11.2: Two-State System
- Page ID
- 15793
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider the simplest possible non-trivial quantum mechanical system. In such a system, there are only two independent eigenstates of the unperturbed Hamiltonian: that is,
\[\begin{align} \label{e12.21} H_0\,\psi_1 &= E_1\,\psi_1,\\[4pt] H_0\,\psi_2&=E_2\,\psi_2.\label{e12.22}\end{align} \]
It is assumed that these states, and their associated eigenvalues, are known. We also expect the states to be orthonormal, and to form a complete set.
Let us now try to solve the modified energy eigenvalue problem
\[ (H_0+H_1)\,\psi_E = E\,\psi_E. \label{e12.23} \]
We can, in fact, solve this problem exactly. Because the eigenstates of \(H_0\) form a complete set, we can write [see Equation \ref{e12.13a}]
\[ \psi_E = \langle 1|E\rangle\,\psi_1 + \langle 2|E\rangle\,\psi_2. \label{e12.24} \]
It follows from Equation \ref{e12.23} that
\[ \langle i|H_0 + H_1|E\rangle = E\,\langle i|E\rangle, \label{e12.25} \]
where \(i=1\) or \(2\). Equations \ref{e12.21}, \ref{e12.22}, \ref{e12.24}, \ref{e12.25}, and the orthonormality condition
\[\langle i|j\rangle = \delta_{ij}, \nonumber \]
yield two coupled equations that can be written in matrix form:
\[\begin{align} \label{e12.27} \left(\begin{array}{cc}E_1-E+e_{11},& e_{12}\\[4pt] e_{12}^\ast, & E_2-E+e_{22}\end{array}\right)\left( \begin{array}{c}\langle 1|E\rangle\\[4pt] \langle 2|E\rangle\end{array}\right)=\left( \begin{array}{c} 0\\[4pt] 0\end{array}\right),\end{align} \]
where
\[\begin{align} e_{11} &= \langle 1|H_1|1\rangle,\\[4pt] e_{22}&=\langle 2|H_1|2\rangle,\\[4pt] e_{12}&=\langle 1|H_1|2\rangle = \langle 2|H_1|1\rangle^\ast.\end{align} \nonumber \]
Here, use has been made of the fact that \(H_1\) is an Hermitian operator.
Consider the special (but not uncommon) case of a perturbing Hamiltonian whose diagonal matrix elements are zero, so that
\[e_{11}= e_{22} = 0. \nonumber \]
The solution of Equation \ref{e12.27} (obtained by setting the determinant of the matrix to zero) is
\[E = \frac{(E_1+E_2)\pm\sqrt{(E_1-E_2)^2 + 4\,|e_{12}|^2}} {2}. \nonumber \]
Let us expand in the supposedly small parameter
\[\epsilon = \frac{|e_{12}|}{|E_1-E_2|}. \nonumber \]
We obtain
\[E \simeq \frac{1}{2}\,(E_1+E_2) \pm \frac{1}{2}\,(E_1-E_2)(1+2\,\epsilon^2+\cdots). \nonumber \]
The previous expression yields the modification of the energy eigenvalues due to the perturbing Hamiltonian:
\[\begin{align} E_1' &= E_1 + \frac{|e_{12}|^2}{E_1-E_2}+ \cdots,\\[4pt] E_2' &= E_2 - \frac{|e_{12}|^2}{E_1-E_2}+\cdots.\end{align} \nonumber \]
Note that \(H_1\) causes the upper eigenvalue to rise, and the lower to fall. It is easily demonstrated that the modified eigenstates take the form
\[\begin{align} \psi_1' &=\psi_1+ \frac{e_{12}^\ast}{E_1-E_2}\,\psi_2+ \cdots,\\[4pt] \psi_2'&= \psi_2 - \frac{e_{12}}{E_1-E_2}\,\psi_1+ \cdots.\end{align} \nonumber \]
Thus, the modified energy eigenstates consist of one of the unperturbed eigenstates, plus a slight admixture of the other. Now, our expansion procedure is only valid when \(\epsilon\ll 1\). This suggests that the condition for the validity of the perturbation method as a whole is
\[|e_{12}|\ll |E_1-E_2|. \nonumber \]
In other words, when we say that \(H_1\) needs to be small compared to \(H_0\), what we are really saying is that the previous inequality must be satisfied.