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物理代写|量子力学代写Quantum mechanics代考|PHYSICS3544 Time-Independent Perturbation Theory: Nondegenerate Case

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物理代写|量子力学代写Quantum mechanics代考|Statement of the Problem

The approximation method we consider here is time-independent perturbation theory, sometimes known as the Rayleigh-Schrödinger perturbation theory. We consider a timeindependent Hamiltonian $H$ such that it can be split into two parts, namely,
$$H=H_{0}+V,$$
where the $V=0$ problem is assumed to have been solved in the sense that both the exact energy eigenkets $\left|n^{(0)}\right\rangle$ and the exact energy eigenvalues $E_{n}^{(0)}$ are known:
$$H_{0}\left|n^{(0)}\right\rangle=E_{n}^{(0)}\left|n^{(0)}\right\rangle .$$
We are required to find approximate eigenkets and eigenvalues for the full Hamiltonian problem
$$\left(H_{0}+V\right)|n\rangle=E_{n}|n\rangle,$$
where $V$ is known as the perturbation; it is not, in general, the full-potential operator. For example, suppose we consider the hydrogen atom in an external electric or magnetic field. The unperturbed Hamiltonian $H_{0}$ is taken to be the kinetic energy $\mathbf{p}^{2} / 2 \mathrm{~m}$ and the Coulomb potential due to the presence of the proton nucleus $-e^{2} / r$. Only that part of the potential due to the interaction with the external $\mathbf{E}$ or $\mathbf{B}$ field is represented by the perturbation $V$.

物理代写|量子力学代写Quantum mechanics代考|The Two-State Problem

Before we embark on a systematic presentation of the basic method, let us see how the expansion in $\lambda$ might indeed be valid in the exactly soluble two-state problem we have encountered many times already. Suppose we have a Hamiltonian that can be written as
$$H=E_{1}^{(0)}\left|1^{(0)}\right\rangle\left\langle 1^{(0)}\left|+E_{2}^{(0)}\right| 2^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{12}\right| 1^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{21}\right| 2^{(0)}\right\rangle\left\langle 1^{(0)}\right|,$$
where $\left|1^{(0)}\right\rangle$ and $\left|2^{(0)}\right\rangle$ are the energy eigenkets for the $\lambda=0$ problem, and we consider the case $V_{11}=V_{22}=0$. In this representation the $H$ may be represented by a square matrix as follows:
$$H=\left(\begin{array}{ll} E_{1}^{(0)} & \lambda V_{12} \ \lambda V_{21} & E_{2}^{(0)} \end{array}\right),$$
where we have used the basis formed by the unperturbed energy eigenkets. The $V$ matrix must, of course, be Hermitian; let us solve the case when $V_{12}$ and $V_{21}$ are real:
$$V_{12}=V_{12}^{}, \quad V_{21}=V_{21}^{} ;$$
hence, by Hermiticity
$$V_{12}=V_{21} \text {. }$$

物理代写|量子力学代写Quantum mechanics代考|Statement of the Problem

$$H=H_{0}+V,$$

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

$$\left(H_{0}+V\right)|n\rangle=E_{n}|n\rangle,$$

物理代写|量子力学代写Quantum mechanics代考|The Two-State Problem

$$H=E_{1}^{(0)}\left|1^{(0)}\right\rangle\left\langle 1^{(0)}\left|+E_{2}^{(0)}\right| 2^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{12}\right| 1^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{21}\right| 2^{(0)}\right\rangle\left\langle 1^{(0)}\right|,$$

$$H=\left(\begin{array}{lll} E_{1}^{(0)} & \lambda V_{12} \lambda V_{21} & E_{2}^{(0)} \end{array}\right)$$

$$V_{12}=V_{12}, \quad V_{21}=V_{21}$$

$$V_{12}=V_{21} .$$

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