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# 物理代写|量子力学代写Quantum mechanics代考|PHYS402 Formal Development of Perturbation Expansion

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## 物理代写|量子力学代写Quantum mechanics代考|Formal Development of Perturbation Expansion

We now state in more precise terms the basic problem we wish to solve. Suppose we know completely and exactly the energy eigenkets and energy eigenvalues of the unperturbed Hamiltonian $H_{0}$, that is
$$H_{0}\left|n^{(0)}\right\rangle=E_{n}^{(0)}\left|n^{(0)}\right\rangle .$$
The set $\left{\left|n^{(0)}\right\rangle\right}$ is complete in the sense that the closure relation $1=\sum_{n}\left|n^{(0)}\right\rangle\left\langle n^{(0)}\right|$ holds. Furthermore, we assume here that the energy spectrum is nondegenerate; in the next section we will relax this assumption. We are interested in obtaining the energy eigenvalues and eigenkets for the problem defined by (5.4). To be consistent with (5.18) we should write $(5.4)$ as
$$\left(H_{0}+\lambda V\right)|n\rangle_{\lambda}=E_{n}^{(\lambda)}|n\rangle_{\lambda}$$
to denote the fact that the energy eigenvalues $E_{n}^{(\lambda)}$ and energy eigenkets $|n\rangle_{\lambda}$ are functions of the continuous parameter $\lambda$; however, we will usually dispense with this correct but more cumbersome notation.

As the continuous parameter $\lambda$ is increased from zero, we expect the energy eigenvalue $E_{n}$ for the $n$th eigenket to depart from its unperturbed value $E_{n}^{(0)}$, so we define the energy shift for the $n$th level as follows:
$$\Delta_{n} \equiv E_{n}-E_{n}^{(0)} .$$
The basic Schrödinger equation to be solved (approximately) is
$$\left(E_{n}^{(0)}-H_{0}\right)|n\rangle=\left(\lambda V-\Delta_{n}\right)|n\rangle .$$
We may be tempted to invert the operator $E_{n}^{(0)}-H_{0}$; however, in general, the inverse operator $1 /\left(E_{n}^{(0)}-H_{0}\right)$ is ill defined because it may act on $\left|n^{(0)}\right\rangle$. Fortunately in our case $\left(\lambda V-\Delta_{n}\right)|n\rangle$ has no component along $\left|n^{(0)}\right\rangle$, as can easily be seen by multiplying both sides of $(5.21)$ by $\left\langle n^{(0)}\right|$ on the left:
$$\left\langle n^{(0)}\left|\left(\lambda V-\Delta_{n}\right)\right| n\right\rangle=0 .$$

## 物理代写|量子力学代写Quantum mechanics代考|Wave Function Renormalization

We are in a position to look at the normalization of the perturbed ket. Recalling the normalization convention we use, (5.31), we see that the perturbed ket $|n\rangle$ is not normalized in the usual manner. We can renormalize the perturbed ket by defining
$$|n\rangle_{N}=Z_{n}^{1 / 2}|n\rangle,$$
where $Z_{n}$ is simply a constant with ${ }{N}\langle n \mid n\rangle{N}=1$. Multiplying $\left\langle n^{(0)}\right|$ on the left we obtain [because of (5.31)]
$$Z_{n}^{1 / 2}=\left\langle n^{0} \mid n\right\rangle_{N} .$$
What is the physical meaning of $Z_{n}$ ? Because $|n\rangle_{N}$ satisfies the usual normalization requirement (5.30), $Z_{n}$ can be regarded as the probability for the perturbed energy eigenstate to be found in the corresponding unperturbed energy eigenstate. Noting
$${ }{N}\langle n \mid n\rangle{N}=Z_{n}\langle n \mid n\rangle=1,$$
we have
\begin{aligned} Z_{n}^{-1}=&\langle n \mid n\rangle=\left(\left\langle n^{(0)}\right|+\lambda\left\langle n^{(1)}\right|+\lambda^{2}\left\langle n^{(2)}\right|+\cdots\right) \ & \times\left(\left|n^{(0)}\right\rangle+\lambda\left|n^{(1)}\right\rangle+\lambda^{2}\left|n^{(2)}\right\rangle+\cdots\right) \ =& 1+\lambda^{2}\left\langle n^{(1)} \mid n^{(1)}\right\rangle+0\left(\lambda^{3}\right) \ =& 1+\lambda^{2} \sum_{k \neq n} \frac{\left|V_{k n}\right|^{2}}{\left(E_{n}^{(0)}-E_{k}^{(0)}\right)^{2}}+0\left(\lambda^{3}\right), \end{aligned}
so up to order $\lambda^{2}$, we get for the probability of the perturbed state to be found in the corresponding unperturbed state
$$Z_{n} \simeq 1-\lambda^{2} \sum_{k \neq n} \frac{\left|V_{k n}\right|^{2}}{\left(E_{n}^{0}-E_{k}^{0}\right)^{2}} .$$
The second term in (5.48b) is to be understood as the probability for “leakage” to states other than $\left|n^{(0)}\right\rangle$. Notice that $Z_{n}$ is less than 1, as expected on the basis of the probability interpretation for $Z$.

## 物理代写|量子力学代写Quantum mechanics代考|Formal Development of Perturbation Expansion

$$H_{0}\left|n^{(0)}\right\rangle=E_{n}^{(0)}\left|n^{(0)}\right\rangle .$$

$$\left(H_{0}+\lambda V\right)|n\rangle_{\lambda}=E_{n}^{(\lambda)}|n\rangle_{\lambda}$$

$$\Delta_{n} \equiv E_{n}-E_{n}^{(0)} .$$

$$\left(E_{n}^{(0)}-H_{0}\right)|n\rangle=\left(\lambda V-\Delta_{n}\right)|n\rangle .$$

$$\left\langle n^{(0)}\left|\left(\lambda V-\Delta_{n}\right)\right| n\right\rangle=0 .$$

## 物理代写|量子力学代写Quantum mechanics代考|Wave Function Renormalization

$$|n\rangle_{N}=Z_{n}^{1 / 2}|n\rangle,$$

$$Z_{n}^{1 / 2}=\left\langle n^{0} \mid n\right\rangle_{N} .$$

$$N\langle n \mid n\rangle N=Z_{n}\langle n \mid n\rangle=1,$$

$$Z_{n}^{-1}=\langle n \mid n\rangle=\left(\left\langlen ^ { ( 0 ) } \left|+\lambda\left\langle n^{(1)}\left|+\lambda^{2}\left\langle n^{(2)}\right|+\cdots\right) \quad \times\left(\left|n^{(0)}\right\rangle+\lambda\left|n^{(1)}\right\rangle+\lambda^{2}\left|n^{(2)}\right\rangle+\cdots\right)=1+\lambda^{2}\left\langle n^{(1)} \mid n^{(1)}\right\rangle+0\left(\lambda^{3}\right)=\quad 1+\lambda^{2} \sum_{k \neq n}\right.\right.\right.\right.$$

$$Z_{n} \simeq 1-\lambda^{2} \sum_{k \neq n} \frac{\left|V_{k n}\right|^{2}}{\left(E_{n}^{0}-E_{k}^{0}\right)^{2}} .$$
(5.48b) 中的第二项应理解为“泄漏”到除 $\left|n^{(0)}\right\rangle$. 请注意 $Z_{n}$ 小于 1 ，正如在概率解释的基础上所预期的那样 $Z$.

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