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# 物理代写|量子力学代写Quantum mechanics代考|Normalization

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## 物理代写|量子力学代写Quantum mechanics代考|Normalization

We return now to the statistical interpretation of the wave function (Equation 1.3 ), which says that $|\Psi(x, t)|^2$ is the probability density for finding the particle at point $x$, at time $t$. It follows (Equation 1.16 ) that the integral of $|\Psi|^2$ over all $x$ must be 1 (the particle’s got to be somewhere):
$$\int_{-\infty}^{+\infty}|\Psi(x, t)|^2 d x=1$$
Without this, the statistical interpretation would be nonsense.
However, this requirement should disturb you: After all, the wave function is supposed to be determined by the Schrödinger equation-we can’t go imposing an extraneous condition on $\Psi$ without checking that the two are consistent. Well, a glance at Equation 1.1 reveals that if $\Psi(x, t)$ is a solution, so too is $A \Psi(x, t)$, where $A$ is any (complex) constant. What we must do, then, is pick this undetermined multiplicative factor so as to ensure that Equation 1.20 is satisfied. This process is called normalizing the wave function. For some solutions to the Schrödinger equation the integral is infinite; in that case no multiplicative factor is going to make it 1 . The same goes for the trivial solution $\Psi=0$. Such non-normalizable solutions cannot represent particles, and must be rejected. Physically realizable states correspond to the square-integrable solutions to Schrödinger’s equation. 14

But wait a minute! Suppose I have normalized the wave function at time $t=0$. How do I know that it will stay normalized, as time goes on, and $\Psi$ evolves? (You can’t keep renormalizing the wave function, for then $A$ becomes a function of $t$, and you no longer have a solution to the Schrödinger equation.) Fortunately, the Schrödinger equation has the remarkable property that it automatically preserves the normalization of the wave function-without this crucial feature the Schrödinger equation would be incompatible with the statistical interpretation, and the whole theory would crumble.
This is important, so we’d better pause for a careful proof. To begin with,
$$\frac{d}{d t} \int_{-\infty}^{+\infty}|\Psi(x, t)|^2 d x=\int_{-\infty}^{+\infty} \frac{\partial}{\partial t}|\Psi(x, t)|^2 d x$$

## 物理代写|量子力学代写Quantum mechanics代考|Momentum

For a particle in state $\Psi$, the expectation value of $x$ is
$$\langle x\rangle=\int_{-\infty}^{+\infty} x|\Psi(x, t)|^2 d x$$
What exactly does this mean? It emphatically does not mean that if you measure the position of one particle over and over again, $\int x|\Psi|^2 d x$ is the average of the results you’ll get. On the contrary: The first measurement (whose outcome is indeterminate) will collapse the wave function to a spike at the value actually obtained, and the subsequent measurements (if they’re performed quickly) will simply repeat that same result. Rather, $\langle x\rangle$ is the average of measurements performed on particles all in the state $\Psi$, which means that either you must find some way of returning the particle to its original state after each measurement, or else you have to prepare a whole ensemble of particles, each in the same state $\Psi$, and measure the positions of all of them: $\langle x\rangle$ is the average of these results. I like to picture a row of bottles on a shelf, each containing a particle in the state $\Psi$ (relative to the center of the bottle). A graduate student with a ruler is assigned to each bottle, and at a signal they all measure the positions of their respective particles. We then construct a histogram of the results, which should match $|\Psi|^2$, and compute the average, which should agree with $\langle x)$. (Of course, since we’re only using a finite sample, we can’t expect perfect agreement, but the more bottles we use, the closer we ought to come.) In short, the expectation value is the average of measurements on an ensemble of identically-prepared systems, not the average of repeated measurements on one and the same system.

Now, as time goes on, $\langle x\rangle$ will change (because of the time dependence of $\Psi$ ), and we might be interested in knowing how fast it moves. Referring to Equations 1.25 and 1.28 , we see that $\frac{16}{}$
$$\frac{d\langle x\rangle}{d t}=\int x \frac{\partial}{\partial t}|\Psi|^2 d x=\frac{i \hbar}{2 m} \int x \frac{\partial}{\partial x}\left(\Psi^* \frac{\partial \Psi}{\partial x}-\frac{\partial \Psi^*}{\partial x} \Psi\right) d x$$

## 物理代写|量子力学代写Quantum mechanics代考|Normalization

$$\int_{-\infty}^{+\infty}|\Psi(x, t)|^2 d x=1$$

$$\frac{d}{d t} \int_{-\infty}^{+\infty}|\Psi(x, t)|^2 d x=\int_{-\infty}^{+\infty} \frac{\partial}{\partial t}|\Psi(x, t)|^2 d x$$

## 物理代写|量子力学代写Quantum mechanics代考|Momentum

$$\langle x\rangle=\int_{-\infty}^{+\infty} x|\Psi(x, t)|^2 d x$$

$$\frac{d\langle x\rangle}{d t}=\int x \frac{\partial}{\partial t}|\Psi|^2 d x=\frac{i \hbar}{2 m} \int x \frac{\partial}{\partial x}\left(\Psi^* \frac{\partial \Psi}{\partial x}-\frac{\partial \Psi^*}{\partial x} \Psi\right) d x$$

## MATLAB代写

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