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},t) & = \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},t) & =\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

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

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:

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

Where last are quantum operators. Following the same idea we make the identification:

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

The Hamiltonian can be written as

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

COMMENTS

i) 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.

ii) Due to this, the vacuum has to be treated as a quantum dynamical object. Therefore we can define to this object a quantum state.

iii) The perturbation of this quantum field is called photon (it is called the quanta of the electromagnetic field).

iv) Consider the ground state of the vacuum as $|0\rangle$ with energy equal to zero. Following the knowledge of the harmonic oscillator $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$ represent an exited state or the ground state with energy $\hbar \omega _{\mathbf {k} }$. This means that $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ creates a photon with energy $\hbar \omega _{\mathbf {k} }$. Now consider the following normalized state: ${\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$.