Phy5646

Welcome to the Quantum Mechanics B PHY5646 Spring 2009

Schrodinger equation. The most fundamental equation of quantum mechanics which describes the rule according to which a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.

This is the second semester of a two-semester graduate level sequence, the first being PHY5645 Quantum A. Its goal is to explain the concepts and mathematical methods of Quantum Mechanics, and to prepare a student to solve quantum mechanics problems arising in different physical applications. The emphasis of the courses is equally on conceptual grasp of the subject as well as on problem solving. This sequence of courses builds the foundation for more advanced courses and graduate research in experimental or theoretical physics.

The key component of the course is the collaborative student contribution to the course Wiki-textbook. Each team of students (see Phy5646 wiki-groups) is responsible for BOTH writing the assigned chapter AND editing chapters of others.

This course's website can be found here.

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Outline of the course:

Stationary state perturbation theory in Quantum Mechanics

Very often, quantum mechanical problems cannot be solved exactly. We have seen last semester that an approximate technique can be very useful since it gives us quantitative insight into a larger class of problems which do not admit exact solutions. The technique we used last semester was WKB, which holds in the asymptotic limit Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar\rightarrow 0} .

Perturbation theory is another very useful technique, which is also approximate, and attempts to find corrections to exact solutions in powers of the terms in the Hamiltonian which render the problem insoluble.

Typically, the (Hamiltonian) problem has the following structure

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}=\mathcal{H}_0+\mathcal{H}'}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}_0} is exactly soluble and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}'} makes it insoluble.

Raleigh-Shrödinger Peturbation Theory

To do this we first consider an auxiliary problem, parameterized by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H} = \mathcal{H}_0 + \lambda \mathcal{H}^'}

If we attempt to find eigenstates Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N(\lambda)\rangle} and eigenvalues Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n} of the Hermitian operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , and assume that they can be expanded in a power series of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E(\lambda) = E_0 + \lambda E_1 + ... + \lambda^n E_n + ...}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N(\lambda)\rangle = |\Psi_n^{(0)}\rangle + \lambda|\Psi_n^{(1)}\rangle + \lambda^2 |\Psi_n^{(2)}\rangle + ... \lambda^j |\Psi_n^{(j)}\rangle + ...}

then they must obey the equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H} |N(\lambda)\rangle = E(\lambda) |N(\lambda)\rangle } .

Which, upon expansion, becomes:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathcal{H}_0 + \lambda \mathcal{H}')(\Sigma_{j=0}^{\infty}\lambda^j |j\rangle ) = (\Sigma_{l=0}^{\infty} \lambda^l E_l)(\Sigma_{j=0}^{\infty}\lambda^j |j\rangle)}

We shall denote the unperturbed states as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle} . We choose the normalization such that the unperturbed states are normalized, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n | n \rangle = 1} , and that the exact state satisfies Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|N(\lambda)\rangle=1} . Note that in general Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle} will not be normalized.

In order for this method to be useful, the perturbed energies must vary continuously with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} . Knowing this we can see several things about our, as yet undetermined peterbed energies and eigenstates. For one, as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \rightarrow 0, |N(\lambda)\rangle \rightarrow |n\rangle} , for some unperturbed state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle} .

Perturbation correction eigenstates states are orthogonal unperturbed states,

Brillouin-Wigner Peturbation Theory

Interaction of radiation and matter

Quantization of electromagnetic radiation

Additional Reading

• Experimental observation of a Lamb-like shift in a solid state setup Science 322, 1357 (2008).

