,x_2\rangle dx_1\ dx_2$$

The above equation can be reversed to get the wave-function $$\psi(x_1,x_2)$$ of the state $$|\psi\rangle$$: $$\psi(x_1,x_2)=\langle x_1,x_2|\psi\rangle$$