Notice removed Draw attention by Community♦ occurred Oct 9 '17 at 22:22 Bounty Ended with AccidentalFourierTransform's answer chosen by Community♦ occurred Oct 9 '17 at 22:22 4 added 1035 characters in body edited Oct 3 '17 at 7:32 Quantumwhisp 2,850726 The EM-Field Hamiltonian is, in principle, a functional (with a chosen operator ordering) that is defined on operator-fields $$\hat{A}(x)$$ and $$\partial_\mu \hat{A}$$. If you carry out the calculations and use definitions of $$\hat{B}$$ and $$\hat{E}$$, you'll arrive at: $$\hat{H} = \int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I'll take this as the Definition of the Hamiltonian in future calculations. For the free field, the Ansatz $$\hat{\vec{A}} = \vec{e}(\hat{a_{\vec{k}}}e^{i(\vec{k}\vec{x} - \omega_{\vec{k}}t)} + \hat{a_\vec{k}}^\dagger e^{-i(\vec{k}\vec{x} - \omega_{\vec{k}}t)})$$ satisfies this wave equation. Using linearity, one can superimpose all the solutions, plug them into the definition of the hamiltonian, and arrives at: $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ Now my question is: Can I also use this Hamiltonian in an interacting theory? (For example an EM-Field coupled to an atom). I'm asking because the wave equations that the Heisenberg operators do change. Superimposing the creation and annihilation operators, as shown above, is no longer a solution to the field equation, so I can't express $$\vec{E}$$ and $$\vec{B}$$ no longer, just using $$\hat{a}$$ and $$\hat{a}^{\dagger}$$? How can I still motivate $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ to be the right Hamiltonian? EDIT: It is clear to me that in case of Interaction, there will be an additional Interaction Term, for example, something like $$\hat{\vec{x}} \hat{\vec{E}} \frac{e}{\hbar}$$. I'm clear of that fact. I however want to know if one can always expand quantities like $$\hat{\vec{E}}$$ in Terms of creation and annihilation operators. For example: The full Hamiltonian for an electron interacting with EM-Field would be (assuming dipole approximation): $$\hat{\vec{p}} \frac{1}{2m} + V(\hat{\vec{x}}) + \hat{\vec{x}} \hat{\vec{E}}(\vec{x}_{Atom}) \frac{e}{\hbar} +\int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I want to know if this can in general be expressed using creation and annihilation operators instead of $$\hat{\vec{E}}$$ and $$\hat{\vec{B}}$$, for example like: $$\hat{\vec{p}} \frac{1}{2m} + V(\hat{\vec{x}}) + \hat{\vec{x}} \hat{\vec{E}}(\vec{x}_{Atom}) \frac{e}{\hbar} + \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ The EM-Field Hamiltonian is, in principle, a functional (with a chosen operator ordering) that is defined on operator-fields $$\hat{A}(x)$$ and $$\partial_\mu \hat{A}$$. If you carry out the calculations and use definitions of $$\hat{B}$$ and $$\hat{E}$$, you'll arrive at: $$\hat{H} = \int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I'll take this as the Definition of the Hamiltonian in future calculations. For the free field, the Ansatz $$\hat{\vec{A}} = \vec{e}(\hat{a_{\vec{k}}}e^{i(\vec{k}\vec{x} - \omega_{\vec{k}}t)} + \hat{a_\vec{k}}^\dagger e^{-i(\vec{k}\vec{x} - \omega_{\vec{k}}t)})$$ satisfies this wave equation. Using linearity, one can superimpose all the solutions, plug them into the definition of the hamiltonian, and arrives at: $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ Now my question is: Can I also use this Hamiltonian in an interacting theory? (For example an EM-Field coupled to an atom). I'm asking because the wave equations that the Heisenberg operators do change. Superimposing the creation and annihilation operators, as shown above, is no longer a solution to the field equation, so I can't