1
$\begingroup$

Basically, what are all the parameters that completely describe an electron in quantum theory.

In classical physics a complete and fundamental description of an electron is given by its mass, charge, and position. So $(m, q, \vec{x}(t))$ gives you complete information about the electron.

I would like the analogous description of the quantum electron. Typically I only see electrons described by a state vector over position or a state vector over spin. But this page explains how to represent both . I am guessing the quantum picture of an electron is something like $(m, q, \phi(\vec{x},s;t))$ where $\left| \phi\right\rangle$ is the state vector and $(\vec{x},s)$ is its "index". $s\in\{-{1\over2}\hbar,{1\over2}\hbar\}$ is the spin measured along some axis in space. Am I missing anything? Like for instance, lepton number? Does this picture change when relativity is introduced? I don't care about how the state evolves for this question, just what mathematical objects are present in the computations.

$\endgroup$
  • 2
    $\begingroup$ Why did you include $m,q$ to the "information"? These are not dynamical variables. They are constants of Nature. One may enumerate many other such constants, like the electron's magnetic moment. But none of them changes with time. I suppose that if your classical starting point is bizarre and unexplained in this way, any "analogous" thing in the quantum theory will have to be similarly bizarre, right? The dynamical information about the electron is only carried by 3 components of the momentum (or similar 3 variables) and 1 quantum bit about the polarization of the spin, that's it. $\endgroup$ – Luboš Motl Jul 19 '14 at 7:39
  • $\begingroup$ I agree with @LubošMotl. I think $m$ and $q$ belongs to the definition of the system, rather than to the description of its state. Although, I've read somewhere, that in QFT $m$ and $q$ are treated as observables with corresponding operators. And that superpositions of eigenstates of this operators aren't physically realizable since they are forbidden by the superselection rules, whatever they are... $\endgroup$ – Wildcat Jul 19 '14 at 8:04
  • 1
    $\begingroup$ Dear Wildcat, no, $m_e$ and $q_e$ are universal constants - also and especially in quantum field theory - so they are $c$-numbers, not $q$-numbers. They are not observables. Well, the total energy/mass $E=mc^2$ and the total charge of a physical system obviously is an operator, it can have different eigenvalues. But these things evaluated at a single electron are just constants, $c$-numbers. $\endgroup$ – Luboš Motl Jul 19 '14 at 8:21
  • $\begingroup$ For the description of free electron you don't need info about it's charges (leptonic, electric etc.). $\endgroup$ – Andrew McAddams Jul 19 '14 at 11:49
  • $\begingroup$ Dear Luboš Motl, I included constants so that my question does not limit itself. But yes the quantum information needed to describe an electron is what I am trying to get at. Your comment answers my question. $\endgroup$ – danabo Jul 19 '14 at 19:10
3
$\begingroup$

$$ \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\bra}[1]{\langle{#1}|} \newcommand{\braket}[2]{\langle{#1}\,|\,{#2}\rangle} \newcommand{\bracket}[3]{\langle{#1}\,|\,{#2}\,|\,{#3}\rangle} $$

Well, I think by specifying mass $m$ and charge $q$ you simply define your system, a single electron, and then its complete non-relativistic description is indeed given by a wave-function $\Psi(\vec{r}, m_s, t)$ as you mentioned. Wave-function of the form $\Psi(\vec{r}, m_s, t)$, also known as spin-orbital, arises as follows.

There is a postulate of QM which says that the state space $H$ of a system, composed of $n$ subsystems each associated with its own space $H_{i}$, is the tensor product of this spaces, \begin{equation} H = H_{1} \otimes H_{1} \otimes \dotsm \otimes H_{n} \, . \end{equation} The notion of a system composed of subsystems in the postulate above is not to be take literally: the state space can be written as a tensor product of state spaces which are not even associated with real physical systems.

For instance, we can subdivide electron with position in space and spin system into 2 subsystems: electron only with position in space and electron only with spin. Taking spin into account the state space for a particle is the tensor product of \begin{equation} H = H_{\text{space}} \otimes H_{\text{spin}} \, , \end{equation} where $H_{\text{space}}$ is a state space spanned by eigenvectors $\ket{r}$ of the position operator \begin{equation} \widehat{\vec{R}} \ket{\vec{r}} = \vec{r} \ket{\vec{r}} \, , \end{equation} and $H_{\text{spin}}$ is a state space spanned by eigenvectors $\ket{m_{s}}$ of a spin component operator conventionally chosen to be $\widehat{S}_{z}$ \begin{equation} \widehat{S}_{z} \ket{m_{s}} = m_{s} \hbar \ket{m_{s}} \, . \end{equation}

The resulting space $H$ is spanned then by $\ket{\vec{r}, m_{s}} = \ket{\vec{r}} \otimes \ket{m_{s}}$ by the property of a tensor product space and the expansion of a state vector $\ket{\Psi} \in H$ then takes the following form \begin{equation} \ket{\Psi(t)} = \sum_{m_s=-s}^{s} \, \int\limits_{\mathrm{R}^{3n}} \Psi_{m_s}(\vec{r}, t) \ket{\vec{r}, m_s} \mathrm{d}^3 \vec{r} \, , \end{equation} where \begin{equation} \Psi_{m_s}(\vec{r}, t) = \braket{\vec{r}, m_s}{\Psi} \, . \end{equation} It should be clear now that coefficients in the expansion of a state vector $\ket{\Psi}$ over the basis $\ket{\vec{r}, m_{s}}$ are given by $2s + 1$ functions $\Psi_{m_s}(\vec{r}, t)$ and that all them are required for the complete description of a state. Thus, for instance, the state of an electron can be represented by a two-row vector \begin{equation} \ket{\Psi(t)} \longleftrightarrow \begin{pmatrix} \Psi_{+1/2}(\vec{r}, t) \\ \Psi_{-1/2}(\vec{r}, t) \end{pmatrix} \, , \end{equation} known as the two-component wavefunction. Alternatively, $\Psi_{+1/2}(\vec{r}, t)$ and $\Psi_{-1/2}(\vec{r}, t)$ can be combined into single piecewise function $\Psi(\vec{r}, m_s, t)$ which can also be used as the description of the state of an electron \begin{equation} \ket{\Psi(t)} \longleftrightarrow \Psi(\vec{r}, m_s, t) = \begin{cases} \Psi_{+1/2}(\vec{r}, t), & m_s = +1/2 \\ \Psi_{-1/2}(\vec{r}, t), & m_s = -1/2 \end{cases} \, . \end{equation} Another equivalent representation of the same idea is one which is usually used in quantum chemistry, where the spin dependence is separated out by introducing the "spin up" and "spin down" spin functions \begin{equation} \alpha(m_s) = \begin{cases} 1, & m_s = +1/2 \\ 0, & m_s = -1/2 \end{cases} \, , \quad \quad \quad \beta(m_s) = \begin{cases} 0, & m_s = +1/2 \\ 1, & m_s = -1/2 \end{cases} \, , \end{equation} and writing down $\Psi(\vec{r}, m_s, t)$ in the following way \begin{equation} \Psi(\vec{r}, m_s, t) = \Psi_{+1/2}(\vec{r}, t) \alpha(m_s) + \Psi_{-1/2}(\vec{r}, t) \beta(m_s) \, . \end{equation}

And since spin is relativistic phenomenon, such description is, in a sense, already relativistic, though, partly. Fully relativistic description is different. The wave function is four-component, rather than two-component, and although, I couldn't tell you how it arises, the role of two additional components is to describe the "spin-up" and "spin-down" state of associated positron.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.