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# 物理代写|量子力学代写Quantum mechanics代考|Degenerate perturbation theory

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## 物理代写|量子力学代写Quantum mechanics代考|Degenerate perturbation theory

In the previous discussion of perturbation theory we assumed that there was a one to one correspondence between the set of eigenvectors $\left{\left|E_n^0\right\rangle\right}$ and the corresponding set of eigenvalues $\left{E_n^0\right}$; that is, we have postulated no degeneracy in our energy eigenvalues. We also mentioned that if such degeneracy is present we cannot use the non-degenerate perturbation theory because of the appearance of zeroes in the denominator of the expressions of the $E_n$ ‘s or $\left|E_n\right\rangle$ in eqs. $(12.51,12.53)$.

There would also be a problem even if no degeneracy is present, but the condition (12.57) is violated. Then the ratio
$$\frac{\left|H_{n k}^{\prime}\right|}{\left|E_n^0-E_k^0\right|}$$
is not small for some of the terms in the sums in eqs.(12.51,12.53), and the perturbative expansion is no longer a good approximation. It must be emphasized that these problems will depend on the specific eigenvalue $E_n^0$. The pertubative expansion continues to hold for the states labelled by $E_n^0$ that satisfy
$$\left|H_{n k}^{\prime}\right| \ll\left|E_n^0-E_k^0\right| \quad \text { all } \quad k \neq n$$
The problem occurs only for the states $E_n$ which violate this condition.
The cure to the problem is to perform an exact (or almost exact) partial diagonalization of the matrix $H_{m n}=H_{m n}^0+H_{m n}^{\prime}$ in the blocks of degenerate or almost degenerate states. If necessary, one can relabel the states so that the degenerate or almost degenerate states appear in blocks in the form
$$H=\left(\begin{array}{cccccc} E_1^0 & a & # & # & # & # \ a^* & E_1^0+\varepsilon_1 & # & # & # & # \ # & # & E_2^0 & b & c & # \ # & # & b^* & E_2^0+\varepsilon_2 & d & # \ # & # & c^* & d^{#} & E_2^0+\varepsilon_2^{\prime} & # \ # & # & # & # & # & \ddots \end{array}\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|More on degeneracy

There is one more step to discuss if the eigenvalues in the same block are still exactly degenerate after the partial diagonalization. For example, suppose in the first block $\tilde{\varepsilon}1=0$, while ” 0 “=0. This degeneracy continues to create a problem as follows. For simplicity consider a 3-level problem whose Hamiltonian matrix has the form $$H=\left(\begin{array}{lll} A & 0 & C \ 0 & A & D \ C^* & D^* & B \end{array}\right)$$ So that we identify $$H_0=\left(\begin{array}{lll} A & 0 & 0 \ 0 & A & 0 \ 0 & 0 & B \end{array}\right), \quad H^{\prime}=\left(\begin{array}{lll} 0 & 0 & C \ 0 & 0 & D \ C^* & D^* & 0 \end{array}\right)$$ The basis is given by $$\left\langle E_1^0\right|=\left(\begin{array}{lll} 1 & 0 & 0 \end{array}\right),\left\langle E_2^0\right|=\left(\begin{array}{lll} 0 & 1 & 0 \end{array}\right),\left\langle E_3^0\right|=\left(\begin{array}{lll} 0 & 0 & 1 \end{array}\right)$$ and the initial energies are $E_1^0=A, E_2^0=A, E_3^0=B$. There is a problem in the perturbation series due to the degeneracy $E_1^0=E_2^0=A$ if $C$ or $D$ is non-zero, as follows. The expansion of (12.51) up to second order is \begin{aligned} & E_1=E_1^0+\frac{H{12}^{\prime} H_{21}}{E_1^0-E_2^0}+\frac{H_{13}^{\prime} H_{31}^{\prime}}{E_1^0-E_3^0}+\cdots=A+\frac{0}{0}+\frac{|C|^2}{A-B}+\cdots \ & E_2=E_2^0+\frac{H_{21}^{\prime} H_{12}^{\prime}}{E_2^0-E_1^0}+\frac{H_{23}^{\prime} H_{32}^{\prime}}{E_2^0-E_3^0}+\cdots=A+\frac{0}{0}+\frac{|D|^2}{A-B}+\cdots \ & E_3=E_3^0+\frac{H_{31}^{\prime} H_{13}^{\prime}}{E_3^0-E_1^0}+\frac{H_{32}^{\prime} H_{23}^{\prime}}{E_3^0-E_2^0}+\cdots=B+\frac{|C|^2}{B-A}+\frac{|D|^2}{B-A}+\cdots \quad \text { (12.69) } \end{aligned}
The $\frac{0}{0}$ is undetermined, and one must find its meaning. To do so, we apply a unitary transformation in such a way as to create a fully isolated $1 \times 1$ block with all off diagonals zeros
$$U^{\dagger} H U=\left(\begin{array}{lll} # & 0 & 0 \ 0 & # & # \ 0 & #^* & # \end{array}\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|Degenerate perturbation theory

$$\frac{\left|H_{n k}^{\prime}\right|}{\left|E_n^0-E_k^0\right|}$$

$$\left|H_{n k}^{\prime}\right| \ll\left|E_n^0-E_k^0\right| \quad \text { all } \quad k \neq n$$

## 物理代写|量子力学代写Quantum mechanics代考|More on degeneracy

$$H_0=\left(\begin{array}{lllllllll} A & 0 & 0 & 0 & A & 0 & 0 & 0 & B \end{array}\right), \quad H^{\prime}=\left(\begin{array}{lllllll} 0 & 0 & C 0 & 0 & D C^* & D^* & 0 \end{array}\right)$$

$$\left\langle E_1^0\right|=\left(\begin{array}{lll} 1 & 0 & 0 \end{array}\right),\left\langle E_2^0\right|=\left(\begin{array}{lll} 0 & 1 & 0 \end{array}\right),\left\langle E_3^0\right|=\left(\begin{array}{lll} 0 & 0 & 1 \end{array}\right)$$

$$E_1=E_1^0+\frac{H 12^{\prime} H_{21}}{E_1^0-E_2^0}+\frac{H_{13}^{\prime} H_{31}^{\prime}}{E_1^0-E_3^0}+\cdots=A+\frac{0}{0}+\frac{|C|^2}{A-B}+\cdots \quad E_2=E_2^0+\frac{H_{21}^{\prime} H_{12}^{\prime}}{E_2^0-E_1^0}+\frac{H_{23}^{\prime} H_{32}^{\prime}}{E_2^0-E_3^0}+\cdots=A+\frac{0}{0}+\frac{|D|^2}{A-B}+\cdots E_3=$$

$$\left.U \wedge{\backslash \text { dagger }} H U=\backslash \text { leftt( } \backslash \text { begin }{\text { array }}{|} # \& 0 \& 0 \backslash 0 \& # \& # \backslash 0 \& # \wedge^* \& # \text { |end }{\text { array }} \backslash \text { right }\right)$$

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