express $$\vec{E}$$ and $$\vec{B}$$ no longer, just using $$\hat{a}$$ and $$\hat{a}^{\dagger}$$? How can I still motivate $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ to be the right Hamiltonian? The EM-Field Hamiltonian is, in principle, a functional (with a chosen operator ordering) that is defined on operator-fields $$\hat{A}(x)$$ and $$\partial_\mu \hat{A}$$. If you carry out the calculations and use definitions of $$\hat{B}$$ and $$\hat{E}$$, you'll arrive at: $$\hat{H} = \int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I'll take this as the Definition of the Hamiltonian in future calculations. For the free field, the Ansatz $$\hat{\vec{A}} = \vec{e}(\hat{a_{\vec{k}}}e^{i(\vec{k}\vec{x} - \omega_{\vec{k}}t)} + \hat{a_\vec{k}}^\dagger e^{-i(\vec{k}\vec{x} - \omega_{\vec{k}}t)})$$ satisfies this wave equation. Using linearity, one can superimpose all the solutions, plug them into the definition of the hamiltonian, and arrives at: $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ Now my question is: Can I also use this Hamiltonian in an interacting theory? (For example an EM-Field coupled to an atom). I'm asking because the wave equations that the Heisenberg operators do change. Superimposing the creation and annihilation operators, as shown above, is no longer a solution to the field equation, so I can't express $$\vec{E}$$ and $$\vec{B}$$ no longer, just using $$\hat{a}$$ and $$\hat{a}^{\dagger}$$? How can I still motivate $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ to be the right Hamiltonian? EDIT: It is clear to me that in case of Interaction, there will be an additional Interaction Term, for example, something like $$\hat{\vec{x}} \hat{\vec{E}} \frac{e}{\hbar}$$. I'm clear of that fact. I however want to know if one can always expand quantities like $$\hat{\vec{E}}$$ in Terms of creation and annihilation operators. For example: The full Hamiltonian for an electron interacting with EM-Field would be (assuming dipole approximation): $$\hat{\vec{p}} \frac{1}{2m} + V(\hat{\vec{x}}) + \hat{\vec{x}} \hat{\vec{E}}(\vec{x}_{Atom}) \frac{e}{\hbar} +\int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I want to know if this can in general be expressed using creation and annihilation operators instead of $$\hat{\vec{E}}$$ and $$\hat{\vec{B}}$$, for example like: $$\hat{\vec{p}} \frac{1}{2m} + V(\hat{\vec{x}}) + \hat{\vec{x}} \hat{\vec{E}}(\vec{x}_{Atom}) \frac{e}{\hbar} + \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ Tweeted twitter.com/StackPhysics/status/914902986245181441 occurred Oct 2 '17 at 17:20 3 deleted 20 characters in body edited Oct 1 '17 at 21:31 AccidentalFourierTransform 26k1473131 The EM-Field Hamiltonian is, in principle, a functional (with a choosenchosen operator ordering) that is defined on operator-fields $$\hat{A}(x)$$ and $$\hat{\partial_\mu A}$$$$\partial_\mu \hat{A}$$. If you carriecarry out the calculations and use definitions of $$\hat{B}$$ and $$\hat{E}$$, you'll arrive at: $$\hat{H} = \int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I'll take this as the Definition of the Hamiltonian in future calculations. For the free field, the Ansatz $$\hat{\vec{A}} = \vec{e}(\hat{a_{\vec{k}}}e^{i(\vec{k}\vec{x} - \omega_{\vec{k}}t)} + \hat{a_\vec{k}}^\dagger e^{-i(\vec{k}\vec{x} - \omega_{\vec{k}}t)})$$ satisfies this wave equation. Using linearity, one can superimpose all the solutions, plug them into the definition of the hamiltonian, and arrives at: $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda}^{\dagger} \omega_{\vec{k}} \hbar$$$$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ Now my question is: Can I also use this Hamiltonian in an interacting