# 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 :-) – DanielSank Apr 13 at 17:54

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.

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:$$|{x_1,x_2}\rangle= |{x_1}\rangle\otimes |{x_2}\rangle$$

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 Apr 13 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. – Charles Francis Apr 13 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 $$|\psi\rangle$$: $$\psi(x_1,x_2)=\langle x_1,x_2|\psi\rangle$$