Identical Particles in Quantum Mechanics

Question

I'm reading Chapter 5 of Griffiths' Quantum Mechanics book about "Identical Particles". He says that:

The state of a two-particle system is a function of the coordinates of particle one ($r_1$), the coordinates of particle two ($r_2$) and the time: $\Psi(r_1,r_2,t)$

I'm having a little trouble with this. I would have thought that wave function (even for a two-particle system) would just have a value for each point in space and in time and so would just be: $\Psi(r,t)$

I suppose I even have an issue with "the coordinates of particle one" - doesn't that imply that we know the exact position of the particles, when we are trying to find their wave function, which is necessarily spread out.

Could someone explain where my thinking about this is wrong?

Answer 1

Let us first discuss the case of distinguishable particles: To study a system of two distinguishable particles, you have to consider the Hilbert space $\mathscr{H} \equiv \mathscr{H}_1 \otimes \mathscr{H}_2$, where $\mathscr{H}_1$ is the Hilbert space of particle one and $\mathscr{H}_2$ the Hilbert space of particle two. A generic state $|\psi\rangle \in \mathscr{H}$ can be expanded as follows: $$ |\psi\rangle = \sum\limits_{ij} c_{ij}\, |i\rangle\otimes |j\rangle \quad , $$

where $\{|i\rangle\}_i$ and $\{|j\rangle\}_j$ are complete orthonormal bases in $\mathscr{H}_1$ and $\mathscr{H}_2$, respectively. The corresponding position-space representation then reads $$\left(\langle x_1| \otimes \langle x_2 |\right) |\psi\rangle \equiv \psi(x_1,x_2) = \sum\limits_{ij} c_{ij}\, \left(\langle x_1| \otimes \langle x_2 |\right)|i\rangle\otimes |j\rangle \equiv\sum\limits_{ij} c_{ij} \,\phi_i(x_1)\, \varphi_j(x_2) \quad ,$$ with $\langle x|i\rangle \equiv \phi_i(x)$ and $\langle x|j\rangle \equiv \varphi_j(x)$.

As you can see, you really need two position indices, $x_1$ and $x_2$, to write the wave function of the composite system. This is necessary for the mathematical formalism to make sense: In order to calculate all expectation values properly, we need information on both particles, not just one.

Regarding your second question: In the one-particle case we interpret (given that the wave function is normalized) $|\Psi(x)|^2 \mathrm{d} x$ as the probability that you find the particle in the volume element $\mathrm{d}x$ around $x$. Similarly, in the two-particle case, we have that $|\psi(x_1,x_2)|^2 \mathrm{d}x_1\, \mathrm{d}x_2$ corresponds to the probability to find particle one in $\mathrm{d}x_1$ around $x_1$ and particle two in $\mathrm{d}x_2$ around $x_2$.

The whole discussion was about distinguishable particles. In the case of two indistinguishable particles you have to consider the anti-symmetrized and symmetrized subspaces of $\mathscr{H} \equiv \mathscr{H}_1 \otimes \mathscr{H}_1 $: For the fermionic two-particle case you need $\mathscr{H}_{\mathrm{F}} \equiv \mathscr{H}_1 \wedge \mathscr{H}_1$, while the two boson Hilbert space is $\mathscr{H}_{\mathrm{B}} \equiv \mathscr{H}_1 \vee\mathscr{H}_1$. In these cases you still need two position labels, each for one particle. However, in contrast to the distinguishable two-particle case, we have that $|\psi_{\mathrm{F}/\mathrm{B}}(x_1,x_2)|^2 = |\psi_{\mathrm{F}/\mathrm{B}}(x_2,x_1)|^2$, which follows from the symmetry properties of the wave function for (two) identical particles. This leads to the fact that the probability to find one of the two particles at $x_1$ and the other at $x_2 \neq x_1$ is given by

$$2!\, |\psi_{\mathrm{F/B}}(x_1,x_2)|^2\mathrm d x_1\,\mathrm dx_2 \quad .$$

Note that it does not make sense to speak of particle one and particle two anymore, as in the case of distinguishable particles.

Answer 2

I would have thought that wave function (even for a two-particle system) would just have a value for each point in space and in time and so would just be: $Ψ(\mathbf{r},t)$

For a two-particle system in One dimension, You need $2$ coordinate (exclude time for a sec), one specifying the coordinate for one particle and the other for another particle. You can understand it in terms of the degree of freedom.

Note that we are not giving the exact position of particles through the wave function but only the probability for them to be at some point (or rather in some interval).

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Basics

For a system with two particles in one dimension (we further can generalize this to more dimensions), we can define the kets $$|x_1\rangle\otimes |x_2\rangle \longleftrightarrow \begin{cases} \text{particle 1 at} \ x_1\\ \text{particle 2 at} \ x_2 \end{cases}$$ The wave function for two-particle then can be written as $$|\psi\rangle =\sum_{\omega_1}\sum_{\omega_2}C_{\omega_1,\omega_2}|\omega_1\rangle \otimes |\omega_2\rangle $$ where $|\omega_1\rangle $ and $|\omega_2\rangle $ are eigen kets of operator $\Omega_1^{(1)}$ and $\Omega_2^{(2)}$ (an operator on $\mathcal{V}_1$ and $\mathcal{V}_2$).

On a position basis, this would become $$\psi(x_1,x_2)=\sum_{\omega_1,\omega_2}C_{\omega_1,\omega_2}\omega_1(x_1)\omega_2(x_2)$$

Answer 3

I'll answer your questions in reverse order.

I suppose I even have an issue with "the coordinates of particle one" - doesn't that imply that we know the exact position of the particles, when we are trying to find their wave function, which is necessarily spread out.

