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I am confused about pure state or in mixed states. I've seen several threads on this forum, but I still can't get the grasp of it. I only have very little quantum chemistry to know what these means. So instead, I want to know the answer for specific examples so that I can get an idea.
So I would like to know "Yes" or "No" to each numbered question in the following text where I attempt to derive wavefunctions and energies of new states produced by excitonic interaction between two identical molecules.
Say for example, I have two identical molecules aligned close to each other. I can have three possible state:
1) When both molecule is in ground state: [itex] \left | \phi _{0} \right \rangle = \left | \varphi _{1} \right \rangle \otimes \left | \varphi _{2} \right \rangle [/itex]
2) When molecule 1 is in excited state: [itex] \left | \phi _{1} \right \rangle = \left | \varphi _{1}^{*} \right \rangle \otimes \left | \varphi _{2} \right \rangle [/itex]
3) When molecule 2 is in excited state: [itex] \left | \phi _{2} \right \rangle = \left | \varphi _{1} \right \rangle \otimes \left | \varphi _{2}^{*} \right \rangle [/itex]
Where [itex] \left | \varphi _{1} \right \rangle [/itex] and [itex] \left | \varphi _{2} \right \rangle [/itex] represents molecule 1 and 2 in ground state, respectively. The star means excited state. (Q1. Are all of these wavefunctions pure state?) So the total wavefunction of the entire system is:
[tex]\left | \Phi \right \rangle = a_{0}\left | \phi _{0} \right \rangle + a_{1}\left | \phi _{1} \right \rangle + a_{2} \left | \phi _{2} \right \rangle [/tex]
(Q2. Is this a pure state or mixed state?) The "true" hamiltonian H for deriving the energy of this system is given by:
[tex] \textbf{H} = \textbf{H}_{0} + \textbf{H}_{J} [/tex]
Where [itex] \textbf{H}_{0} [/itex] is the Hamiltonian in the absence of excitonic interaction, and [itex] \textbf{H}_{J} [/itex] is the excitonic interaction Hamiltonian. In the case of 2 identical molecules with excitonic interactions strength of J, ground state energy of Egr and excited state energy of Eex, the Hamiltonian is:
[tex]\textbf{H} = \begin{pmatrix}
E_{gr} & 0 & 0\\
0 & E_{ex} & J \\
0 & J & E_{ex}
\end{pmatrix}[/tex]
Solving the Schrodinger equation:
[tex]\textbf{H}\left | \Phi \right \rangle = E\left | \Phi \right \rangle[/tex]
through diagonalization gives eigenvalues of:
[itex]E_{0} = E_{gr}[/itex]
[itex]E_{1} = E_{ex}-J[/itex]
[itex]E_{2} = E_{ex}+J[/itex]
and eigenvectors of:
[itex]\left | \Phi _{0} \right \rangle = \left | \phi _{0} \right \rangle[/itex]
[itex]\left | \Phi _{1} \right \rangle = \frac{1}{\sqrt{2}} \left ( \left | \phi _{1} \right \rangle - \left | \phi _{2} \right \rangle \right )[/itex]
[itex]\left | \Phi _{2} \right \rangle = \frac{1}{\sqrt{2}} \left ( \left | \phi _{1} \right \rangle + \left | \phi _{2} \right \rangle \right )[/itex]
(Q3. Are these pure state or mixed state?) Finally, the wavefunction for the "new" entire system that takes account of excitonic interactions is given by:
[tex]\left | \Psi \right \rangle = b_{0} \left | \Phi _{0} \right \rangle + b_{1} \left | \Phi _{1} \right \rangle + b_{2} \left | \Phi _{2} \right \rangle[/tex]
(Q4. Is this pure state or mixed state?)
Thank you.
So I would like to know "Yes" or "No" to each numbered question in the following text where I attempt to derive wavefunctions and energies of new states produced by excitonic interaction between two identical molecules.
Say for example, I have two identical molecules aligned close to each other. I can have three possible state:
1) When both molecule is in ground state: [itex] \left | \phi _{0} \right \rangle = \left | \varphi _{1} \right \rangle \otimes \left | \varphi _{2} \right \rangle [/itex]
2) When molecule 1 is in excited state: [itex] \left | \phi _{1} \right \rangle = \left | \varphi _{1}^{*} \right \rangle \otimes \left | \varphi _{2} \right \rangle [/itex]
3) When molecule 2 is in excited state: [itex] \left | \phi _{2} \right \rangle = \left | \varphi _{1} \right \rangle \otimes \left | \varphi _{2}^{*} \right \rangle [/itex]
Where [itex] \left | \varphi _{1} \right \rangle [/itex] and [itex] \left | \varphi _{2} \right \rangle [/itex] represents molecule 1 and 2 in ground state, respectively. The star means excited state. (Q1. Are all of these wavefunctions pure state?) So the total wavefunction of the entire system is:
[tex]\left | \Phi \right \rangle = a_{0}\left | \phi _{0} \right \rangle + a_{1}\left | \phi _{1} \right \rangle + a_{2} \left | \phi _{2} \right \rangle [/tex]
(Q2. Is this a pure state or mixed state?) The "true" hamiltonian H for deriving the energy of this system is given by:
[tex] \textbf{H} = \textbf{H}_{0} + \textbf{H}_{J} [/tex]
Where [itex] \textbf{H}_{0} [/itex] is the Hamiltonian in the absence of excitonic interaction, and [itex] \textbf{H}_{J} [/itex] is the excitonic interaction Hamiltonian. In the case of 2 identical molecules with excitonic interactions strength of J, ground state energy of Egr and excited state energy of Eex, the Hamiltonian is:
[tex]\textbf{H} = \begin{pmatrix}
E_{gr} & 0 & 0\\
0 & E_{ex} & J \\
0 & J & E_{ex}
\end{pmatrix}[/tex]
Solving the Schrodinger equation:
[tex]\textbf{H}\left | \Phi \right \rangle = E\left | \Phi \right \rangle[/tex]
through diagonalization gives eigenvalues of:
[itex]E_{0} = E_{gr}[/itex]
[itex]E_{1} = E_{ex}-J[/itex]
[itex]E_{2} = E_{ex}+J[/itex]
and eigenvectors of:
[itex]\left | \Phi _{0} \right \rangle = \left | \phi _{0} \right \rangle[/itex]
[itex]\left | \Phi _{1} \right \rangle = \frac{1}{\sqrt{2}} \left ( \left | \phi _{1} \right \rangle - \left | \phi _{2} \right \rangle \right )[/itex]
[itex]\left | \Phi _{2} \right \rangle = \frac{1}{\sqrt{2}} \left ( \left | \phi _{1} \right \rangle + \left | \phi _{2} \right \rangle \right )[/itex]
(Q3. Are these pure state or mixed state?) Finally, the wavefunction for the "new" entire system that takes account of excitonic interactions is given by:
[tex]\left | \Psi \right \rangle = b_{0} \left | \Phi _{0} \right \rangle + b_{1} \left | \Phi _{1} \right \rangle + b_{2} \left | \Phi _{2} \right \rangle[/tex]
(Q4. Is this pure state or mixed state?)
Thank you.