Prerequisite Introduction to Many-Body Quantum Mechanics, Angular momentum addition (quantum mechanics)
, spin angular momentum
In quantum mechanics, we assume that many elementary particles are indistinguishable, such as electrons, protons, neutrons, etc. We call the indistinguishable particles identical particles. For example, if we measure two electrons at two locations at the same time, we can only know that there is one electron in each location, but not which electron is where. For macroscopic objects, we can easily distinguish them by marking or tracking their trajectories, but particles in quantum mechanics can neither be marked nor have the concept of trajectories, so the only way we can distinguish particles is to use them. Physical properties, such as mass, charge, spin and other physical properties.
Symmetry of wave function
Regardless of spin , how do we use their wave functions to embody "indistinguishability"? For the two-particle wave function $\psi( \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2)$ under the position representation, the probability density function of two electrons is found at any two points $ \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2$ at the same time. The link is not completed
\begin{equation}
f( \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2) = \left\lvert \psi( \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2) \right\rvert ^2
\end{equation}
In order to reflect that the two particles are indistinguishable, we require the probability distribution to satisfy $f( \boldsymbol{\mathbf{r}} _2, \boldsymbol{\mathbf{r}} _1) = f( \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2)$. So the wave function must satisfy one of the following two conditions
\begin{equation}
\psi( \boldsymbol{\mathbf{r}} _2, \boldsymbol{\mathbf{r}} _1) = \pm \psi( \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2)
\end{equation}
If we take a positive sign, we call such a wave function
symmetric (symmetric), and if it takes a negative sign, we call it
antisymmetric. The wave function of two identical particles can only take one of these two situations. If the wave function is neither symmetric nor antisymmetric, then it is called
not symmetric. Note that for the wave function of the identical particle, the index of $ \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2$ no longer represents the number of the particle, but is only used to distinguish two different positions.
We will see later that a wave function with commutative symmetry or antisymmetric will always maintain its symmetry during the evolution process according to the Schrödinger equation.
Symmetry of Spin State
If we only consider the spin state vector space of two particles, we also require that the state vectors of identical particles in this space satisfy exchange symmetry due to probabilistic reasons.
Example 1
Take the common two-electron spin as an example. The single-electron spin dimension is $1/2$, the spin state is in a 2-dimensional space, and the two bases are usually denoted as $\uparrow$ and $\downarrow$, which represent the spin $z$ component $m_s=\pm 1/2$ respectively. The spin state of a two-electron is the tensor product of the space and itself, which is $2\times2 = 4$ dimensional, and the orthogonal normalized basis can naturally be taken as $\uparrow\uparrow, \uparrow\downarrow, \downarrow\uparrow, \downarrow\downarrow$. But the middle two are neither symmetric nor anti-symmetric, so we can slightly transform them to get 4 orthogonal normalized bases with commutative symmetry.
\begin{equation}
\uparrow\uparrow \qquad \frac{1}{\sqrt 2}(\uparrow\downarrow + \downarrow\uparrow) \qquad \downarrow\downarrow
\end{equation}
\begin{equation}
\frac{1}{\sqrt 2}(\uparrow\downarrow - \downarrow\uparrow)
\end{equation}
Among them, the three bases of
eq. 3 are symmetrical, namely
triplet; one of the bases of
eq. 4 is antisymmetric, namely
singlet. Obviously, any linear combination of triplet states is a symmetric spin state
1, they form a 3-dimensional subspace
, and antisymmetric There is only a 1-dimensional subspace with a singlet as the basis.
But how exactly is the symmetry defined here? We can use $\chi_{1,2}$ to represent any vector in this 4-dimensional space. To define $\chi_{2,1}$ is to swap the two arrows (single-event states) in each item of $\chi_{1,2}$. For example, $\downarrow\uparrow$ becomes $\uparrow\downarrow$, or $\downarrow\uparrow - \uparrow\downarrow$ becomes $\uparrow\downarrow - \downarrow\uparrow$. So the (anti)symmetric condition is
\begin{equation}
\chi_{2,1} = \pm\chi_{1,2}
\end{equation}
We can regard $\uparrow\uparrow$ and $\downarrow\downarrow$ as the state where the spin directions of the two electrons are the same, and $(\uparrow\downarrow \pm \downarrow\uparrow)/\sqrt{2}$ as the state where the spin directions of the two electrons are opposite. For the latter, we can only say "one electron spin is $\uparrow$ and the other is $\downarrow$" but we cannot say which is $\uparrow$ and which is $\downarrow$. This just reflects the indistinguishability of identical particles. In contrast, $\uparrow\downarrow$ means that the first electron is $\uparrow$ and the second is $\downarrow$, so it cannot be used for two electrons.
Generally, for two identical particles with a spin quantum number of $s$, a single particle is in a $N = 2s+1$-dimensional vector space, and the two-particle state space is $N^2$-dimensional. The symmetric subspace is $N(N+1)/2$ dimensional, and the antisymmetric subspace is $N(N-1)/2$ dimensional, and the sum is exactly equal to $N^2$. So they are two complementary orthogonal subspaces . We leave the derivation in the "particle exchange operator ".
Bosons and Fermions
When we consider the total state of two particles, we can regard the total space as the tensor product space of the two-particle wave function space and the two-particle spin state space. The two-particle state vector can always be decomposed into a linear combination of the wave function and the tensor product of the spin.
\begin{equation}
\left\lvert \Psi_{1,2} \right\rangle = \sum_i \psi_i( \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2) \chi^{(i)}_{1,2}
\end{equation}
What we often discuss is just a simple state that can be recorded as one item
\begin{equation}
\left\lvert \Psi_{1,2} \right\rangle = \psi( \boldsymbol{\mathbf{r}} _1, \boldsymbol{\mathbf{r}} _2) \chi_{1,2}
\end{equation}
Similar to
eq. 2 , identical particles need to satisfy
\begin{equation}
\left\lvert \Psi_{1,2} \right\rangle = \pm \left\lvert \Psi_{2,1} \right\rangle
\end{equation}
For
eq. 7 , $ \left\lvert \Psi_{1,2} \right\rangle $ is symmetric if and only if the wave function and spin have the same symmetry (both symmetric or both are antisymmetric), and $ \left\lvert \Psi_{1,2} \right\rangle $ is antisymmetric if and only if the wave function And spin have opposite symmetry (one symmetry and one antisymmetric).
The two exchange symmetries of wave function or spin will bring experimentally observable results. Experiments show that we can divide all elementary particles into two categories. Those with an integer spin are called bosons (boson), which have a symmetric state vector; those with a half-integer spin are called 费 Fermion, has an antisymmetric state vector. This is the basic hypothesis of quantum mechanics regarding identical particles. As for why spin can determine the symmetry of the wave function, quantum field theory is needed to explain it.
For example, an electron is a fermion with a spin of 1/2. In order to ensure the antisymmetric of the total state, if the spins of the two electrons are singlet (antisymmetric), then the wave function must be symmetric; otherwise, if the two electrons are in triplet State (symmetric), then the wave function must be antisymmetric.
The case of multiple particles
Above we have only discussed two identical particles. If there are more, then fermions and bosons respectively require the state vector of the system to maintain symmetry and antisymmetric when exchanging any two particles.
1. ^ when we say so, it is not normalized by default