Classical view

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Let's use transfer gauge (sometimes called Coulomb gauge):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi (\mathbf{r},t)=0 }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \mathbf{A}=0}

In this gauge the electromagnetic fields are given by:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{E}(\mathbf{r},t)=-\frac{1}{c}\frac{\partial \mathbf{A} }{\partial t}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{B}(\mathbf{r},t)=\nabla \times \mathbf{A}}

The energy in this radiation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon = \frac{1}{8\pi} \int d^{3}\mathbf{r} (\mathbf{E}^{2}+\mathbf{B}^{2})}

The rate and direction of energy transfer are given by poynting vector

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{P} = \frac{c}{4\pi} \mathbf{E} \times \mathbf{B} }

The radiation generated by classical current is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Box \mathbf{A} = -\frac{4\pi}{c} \mathbf{j}}

Where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Box} is the d'Alembert operator. Solutions in the region where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{j}=0} are given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{A}(\mathbf{r},t) = \alpha \boldsymbol{\lambda}\frac{e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}}{\sqrt{V}}+\alpha^{*} \boldsymbol{\lambda}^{*} \frac{e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}}{\sqrt{V}} }

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega=c|\mathbf{k}|} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{\lambda}\cdot \mathbf{k}=0 } in order to satisfy the transversality. Here the plane waves are normalized respect to a volume Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} . This is just for convenience and the physics wont change. We can choose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{\lambda}\cdot\boldsymbol{\lambda}^{*}=1} . Notice that in this writing Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{A}} is a real vector.

Let's compute Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varepsilon} . For this

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{E}(\mathbf{r},t) & =-\frac{1}{c}\frac{\partial \mathbf{A} }{\partial t} \\ & =-\frac{1}{c\sqrt{V}}\frac{\partial}{\partial t}\left[\alpha \boldsymbol{\lambda}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}+\alpha^{*} \boldsymbol{\lambda}^{*} e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right] \\ & =-\frac{i\omega}{c\sqrt{V}}\left[-\alpha \boldsymbol{\lambda} e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}+\alpha^{*} \boldsymbol{\lambda}^{*} e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right] \\ \mathbf{E}^{2}(\mathbf{r},t) & = -\frac{\omega^{2}}{c^{2}V}\left[\alpha^{2} \boldsymbol{\lambda}^{2} e^{2i(\mathbf{k}\cdot\mathbf{r}-\omega t)}-\alpha\alpha^{*}\boldsymbol{\lambda}\cdot\boldsymbol{\lambda}^{*}-\alpha^{*}\alpha\boldsymbol{\lambda}^{*}\cdot\boldsymbol{\lambda}+\alpha^{*2} \boldsymbol{\lambda}^{*2} e^{-2i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right] \\ \end{align} }

Taking the average, the oscillating terms will disappear. Then we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{E}^{2}(\mathbf{r}) & = \frac{\omega^{2}}{c^{2}V}\left[\alpha\alpha^{*}+\alpha^{*}\alpha\right] \\ &=2\frac{\omega^{2}}{c^{2}V}|\alpha|^2 \\ \end{align} }

It is well known that for plane waves Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{B}=\mathbf{n}\times \mathbf{E} } , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{n}} is the direction of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{k}} . This clearly shows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{B}^{2}=\mathbf{E}^{2}} . However let's see this explicitly:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{B}(\mathbf{r},t) & =\nabla \times\mathbf{A}\\ & =\nabla \left[\alpha \boldsymbol{\lambda}\frac{e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}}{\sqrt{V}}+\alpha^{*} \boldsymbol{\lambda}^{*} \frac{e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}}{\sqrt{V}}\right] \\ \end{align} }

Each component is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{B}_{i}(\mathbf{r},t)& =\frac{1}{{\sqrt{V}}}\left[\alpha \varepsilon _{ijk}\partial_{j} \left(\boldsymbol{\lambda}_{k}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right)+\alpha^{*} \varepsilon _{ijk}\partial_{j} \left(\boldsymbol{\lambda}^{*}_{k}e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right)\right] \\ & =\frac{i}{{\sqrt{V}}}\left[\alpha \varepsilon _{ijk}\mathbf{k}_{j} \boldsymbol{\lambda}_{k}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}-\alpha^{*} \varepsilon _{ijk}\mathbf{k}_{j} \boldsymbol{\lambda}^{*}_{k}e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right] \\ \end{align} }

