14 Two-particle systems

14.1 Two-particle wave functions

If we have two particles in one dimension, our system is described by two positions (and classically, two momenta). So instead of having a wave function $\psi (x)$, we will now need a wave function $\psi (x_1,x_2)$, and a probability density $P(x_1,x_2)=|\psi (x_1,x_2){|}^2$. It is important to understand what this probability density describes. It gives us, for any pair of positions of the first and second particle, the probability density of finding the system in that particular situation (state). Whereas for a single particle we had $$\begin{array}{}\text{(14.1)}& \text{probability to find particle in }a<x<b={\int }_a^{b}P(x)\mathrm{d}x,\end{array}$$ we now have a $$\begin{array}{}\text{(14.2)}& \begin{array}{r}\text{probability to find particle 1 in }a<x_1<b\\ \text{and particle 2 in }c<x_2<d\end{array}={\int }_a^b{\int }_c^{d}P(x_1,x_2)\mathrm{d}{x}_1\mathrm{d}{x}_2.\end{array}$$ Note that this is probabilitity density is constructed from one function of two variables. You may have thought that a system of two particles requires two wave functions, one for the first and one for the second particle. But that is not how things work. You have one wave function, which maps any point in the space of classical configurations (with points labelled by $(x_1,x_2)$) to a single complex number.

If you are only interested in the probability density of one of the particles, we need to integrate the density over the position of the other. So we can write $$\begin{array}{}\text{(14.3)}& P(x_1)=\int P(x_1,x_2)\mathrm{d}{x}_2,\end{array}$$ and similar for $P(x_2)$. Intuitively this should make sense: if you do not care about where particle 2 is located, you need to ‘collect’ all situations which lead to particle 1 being at position $x_1$.

By analogy with the one-particle situation, you will not be surprised to learn that if you measure both the positions of particle 1 and particle 2 to be ${\tilde{x}}_1$ and ${\tilde{x}}_2$, the wave function collapses to the product of position eigenstates of the two particles, $$\begin{array}{}\text{(14.4)}& {\psi }_{\text{before}}(x_1,x_2)\to {\psi }_{\text{after}}(x_1,x_2)\propto \delta (x_1-{\tilde{x}}_1)\delta (x_2-{\tilde{x}}_2)\end{array}$$ What happens if you decide to only measure the position of, say, particle 1, but not measure particle 2? Well, in that case the wave function collapses according to $$\begin{array}{}\text{(14.5)}& \begin{array}{rl}{\psi }_{\text{before}}(x_1,x_2)\to {\psi }_{\text{after}}(x_1,x_2)& \propto \delta (x_1-{\tilde{x}}_1){\psi }_{\text{before}}(x_1,x_2)\\ & =\delta (x_1-{\tilde{x}}_1){\psi }_{\text{before}}({\tilde{x}}_1,x_2).\end{array}\end{array}$$ That is to say, the wave function now is a ‘slice’ of the original wave function, taken at the position where we found particle 1.

