# 物理代写|量子力学作业代写Quantum Mechanics代考|Schrödinger’s Wave Equation

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## 物理代写|量子力学作业代写Quantum Mechanics代考|Time-Dependent Wave Equation

We now turn to the Schrödinger picture and examine the time evolution of $\left|\alpha, t_{0} ; t\right\rangle$ in the $x$-representation. In other words, our task is to study the behavior of the wave function
$$\psi\left(\mathbf{x}^{\prime}, t\right)=\left\langle\mathbf{x}^{\prime} \mid \alpha, t_{0} ; t\right\rangle$$
as a function of time, where $\left|\alpha, t_{0} ; t\right\rangle$ is a state ket in the Schrödinger picture at time $t$, and $\left\langle\mathbf{x}^{\prime}\right|$ is a time-independent position eigenbra with eigenvalue $\mathbf{x}^{\prime}$. The Hamiltonian operator is taken to be
$$H=\frac{\mathbf{p}^{2}}{2 m}+V(\mathbf{x})$$

## 物理代写|量子力学作业代写Quantum Mechanics代考|The Time-Independent Wave Equation

We now derive the partial differential equation satisfied by energy eigenfunctions. We showed in Section $2.1$ that the time dependence of a stationary state is given by $\exp \left(-i E_{d^{\prime}} t / \hbar\right)$. This enables us to write its wave function as
$$\left\langle\mathbf{x}^{\prime} \mid a^{\prime}, t_{0} ; t\right\rangle=\left\langle\mathbf{x}^{\prime} \mid a^{\prime}\right\rangle \exp \left(\frac{-i E_{a^{\prime}} t}{\hbar}\right),$$
where it is understood that initially the system is prepared in a simultaneous eigenstate of $A$ and $H$ with eigenvalues $a^{\prime}$ and $E_{\alpha^{\prime}}$, respectively. Let us now substitute (2.184) into the time-dependent Schrödinger equation (2.182). We are then led to
$$-\left(\frac{\hbar^{2}}{2 m}\right) \nabla^{\prime 2}\left\langle\mathbf{x}^{\prime} \mid a^{\prime}\right\rangle+V\left(\mathbf{x}^{\prime}\right)\left\langle\mathbf{x}^{\prime} \mid a^{\prime}\right\rangle=E_{a^{\prime}}\left\langle\mathbf{x}^{\prime} \mid a^{\prime}\right\rangle$$

## 物理代写|量子力学作业代写QUANTUM MECHANICS代考|Interpretations of the Wave Function

We now turn to discussions of the physical interpretations of the wave function. In Section $1.7$ we commented on the probabilistic interpretation of $|\psi|^{2}$ that follows from the fact that $\left\langle\mathbf{x}^{\prime} \mid \alpha, t_{0} ; t\right\rangle$ is to be regarded as an expansion coefficient of $\left|\alpha, t_{0} ; t\right\rangle$ in terms of the position eigenkets $\left{\left|\mathbf{x}^{\prime}\right\rangle\right}$. The quantity $\rho\left(\mathbf{x}^{\prime}, t\right)$ defined by
$$\rho\left(\mathbf{x}^{\prime}, t\right)=\left|\psi\left(\mathbf{x}^{\prime}, t\right)\right|^{2}=\left|\left\langle\mathbf{x}^{\prime} \mid \alpha, t_{0} ; t\right\rangle\right|^{2}$$
is therefore regarded as the probability density in wave mechanics. Specifically, when we use a detector that ascertains the presence of the particle within a small volume element $d^{3} x^{\prime}$ around $\mathbf{x}^{\prime}$, the probability of recording a positive result at time $t$ is given by $\rho\left(\mathbf{x}^{\prime}, t\right) d^{3} x^{\prime}$.
In the remainder of this section we use $\mathbf{x}$ for $\mathbf{x}^{\prime}$ because the position operator will not appear. Using Schrödinger’s time-dependent wave equation, it is straightforward to derive the continuity equation
$$\frac{\partial \rho}{\partial t}+\nabla \cdot \mathbf{j}=0$$
where $\rho(\mathbf{x}, t)$ stands for $|\psi|^{2}$ as before, and $\mathbf{j}(\mathbf{x}, t)$, known as the probability flux, is given by
\begin{aligned} \mathbf{j}(\mathbf{x}, t) &=-\left(\frac{i \hbar}{2 m}\right)\left[\psi^{} \nabla \psi-\left(\nabla \psi^{}\right) \psi\right] \ &=\left(\frac{\hbar}{m}\right) \operatorname{Im}\left(\psi^{*} \nabla \psi\right) . \end{aligned}

## 物理代写|量子力学作业代写QUANTUM MECHANICS代考|TIME-DEPENDENT WAVE EQUATION

ψ(X′,吨)=⟨X′∣一种,吨0;吨⟩

H=p22米+在(X)

## 物理代写|量子力学作业代写QUANTUM MECHANICS代考|THE TIME-INDEPENDENT WAVE EQUATION

⟨X′∣一种′,吨0;吨⟩=⟨X′∣一种′⟩经验⁡(−一世和一种′吨⁇),

−(⁇22米)∇′2⟨X′∣一种′⟩+在(X′)⟨X′∣一种′⟩=和一种′⟨X′∣一种′⟩

## 物理代写|量子力学作业代写QUANTUM MECHANICS代考|INTERPRETATIONS OF THE WAVE FUNCTION

ρ(X′,吨)=|ψ(X′,吨)|2=|⟨X′∣一种,吨0;吨⟩|2

∂ρ∂吨+∇⋅j=0

\begin{aligned} \mathbf{j}(\mathbf{x}, t) &=-\left(\frac{i \hbar}{2 m}\right)\left[\psi^{} \nabla \psi-\left(\nabla \psi^{}\right) \psi\right] \ &=\left(\frac{\hbar}{m}\right) \operatorname{Im}\left(\psi^{*} \nabla \psi\right) . \end{aligned}

## Matlab代写

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