Just think about a one particle case. We can talk about 'the coordinate of the particle $x$', and the wavefunction $\Psi(x)$ is a complex-valued function of that coordinate. The coordinate is just a label for possible positions -- it does not imply we know that the particle is located at any particular position, since $\Psi(x)$ doesn't need to be a delta function.

I'm having a little trouble with this. I would have thought that wave function (even for a two-particle system) would just have a value for each point in space and in time and so would just be: Ψ(𝑟,𝑡)

This is indeed a subtle point and this is really one of the key points at the heart of quantum mechanics.

When you are first introduced to quantum mechanics, you tend to think of the wavefunction as being "like" the electric field, but instead of a vector at every point in space, you have a complex number. Then you might think the natural generalization to having two particles is to have two fields. For example, by analogy, you might think of having two fields lke electric and magnetic field, and so you have two vectors at every point in space.

In fact this is very much not the correct generalization. A better analogy for quantum mechanics is probability theory. Let's say I want to know the probability of it raining on Monday and Tuesday. You might say, "well there's a 10% chance of rain on Monday and a 10% chance on Tuesday, and I multiply them together." But this is not a good model because these events are correlated. If it rains on Monday, then there is a higher chance that it will rain on Tuesday.

In quantum mechanics, we need to associate one complex number (called a probability amplitude) with every state of the system. For two particles, a state of the system would be "Particle 1 is at location $x_1$ and particle 2 is at position $x_2$." We can write the complex number associated with this state as $\Psi(x_1,x_2)$. You might then think, by analogy with the electric/magnetic field situation, that you could decompose this complex number into a complex number associated with particle 1, and one for particle 2: $\Psi(x_1,x_2)=\Psi_1(x_1)\Psi_2(x_2)$. In general this is incorrect, because we want to allow for the case that particle 1's position is correlated with particle 2's position. Therefore we should really think of the two particle wavefunction on living on a space which is 6 dimensional, 3 dimensions for particle 1 and 3 for particle 2.

Having said that, there are specific situations where it is ok to think of particles 1 and 2 as being uncorrelated, in which case it is fine to decompose $\Psi(x_1,x_2)=\Psi_1(x_1)\Psi_2(x_2)$. This is also true in probability theory -- if we have two dice and want to know the probability that both are a 6, then we can in fact multiply the probability that dice 1 comes up with a 6 and dice 2 comes up with a 6, since these are two independent events. However, it is not true in general.

For what it is worth, I personally feel that this subject is made more confusing because of how it is taught, which is often to introduce the topic with two particle problems where the particles are uncorrelated and the wavefunctions for each particle can be separated. As I've tried to emphasize this is only a special case, and if you aren't careful you can get the wrong idea about how to generalize quantum mechanics from 1 to 2 particles.

Answer 4

@Jacob's perfect answer is hard to improve. I will just try to explicitly address one point of confusion that you allude to in your question.

When we speak of $\psi(x_1)$, it does not mean that $x_1$ is the position of particle $1$. It means that $\psi(x_1)$ is the probability amplitude of finding particle $1$ at a position $x_1$. Since there are two particles, there are two positions you can measure simultaneously, and thus, $\psi(x_1)$ is the probability amplitude that particle $1$ is found at $x_1$ and $\phi(x_2)$ is the probability amplitude that particle $2$ is found at $x_2$.

In other words, let's imagine that you didn't even know that you had two particles, what you would still find, when you go out and do the experiment, is that there are two positions that you can simultaneously measure -- so, we need to describe these two probability amplitudes. Just like when we have one position to measure and we speak of $\psi(x)$, we do not mean that the particle is at a particular position $x$ but rather we mean that we are speaking of the probability amplitude of finding the particle at $x$ upon measurement, similarly, when we have two positions, we can speak of $\psi(x_1)$ and $\phi(x_2)$.

As @Jacob's answer explains in mathematical detail, the full wavefunction of the system would be either Fermionic/Bosonic product of $\psi(x_1)$ and $\phi(x_2)$ for identical particles and it would be a simple outer product for distinguishable particles. Intuitively, for distinguishable particles, the full wavefunction can simply be $\psi(x_1)\phi(x_2)$ (very roughly, just like the probabilities of independent events multiply). For indistinguishable particles, you have to make sure that the wavefunction is invariant (up to an overall phase) under exchanging the two particles (arbitrary number of times) -- this leads to two options, the symmetric/anti-symmetric $\psi(x_1)\phi(x_2)\pm\psi(x_2)\phi(x_1)$.

Answer 5

I would have thought that wave function (even for a two-particle system) would just have a value for each point in space and in time and so would just be: $\Psi(r,t)$.

Well that is not how the Schroedinger equation for systems of many particles works. It is hard to see how could it work - what would the Hamiltonian to use for interacting two, three or more electrons that only refers to single position vector $\mathbf r$ be? We need more vectors to describe configuration and interaction of many-particle systems. The standard way to describe interaction of electrons is via Colombic terms

$$ \frac{Kq_iq_k}{|\mathbf r_i-\mathbf r_k|} $$ where obviously we need different coordinates $\mathbf r_i$ $\mathbf r_k$. If they were the same, the interaction terms would be always infinite and meaningless.

One can get away with using a single vector $\mathbf r$ for describing single electron atoms, because the nucleus can be fixed, but with $n$ electrons, we need $n$ position vectors.

I suppose I even have an issue with "the coordinates of particle one" - doesn't that imply that we know the exact position of the particles, when we are trying to find their wave function, which is necessarily spread out.

No, it doesn't. It is just a coordinate in configuration space, we do not know its "value". It is just a mathematical variable to help us make calculations of expected averages of various quantities (including the coordinates). But there is no need to know or to distinguish the particles to use the coordinates and the rules of quantum theory.