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{B}(\mathbf{r},t) & =\frac{i}{{\sqrt{V}}}\left[\alpha \mathbf{k}\times\boldsymbol{\lambda}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}-\alpha^{*} \mathbf{k}\times\boldsymbol{\lambda}^{*} e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right] \\ \mathbf{B}^{2}(\mathbf{r},t) & =\frac{-1}{{V}}\left[\alpha \mathbf{k}\times\boldsymbol{\lambda}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}-\alpha^{*} \mathbf{k}\times\boldsymbol{\lambda}^{*} e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right] \\ & =\frac{-1}{{V}}\left[\alpha^{2} (\mathbf{k}\times\boldsymbol{\lambda})^{2}e^{2i(\mathbf{k}\cdot\mathbf{r}-\omega t)}-\alpha \alpha^{*}(\mathbf{k}\times\boldsymbol{\lambda})\cdot(\mathbf{k}\times\boldsymbol{\lambda}^{*})-\alpha^{*} \alpha(\mathbf{k}\times\boldsymbol{\lambda}^{*})\cdot(\mathbf{k}\times\boldsymbol{\lambda})+\alpha^{*2} (\mathbf{k}\times\boldsymbol{\lambda}^{*})^{2} e^{-2i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\right] \\ \end{align} }

Again taking the average the oscillating terms vanish. Then we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathbf{B}^{2}(\mathbf{r}) & =\frac{1}{{V}}\left[\alpha \alpha^{*}+\alpha^{*} \alpha\right](\mathbf{k}\times\boldsymbol{\lambda})\cdot(\mathbf{k}\times\boldsymbol{\lambda}^{*}) \\ & =\frac{1}{{V}}\left[\alpha \alpha^{*}+\alpha^{*} \alpha\right][\mathbf{k}^{2}(\boldsymbol{\lambda}\cdot\boldsymbol{\lambda^{*}})-(\mathbf{k}\cdot\boldsymbol{\lambda^{*}})\cdot(\mathbf{k}\cdot\boldsymbol{\lambda^{*}})] \\ & =\frac{2}{{V}}|\alpha|^{2}\mathbf{k}^{2}\\ &=2\frac{\omega^{2}}{c^{2}V}|\alpha|^2 \\ &= \mathbf{E}^{2}(\mathbf{r},t)\\ \end{align} }

Finally the energy of this radiation is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \varepsilon &= \frac{1}{8\pi} \int d^{3}\mathbf{r} (\mathbf{E}^{2}+\mathbf{B}^{2}) \\ &=\frac{1}{4\pi} \int d^{3}\mathbf{r}\; \mathbf{E}^{2}\\ &=\frac{1}{4\pi} \int d^{3}\mathbf{r} \left(2\frac{\omega^{2}}{c^{2}V}|\alpha|^2\right)\\ &=\frac{\omega^{2}}{2\pi c^{2}}|\alpha|^2\\ \end{align}}

So far we have treated the potential Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(\mathbf{r},t)} as a combination of two waves with the same frequency. Now let's extend the discussion to any form of $A(\mathbf {r} ,t)$. To do this we can expand $A(\mathbf {r} ,t)$ in a infinite series of Fourier:

${\begin{aligned}A(\mathbf {r} ,t)=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\\\end{aligned}}$

To calculate the energy with use the fact that any exponential time-dependent term is in average zero. Therefore in the previous sum all cross terms with different $\mathbf {k}$ vanishes. Then it is clear that

${\begin{aligned}\mathbf {E} ^{2}(\mathbf {r} )&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\omega ^{2}}{c^{2}V}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\\mathbf {B} ^{2}(\mathbf {r} )&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\mathbf {k} ^{2}}{V}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\\end{aligned}}$