14.2 Hamiltonian eigenfunctions in a box

To keep things concrete, let us now assume that our two particles are put in a box of size $L$, so that the positions $x_1$ and $x_2$ satisfy $0<x_1<L$ and $0<x_2<L$. We will also assume that the potential vanishes. The Hamiltonain for two free, or non-interacting, particles is simply the sum of two single-particle Hamiltonians. If they have equal masses, then we have $$\begin{array}{}\text{(14.6)}& \hat{H}=\frac{1}{2m}{\widehat{p_1}}^2+\frac{1}{2m}{\widehat{p_2}}^2=-\frac{{\hslash }^2}{2m}\frac{\partial }{\partial x_1^2}-\frac{{\hslash }^2}{2m}\frac{\partial }{\partial x_2^2}.\end{array}$$ It is therefore easy to find eigenfunctions: they are simply products of single-particle eigenfunctions. So $$\begin{array}{}\text{(14.7)}& \varphi (x_1,x_2)=\frac{2}{L}\sin \left(\frac{n\pi x_1}{L}\right)\sin \left(\frac{m\pi x_2}{L}\right)\end{array}$$ is a unit-normalised eigenfunction of $\text{(14.6)}$ for any two integers $m$ and $n$. The time-dependence can be found easily by using our knowledge of stationary states. We simply need to find the eigenvalue of this wave function, and then the time-dependence is a simple factor $\exp (-iEt/\hslash )$. The Hamiltonian acting on the wave function above gives $$\begin{array}{}\text{(14.8)}& \hat{H}\varphi (x_1,x_2)=\frac{{\hslash }^2}{2m}\frac{{\pi }^2}{L^2}(n^2+m^2)\varphi (x_1,x_2).\end{array}$$ and the energy eigenvalue is simply the sum of the eigenvalues of the individual particle wave functions.

A wave function of the type $\text{(14.7)}$ is called ‘separable’, as it separates into a product of an $x_1$-dependent function and an $x_2$-dependent function.

14.3 Non-separable wave functions: entanglement

Interesting things happen when we add two basis functions $\text{(14.7)}$ together (remember, Schrödinger’s equation is linear, so we can do that). An example is $$\begin{array}{}\text{(14.9)}& \psi (x_1,x_2)=\sqrt{\frac{18}{5}}\frac{1}{L}[\sin \left(\frac{\pi x_1}{L}\right)\sin \left(\frac{3\pi x_2}{L}\right)+\frac{1}{3}\sin \left(\frac{3\pi x_1}{L}\right)\sin \left(\frac{2\pi x_2}{L}\right)].\end{array}$$ This is no longer a separable wave function; you cannot write it as the product of one function of only $x_1$ and another one of only $x_2$. The probability density is plotted in the figure below.

The probability density of the position of particle 1 is obtained as above by integrating over $x_2$. This computation gives $$\begin{array}{}\text{(14.10)}& P_{\text{before }{x}_2\text{ measurement}}(x_1)={\int }_0^L|\psi (x_1,x_2){|}^2\mathrm{d}{x}_2.\end{array}$$ However, if you first measure the position of particle $x_2$ to be ${\tilde{x}}_2$, the wave function collapses to $\psi (x_1,{\tilde{x}}_2)$. In this case the probability density is $$\begin{array}{}\text{(14.11)}& P_{\text{after }{x}_2\text{ measurement}}(x_1)\propto |\psi (x_1,{\tilde{x}}_2){|}^2.\end{array}$$ The probability density $\text{(14.10)}$ integrates, for every value of $x_1$ the density along a vertical line in the plot. The density $\text{(14.11)}$, on the other hand, simply takes a horizontal slice through the plot. These clearly do not have to agree. To make this concrete, the density for our example state $\text{(14.9)}$ before the measurement is $$\begin{array}{}\text{(14.12)}& P_{\text{before }{x}_2\text{ measurement}}(x_1)=\frac{2}{5L}\sin (\pi x_1{)}^2[6+2\cos (2\pi x_1)+\cos (4\pi x_1)],\end{array}$$ On the other hand, if we measure, for example, the position of particle 2 to be $x_2=L/3$, then the density for the other particle after that measurement will be $$\begin{array}{}\text{(14.13)}& P_{\text{after }{x}_2=L/3\text{ measurement}}(x_1)=\frac{2}{L}\sin (3\pi x_1{)}^2.\end{array}$$ This clearly is not the same.

We thus see that the measurement of one particule influences the subsequent measurement of the other particle. Quantum states with that property are called entangled states. Non-separable wave functions thus describe entangled particles.

