📂Quantum Mechanics

규격화된 파동함수의 상태는 시간의 변화에 무관하다

Theorem¹

A normalized wave function remains in a normalized state even as time changes.

Explanation

Let us assume the wave function is normalized at time $t=0$. According to the theorem, it is guaranteed to remain in a normalized state as time progresses. This is a very crucial fact that allows us to treat the wave function as a probability density function.

Proof

Strategy: To show that it remains constant over time, we will differentiate $\displaystyle{ \int_{-\infty}^{+\infty}\psi ^{\ast} (x,t) \psi (x,t) dx }$ with respect to time and show that the result is $0$. This means we need to confirm that ${ \frac{d}{dt} \left( \int_{-\infty}^{+\infty}\psi ^{\ast} \psi dx \right)=0}$ to complete the proof.

---

$$ \begin{equation} \frac{d}{dt} \left( \int_{-\infty}^{\infty} \psi ^{\ast} (x,t) \psi ^\ (x,t) dx \right) = \int \left( \frac{\partial \psi ^{\ast}}{\partial t}\psi + \psi ^{\ast} \frac{\partial \psi}{\partial t} \right) dx \end{equation} $$

At this point, it follows from the Schrödinger equation that:

$$ \begin{align*} && i\hbar \frac{\partial \psi }{\partial t} &= -\frac{\hbar ^2}{2m}\frac{\partial ^2 \psi}{\partial x ^2} +U\psi \\[1em] \implies && \frac{\partial \psi }{\partial t} &= -\frac{\hbar}{2mi}\frac{\partial ^2 \psi}{\partial x ^2} +\frac{U\psi }{i \hbar} \end{align*} $$

Then the complex conjugate is as follows:

$$ \frac{\partial \psi ^{\ast} }{\partial t} = \frac{\hbar}{2mi}\frac{\partial ^2 \psi ^{\ast}}{\partial x ^2} -\frac{U^{\ast} \psi ^{\ast} }{i \hbar} $$

Substituting this into $(1)$ and separating terms with and without the potential, we get:

$$ \int \frac{\hbar}{2mi} \left( \frac{\partial ^2 \psi ^{\ast}}{\partial x^2}\psi - \psi ^{\ast} \frac{\partial ^2 \psi}{\partial x^2} \right)dx + \int \frac{1}{i\hbar}\left( \psi ^{\ast} U \psi - U^{\ast} \psi ^{\ast} \psi \right) dx $$

Here, the second term (the term including potential) equals $0$ for the following reason:

$$ \braket{ U } = \braket{ \psi \vert U\psi } = \int \psi ^{\ast} U \psi dx $$

$$ \braket{U} ^{\ast} = \braket{ \psi \vert U\psi }^{\ast} = \braket{ U\psi \vert \psi } = \int U^{\ast} \psi ^{\ast} \psi dx $$

Since the expected value of the potential $U$ is real,

$$ { \begin{align*} \braket{U} - \braket{U} ^{\ast} &= \int \psi ^{\ast} U \psi dx - \int U^{\ast} \psi ^{\ast} \psi dx \\ &= \int \left( \psi ^{\ast} U \psi - U^{\ast} \psi ^{\ast} \psi \right)dx = 0 \end{align*}} $$

we modify the form of the first term (the term without potential) using the following equation.

$$ \begin{align*} \frac{\partial }{\partial x} \left( \frac{\partial \psi ^{\ast}}{\partial x} \psi - \psi ^{\ast} \frac{\partial \psi }{\partial x} \right) &= \left( \frac{\partial ^2 \psi ^{\ast}}{\partial x^2}\psi + \frac{\partial \psi ^{\ast}}{\partial x} \frac{\partial \psi }{\partial x} -\frac{\partial \psi ^{\ast}}{\partial x} \frac{\partial \psi }{\partial x} - \psi ^{\ast} \frac{\partial ^2 \psi}{\partial x^2} \right) \\ &= \left( \frac{\partial ^2 \psi ^{\ast}}{\partial x^2}\psi - \psi ^{\ast} \frac{\partial ^2 \psi}{\partial x^2} \right) \end{align*} $$

Substituting this, because the function values of the wave function at both ends are $0$, we obtain the following:

$$ \begin{align*} \int \frac{\hbar}{2mi} \left( \frac{\partial ^2 \psi ^{\ast}}{\partial x^2}\psi - \psi ^{\ast} \frac{\partial ^2 \psi }{\partial x^2} \right) dx &= \frac{\hbar}{2mi} \int_{-\infty}^{\infty} \frac{\partial }{\partial x} \left( \frac{\partial \psi ^{\ast}}{\partial x} \psi - \psi ^{\ast} \frac{\partial \psi }{\partial x} \right) dx \\ &= \frac{\hbar}{2mi} \left[ \frac{\partial \psi ^{\ast}}{\partial x} \psi - \psi ^{\ast} \frac{\partial \psi }{\partial x} \right]_{-\infty}^{\infty} \\ &= 0 \end{align*} $$

Thus, we reach the following conclusion:

$$ \frac{d}{dt} \left( \int_{-\infty}^{\infty} \psi ^{\ast} (x,t) \psi ^\ (x,t) dx \right) = 0 $$

■