Then the energy is given by

${\begin{aligned}\varepsilon &={\frac {1}{8\pi }}\int d^{3}\mathbf {r} (\mathbf {E} ^{2}+\mathbf {B} ^{2})\\&={\frac {1}{4\pi }}\int d^{3}\mathbf {r} \;\mathbf {E} ^{2}\\&={\frac {1}{4\pi }}\int d^{3}\mathbf {r} \sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\omega ^{2}}{c^{2}V}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\&={\frac {1}{4\pi }}\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\omega ^{2}}{c^{2}}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\\end{aligned}}$

Let's define the following quantities:

${\begin{aligned}Q_{\mathbf {k} {\boldsymbol {\lambda }}}&={\frac {1}{{\sqrt {4\pi }}c}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*})\\P_{\mathbf {k} {\boldsymbol {\lambda }}}&={\frac {-i\omega }{{\sqrt {4\pi }}c}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}-A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*})\\\end{aligned}}$

Notice that

${\begin{aligned}\omega ^{2}Q_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}&={\frac {\omega ^{2}}{4\pi c^{2}}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}+A_{\mathbf {k} {\boldsymbol {\lambda }}}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*2})\\P_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}&={\frac {-\omega ^{2}}{4\pi c^{2}}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}-A_{\mathbf {k} {\boldsymbol {\lambda }}}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}-A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*2})\\\end{aligned}}$

Adding

${\begin{aligned}P_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}+\omega ^{2}Q_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}&={\frac {\omega ^{2}}{2\pi c^{2}}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}})\\\end{aligned}}$

Then the energy (in this case the Hamiltonian) can be written as

${\begin{aligned}H={\frac {1}{2}}\sum _{\mathbf {k} {\boldsymbol {\lambda }}}P_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}+\omega ^{2}Q_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}\end{aligned}}$

We end up with a collection of harmonic oscillators, each labeled by $\mathbf {k}$ adn ${\boldsymbol {\lambda }}$, whose frequencies depends on $|\mathbf {k} |$.

From classical mechanics to quatum mechanics for radiation

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As usual we proceed to do the canonical quantization:

${\begin{aligned}P_{\mathbf {k} {\boldsymbol {\lambda }}}&\to \mathbf {P} _{\mathbf {k} {\boldsymbol {\lambda }}}\\Q_{\mathbf {k} {\boldsymbol {\lambda }}}&\to \mathbf {Q} _{\mathbf {k} {\boldsymbol {\lambda }}}\\\end{aligned}}$

${\begin{aligned}A_{\mathbf {k} {\boldsymbol {\lambda }}}&\to {\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\;\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\;,\;[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}},\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }]=\delta _{\mathbf {kk'} }\delta _{\boldsymbol {\lambda \lambda '}}\\\end{aligned}}$

Where last are quantum operators. The Hamiltonian can be written as

${\begin{aligned}\mathbf {H} _{radiation}&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\hbar \omega _{\mathbf {k} }(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}+{\frac {1}{2}})&={\frac {1}{2}}\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\hbar \omega _{\mathbf {k} }(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger })\\\end{aligned}}$

The classical potential can be written as

$\underbrace {A(\mathbf {r} ,t)=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]} _{\textrm {ClassicalVectorpotential}}\;\;\;\longrightarrow \;\;\;\underbrace {\mathbf {A} _{int}(\mathbf {r} ,t)={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]} _{\textrm {QuantumOperator}}$

Notice that the quantum operator is time dependent. Therefor we can identify it as the field operator in interaction representation. (That's the reason to label it with int). Let's find the Schrodinger representation of the field operator:

${\begin{aligned}\mathbf {A} (\mathbf {r} )&=e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {A} _{int}(\mathbf {r} ,t)e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\\&=e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\left[{\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\right]e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\\&={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\left[e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\right]{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\left[e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\right]{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\\&={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}e^{i\omega t}\right]{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }e^{-i\omega t}\right]{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\\&={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i\mathbf {k} \cdot \mathbf {r} }}{\sqrt {V}}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }{\boldsymbol {\lambda }}^{*}{\frac {e^{-i\mathbf {k} \cdot \mathbf {r} }}{\sqrt {V}}}\right]\\\end{aligned}}$

COMMENTS

• The meaning of $\mathbf {H} _{radiation}$ is as following: The classical electromagnetic field is quantized. This quantum field exist even if there is not any source. This means that the vacuum is a physical object who can interact with matter. In classical mechanics this doesn't occur because, fields are created by sources.
• Due to this, the vacuum has to be treated as a quantum dynamical object. Therefore we can define to this object a quantum state.
• The perturbation of this quantum field is called photon (it is called the quanta of the electromagnetic field).