Now that you have seen how one-particle wave functions can be used to build two-particle wave functions, you can of course apply knowledge from previous chapters to construct more interesting two-particle states. One useful example is the combination of Gaussian wave functions. A Gaussian for two particles is given by $$\begin{array}{}\text{(14.14)}& \psi (x_1,x_2)=\frac{1}{\sqrt{2\pi {\mathrm{\Delta }}^2}}\exp [-\frac{x_1^2+x_2^2}{4{\mathrm{\Delta }}^2}].\end{array}$$ This is clearly a separable state. It can be considered an initial wave function, and its time evolution then follows by using the results computed in the previous chapter. More complicated wave functions can be obtained by linear superposition. A separable and a non-separable example are given in the figure below.

More complicated things happen when we consider interacting particles, that is, systems for which $V(x_1,x_2)\ne 0$. Needless to say, solving the Schrödinger equation for such systems is even more complicated than for a single particle with a non-zero potential, and this almost always requires numerical techniques. This goes beyond the scope of the current module. We may touch on these briefly in a problem session later.

For further reading on the topic in this chapter, see Jon J. V. Maestri, Rubin H. Landau, and Manuel J. Páez. Two-particle schrödinger equation animations of wave packet–wave packet scattering. American Journal of Physics, 68(12):1113–1119, 2000. URL: http://dx.doi.org/10.1119/1.1286310, doi:10.1119/1.1286310., and also see Schroeder’s book.

14.4 Problems

1. Verify the normalisation constant for the example wave function used in the notes, $$\begin{array}{}\text{(14.15)}& \psi (x_1,x_2)=\sqrt{\frac{18}{5}}\frac{1}{L}[\sin \left(\frac{\pi x_1}{L}\right)\sin \left(\frac{3\pi x_2}{L}\right)+\frac{1}{3}\sin \left(\frac{3\pi x_1}{L}\right)\sin \left(\frac{2\pi x_2}{L}\right)].\end{array}$$

   Solution ▶ To avoid having to do integrals, write this function in terms of single-particle wave functions, $$\begin{array}{}\text{(14.16)}& \psi (x_1,x_2)=\sqrt{\frac{18}{5}}\frac{1}{2}[{\psi }_1(x_1){\psi }_3(x_2)+\frac{1}{3}{\psi }_3(x_1){\psi }_2(x_2)].\end{array}$$ We can now use orthonormality of the ${\psi }_n(y)$ to compute the norm of the wave function, $$\begin{array}{}\text{(14.17)}& \begin{array}{c}⟨\psi ,\psi ⟩=\frac{18}{5}\frac{1}{4}{\int }_0^L\mathrm{d}{x}_1{\int }_0^L\mathrm{d}{x}_2({\psi }_1^2(x_1){\psi }_3^2(x_2)+\frac{1}{9}{\psi }_3^2(x_1){\psi }_2^2(x_2))\\ =\frac{9}{10}(1+\frac{1}{9})=1,\end{array}\end{array}$$ where the cross-terms drop out because of orthonormality of the single-particle wave functions. The wave function is thus correctly normalised.

2. How does the wave function in the previous problem evolve in time?

   Solution ▶ The two terms in the wave function are each eigenfunctions of the Hamiltonian, so we know their time evolution is simply multiplication with $\exp (-iEt/\hslash )$ with $E$ the energy eigenvalue for each term. So we get $$\begin{array}{}\text{(14.18)}& \begin{array}{c}\psi (x_1,x_2,tt)=\sqrt{\frac{18}{5}}\frac{1}{L}[e^{-i{E}_{1,3}t/\hslash }\sin \left(\frac{\pi x_1}{L}\right)\sin \left(\frac{3\pi x_2}{L}\right)\\ +\frac{1}{3}{e}^{-i{E}_{1,3}t/\hslash }\sin \left(\frac{3\pi x_1}{L}\right)\sin \left(\frac{2\pi x_2}{L}\right)].\end{array}\end{array}$$ with the energy eigenvalues $$\begin{array}{}\text{(14.19)}& E_{1,3}=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}(1+3^2),E_{3,2}=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}(3^2+2^2).\end{array}$$