theory? (For example an EM-Field coupled to an atom). I'm asking because the wave equations that the Heisenberg operators do change. Superimposing the creation and annihilation operators, as shown above, is no longer a solution to the field equation, so I can't express $$\vec{E}$$ and $$\vec{B}$$ no longer, just using $$\hat{a}$$ and $$\hat{a}^{\dagger}$$? How can I still motivate $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda}^{\dagger} \omega_{\vec{k}} \hbar$$$$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ to be the right Hamiltonian? The EM-Field Hamiltonian is, in principle, a functional (with a choosen operator ordering) that is defined on operator-fields $$\hat{A}(x)$$ and $$\hat{\partial_\mu A}$$. If you carrie out the calculations and use definitions of $$\hat{B}$$ and $$\hat{E}$$, you'll arrive at: $$\hat{H} = \int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I'll take this as the Definition of the Hamiltonian in future calculations. For the free field, the Ansatz $$\hat{\vec{A}} = \vec{e}(\hat{a_{\vec{k}}}e^{i(\vec{k}\vec{x} - \omega_{\vec{k}}t)} + \hat{a_\vec{k}}^\dagger e^{-i(\vec{k}\vec{x} - \omega_{\vec{k}}t)})$$ satisfies this wave equation. Using linearity, one can superimpose all the solutions, plug them into the definition of the hamiltonian, and arrives at: $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda}^{\dagger} \omega_{\vec{k}} \hbar$$ Now my question is: Can I also use this Hamiltonian in an interacting theory? (For example an EM-Field coupled to an atom). I'm asking because the wave equations that the Heisenberg operators do change. Superimposing the creation and annihilation operators, as shown above, is no longer a solution to the field equation, so I can't express $$\vec{E}$$ and $$\vec{B}$$ no longer, just using $$\hat{a}$$ and $$\hat{a}^{\dagger}$$? How can I still motivate $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda}^{\dagger} \omega_{\vec{k}} \hbar$$ to be the right Hamiltonian? The EM-Field Hamiltonian is, in principle, a functional (with a chosen operator ordering) that is defined on operator-fields $$\hat{A}(x)$$ and $$\partial_\mu \hat{A}$$. If you carry out the calculations and use definitions of $$\hat{B}$$ and $$\hat{E}$$, you'll arrive at: $$\hat{H} = \int d^3x \frac{\epsilon_0}{2} (\hat{\vec{E}}^2 + c^2 \hat{\vec{B}}^2)$$ I'll take this as the Definition of the Hamiltonian in future calculations. For the free field, the Ansatz $$\hat{\vec{A}} = \vec{e}(\hat{a_{\vec{k}}}e^{i(\vec{k}\vec{x} - \omega_{\vec{k}}t)} + \hat{a_\vec{k}}^\dagger e^{-i(\vec{k}\vec{x} - \omega_{\vec{k}}t)})$$ satisfies this wave equation. Using linearity, one can superimpose all the solutions, plug them into the definition of the hamiltonian, and arrives at: $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ Now my question is: Can I also use this Hamiltonian in an interacting theory? (For example an EM-Field coupled to an atom). I'm asking because the wave equations that the Heisenberg operators do change. Superimposing the creation and annihilation operators, as shown above, is no longer a solution to the field equation, so I can't express $$\vec{E}$$ and $$\vec{B}$$ no longer, just using $$\hat{a}$$ and $$\hat{a}^{\dagger}$$? How can I still motivate $$\hat{H} = \sum_{\vec{k}, \vec{\lambda}} \hat{a}_{\vec{k}, \lambda}^{\dagger}\hat{a}_{\vec{k}, \lambda} \omega_{\vec{k}} \hbar$$ to be the right Hamiltonian? Notice added Draw attention by Quantumwhisp occurred Oct 1 '17 at 21:11 Bounty Started worth 50 reputation by Quantumwhisp occurred Oct 1 '17 at 21:11 2 edited tags | link edited Oct 1 '17 at 18:26 Quantumwhisp 2,850726 1 asked Sep 29 '17 at 10:01 Quantumwhisp 2,850726