ANALYSIS OF THE VACUUM AT GROUND STATE

Let's call $|0\rangle$ the ground state of the vacuum. The following can be stated:

• The energy of the ground state is infinite. To see this notice that for ground state we have ${\begin{aligned}\mathbf {H} _{radiation}&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {1}{2}}\hbar \omega _{\mathbf {k} }=\infty \end{aligned}}$
• The state $\;\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$ represent an exited state of the vacuum with energy $\hbar \omega _{\mathbf {k} }(1+1/2)$. This means that the extra energy $\hbar \omega _{\mathbf {k} }$is carried by a single photon. Therefore $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ represent the creation operator of one single photon with energy $\hbar \omega _{\mathbf {k} }$. In the same reasoning, $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}$ represent the annihilation operator of one single photon.
• Consider the following normalized state of the vacuum: ${\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$. At the first glance we may think that $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ creates a single photon with energy $2\hbar \omega _{\mathbf {k} }$. However this interpretation is forbidden in our model. Instead, this operator will create two photons each of the carryng the energy $\hbar \omega _{\mathbf {k} }$.

  Proof

  Suppose that $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ creates a single photon with energy $2\hbar \omega _{\mathbf {k} }$. We can find an operator $\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }$ who can create a photon with the same energy $2\hbar \omega _{\mathbf {k} }$. This means that

  ${\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle =\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle \;\;\;\longrightarrow \;\;\;{\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }=\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\;\;\;\longrightarrow \;\;\;{\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}=\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}$

  using this we can see that

  ${\begin{aligned}\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}},\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\right]&=0\\\left[\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}},\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\right]&=0\\\end{aligned}}$

  Since $\left[\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}},\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\right]=1$, the initial assumption is wrong, namely:

  ${\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle \neq \mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$

  This means that $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ cannot create a single photon with energy $2\hbar \omega _{\mathbf {k} }$. Instead it will create two photons each of them with energy $\hbar \omega _{\mathbf {k} \blacksquare }$

ALGEBRA OF VACUUM STATES

A general vacuum state can be written as

$|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle$

where $n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}$ is the number of photons in the state $\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}$ which exist in the vacuum. Using our knowledge of harmonic oscillator we conclude that this state can be written as

$|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle =\prod _{\mathbf {k_{j}} {\boldsymbol {\lambda _{j}}}}{\frac {(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger })^{n_{\mathbf {k_{j}} {\boldsymbol {\lambda _{j}}}}}}{\sqrt {n_{\mathbf {k_{j}} {\boldsymbol {\lambda _{j}}}}!}}}|0\rangle$

Also it is clear that

$\mathbf {a} _{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}^{\dagger }|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle ={\sqrt {n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}+1}}|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}+1;...\rangle$

$\mathbf {a} _{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle ={\sqrt {n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}}}|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}-1;...\rangle$

Matter + Radiation: Electron bounded to a nucleus with transverse radiation field

---

The Hamiltonian of the system can be written as:

${\begin{aligned}\mathbf {H} &=\mathbf {H} _{particle+radiation}+\mathbf {H} _{radiation}\\&={\frac {(\mathbf {P} -{\frac {e}{c}}\mathbf {A} (\mathbf {r} ))^{2}}{2m}}+\mathbf {V(\mathbf {r} )} +\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\hbar \omega _{\mathbf {k} }(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}+{\frac {1}{2}})\end{aligned}}$

Notice that in this writing the Hamiltonian is time independent. The field operator $\mathbf {A} (\mathbf {r} )$ must be time independent. The state of whole system can be written as

$|\psi \rangle =|m;\{n_{\mathbf {k} {\boldsymbol {\lambda }}}\}\rangle =|m\rangle \otimes |\{n_{\mathbf {k} {\boldsymbol {\lambda }}}\}\rangle$

The Schrodinger equation can be transformed to interaction representation (all operators with be time dependent). This transformation is give by:

${\begin{aligned}|\psi \rangle &=e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\|\chi \rangle &=e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \;\;\;\longleftarrow \;\;\;\mathbf {H} _{par}={\frac {\mathbf {P} ^{2}}{2m}}+\mathbf {V(\mathbf {r} )} \\\end{aligned}}$

Then

${\begin{aligned}i\hbar {\frac {\partial }{\partial t}}|\psi \rangle &=[\mathbf {H} _{par+rad}+\mathbf {H} _{rad}]|\psi \rangle \\i\hbar {\frac {\partial }{\partial t}}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle &=[\mathbf {H} _{par+rad}+\mathbf {H} _{rad}]e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\\mathbf {H} _{rad}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle +i\hbar e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}{\frac {\partial }{\partial t}}|\chi \rangle &=[\mathbf {H} _{par+rad}+\mathbf {H} _{rad}]e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {H} _{par+rad}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\left[{\frac {(\mathbf {P} -{\frac {e}{c}}\mathbf {A} (\mathbf {r} ))^{2}}{2m}}+\mathbf {V(\mathbf {r} )} \right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=\left[{\frac {e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}(\mathbf {P} -{\frac {e}{c}}\mathbf {A} (\mathbf {r} ))^{2}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}}{2m}}+\mathbf {V} _{int}(\mathbf {r} )\right]|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=\left[{\frac {(\mathbf {P} -{\frac {e}{c}}\mathbf {A} _{int}(\mathbf {r} ,t))^{2}}{2m}}+\mathbf {V} (\mathbf {r} )\right]|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=\left[\left({\frac {\mathbf {P} ^{2}}{2m}}+\mathbf {V} (\mathbf {r} )\right)-2{\frac {e}{2mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle &=\left[\left({\frac {\mathbf {P} ^{2}}{2m}}+\mathbf {V} (\mathbf {r} )\right)-2{\frac {e}{2mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\\mathbf {H} _{par}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\varphi \rangle +i\hbar e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}{\frac {\partial }{\partial t}}|\varphi \rangle &=\left[\mathbf {H} _{par}-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\i\hbar e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}{\frac {\partial }{\partial t}}|\varphi \rangle &=\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\varphi \rangle &=e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\varphi \rangle &=\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}re^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}} ,t)\mathbf {P} _{int}(t)+{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}re^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}} ,t)\right]|\varphi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\varphi \rangle &=\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} _{int}(t),t)\mathbf {P} _{int}(t)+{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} _{int}(t),t)\right]|\varphi \rangle \\\end{aligned}}$

Spin

Spin 1/2 Angular Momentum

The angular momentum of a stationary spin 1/2 particle is found to be quantized to the $\pm {\frac {\hbar }{2}}$ regardless of the direction of the axis chosen to measure the angular momentum. There is a vector operator ${\textbf {S}}=(S_{x},S_{y},S_{z})$ when projected along an arbitrary axis satisfies the following equations:

${\textbf {S}}\cdot {\hat {m}}|{\hat {m}}\uparrow \rangle ={\dfrac {\hbar }{2}}|{\hat {m}}\uparrow \rangle$

${\textbf {S}}\cdot {\hat {m}}|{\hat {m}}\downarrow \rangle ={\dfrac {-\hbar }{2}}|{\hat {m}}\downarrow \rangle$

$|{\hat {m}}\uparrow \rangle$ and $|{\hat {m}}\downarrow \rangle$ form a complete basis, which means that any state $|{\hat {n}}\uparrow \rangle$ and $|{\hat {n}}\downarrow \rangle$ can be expanded as a linear combination of $|{\hat {m}}\uparrow \rangle$ and $|{\hat {m}}\downarrow \rangle$.

The spin angular momentum operator obeys the standard commutation relations

$[S_{\mu },S_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }S_{\lambda }\Rightarrow [S_{x},S_{z}]=-i\hbar S_{y}$

The most commonly used basis is the one which diagonalizes ${\textbf {S}}\cdot {\hat {z}}=S_{z}$.

By acting on the states $|{\hat {z}}\uparrow \rangle$ and $|{\hat {z}}\downarrow \rangle$ with $S_{z}$, we find $S_{z}|{\hat {z}}\uparrow \rangle ={\dfrac {\hbar }{2}}|{\hat {z}}\uparrow \rangle$ $S_{z}|{\hat {z}}\downarrow \rangle ={\dfrac {-\hbar }{2}}|{\hat {z}}\downarrow \rangle$

Now by acting to the left with another state, we can form a 2x2 matrix.

$S_{z}=\left({\begin{array}{ll}\langle {\hat {z}}\uparrow |S_{z}|{\hat {z}}\uparrow \rangle &\langle {\hat {z}}\uparrow |S_{z}|{\hat {z}}\downarrow \rangle \\\langle {\hat {z}}\downarrow |S_{z}|{\hat {z}}\uparrow \rangle &\langle {\hat {z}}\downarrow |S_{z}|{\hat {z}}\downarrow \rangle \end{array}}\right)=\left({\begin{array}{ll}\hbar /2&0\\0&-\hbar /2\end{array}}\right)={\dfrac {\hbar }{2}}\left({\begin{array}{ll}1&0\\0&-1\end{array}}\right)={\dfrac {\hbar }{2}}\sigma _{z}$

where $\sigma _{z}$ is the z Pauli spin matrix. Repeating the steps (or commutator relations), we can solve for the x and y directions.

$S_{x}={\dfrac {\hbar }{2}}\left({\begin{array}{ll}0&1\\1&0\end{array}}\right)={\dfrac {\hbar }{2}}\sigma _{x}$

$S_{y}={\dfrac {\hbar }{2}}\left({\begin{array}{ll}0&-i\\i&0\end{array}}\right)={\dfrac {\hbar }{2}}\sigma _{y}$

Properties of the Pauli Spin Matrices

Each Pauli matrix squared produces the unity matrix

$\sigma _{x}^{2}=\sigma _{y}^{2}=\sigma _{z}^{2}=\left({\begin{array}{ll}1&0\\0&1\end{array}}\right)$

The commutation relation is as follows

$[\sigma _{\mu },\sigma _{\nu }]=2i\epsilon _{\mu \nu \lambda }\sigma _{\lambda }$

and the anticommutator relation

${\sigma _{\mu },\sigma _{\nu }}=\sigma _{\mu }\sigma _{\nu }+\sigma _{\nu }\sigma _{\mu }=2\delta _{\mu \nu }\left({\begin{array}{ll}0&1\\1&0\end{array}}\right)$

For example, if $\sigma _{\mu }\sigma _{\nu }=-\sigma _{\nu }\sigma _{\mu }$

Then

$\sigma _{\mu }\sigma _{\nu }={\dfrac {1}{2}}[\sigma _{\mu },\sigma _{\nu }]+{\dfrac {1}{2}}{\sigma _{\mu },\sigma _{\nu }}=i\epsilon _{\mu \nu \lambda }\sigma _{\lambda }+\delta _{\mu \nu }$

$S_{\mu }S_{\nu }={\dfrac {\hbar ^{2}}{4}}\delta _{\mu \nu }+{\dfrac {i\hbar }{2}}\epsilon _{\mu \nu \lambda }S_{\lambda }$

The above equation is true for 1/2 spins only!!