# # The Methods of Quantum Mechanics

As er have seen, if we want to describe a free particle, which is partially localized, we could use a wave packet. De Broglie's hypothesis does not tell us what type of wave one can associate with a particle that is not free and that is acted on by a force. If a particle is acted on by a force, its momentum and its energy will not be constant. Therefore, it is meaningless to talk about a $\lambda$ and $\nu$, because these quantities are changing.

The Schrödinger theory tells us how to obtain the wavefunction $\psi(x,t)$ associated with a particle, when we specify the forces acting on the particle, by giving the potential energy associated with the forces.

Potential energy

In quantum mechanics the potential energy is often refered to simply as the potential.

TIP

## # Schrödinger Equation

$e^{\pm i \theta} = \cos \theta \pm i \sin \theta$

then any 'point' in the conplex plane can be described as

$A e^{\pm i \theta} = A (\cos \theta \pm i \sin \theta)$

A free particle wave function can be described as

$\psi (x, t) = A e^{i(kx-\omega t)} = A [\cos (kx - \omega t) + i \sin(kx - \omega t)]$

This complex form of wave function has real and imaginary parts, both of which have the same wavelength and frequency:

$\lambda = \frac{2\pi}{k}; \nu = \frac{\omega}{2\pi}$

Hence, the corresponding momentum and energy are:

$p = h/\lambda; E = h\nu$

It is $|\psi|^2$ that has physical significance.

$|\psi|^2= \psi^{*} \psi = A^{*} e^{-i(kx-\omega t)} A e^{i(kx-\omega t)} = A^{*} A$

where $\psi^{*}$ is the complex conjugate of $\psi$.

Operators

In mathematics an operator is represented by a symbol or group of symbols and indecates an operation to b performed.

Thus, for example, when the operator $-i\hbar \frac{\partial}{\partial x}$ is placed in front of a function, it indicates that the function is to be differentiated with respect to $x$ and then multiplied by $-i\hbar$, where $i$ is the imaginary number $\sqrt {-1}$. Another operator is $i \hbar \frac{\partial}{\partial t}$.

Applying the operator $-i\hbar \frac{\partial}{\partial x}$ to the complex form of wave function $\psi (x, t) = A e^{i(kx-\omega t)}$.

$-i\hbar \frac{\partial \psi }{\partial x} = -i \hbar iAke^{i(kx - \omega t)} = \frac{h}{2\pi} \frac{2\pi}{\lambda} A e^{i(kx - \omega t)} = p \psi$

Silimarly, applying $i \hbar \frac{\partial}{\partial t}$ to $\psi$, one can obtain

$i \hbar \frac{\partial \psi}{\partial t} = i \hbar (-i\omega) A e^{i(kx - \omega t)} = \frac{h}{2\pi} 2\pi \nu A e^{i(kx - \omega t)} = E \psi$

The total energy of a particle is equal to the kinetic energy plus the potential energy,

$E = \frac{1}{2} mv^2 + E_p = \frac{p^2}{2m} + E_p$

Multiplying both sides of this equation by the wavefunction $\psi$ we obtain

$E\psi = \frac{p^2}{2m} \psi + E_p \psi$

WARNING

When the particle is not free, the relationship cannot be proved, but we postulate that it still holds.

Then we get

$i \hbar \frac{\partial \psi}{\partial t} = \frac{1}{2m} p p \psi + E_p \psi = \frac{1}{2m} \left(-i\hbar \frac{\partial \psi }{\partial x}\right) \left(-i\hbar \frac{\partial \psi }{\partial x}\right) + E_p\psi = - \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} + E_p \psi$

which is conventionally written as

$- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} + E_p \psi = i \hbar \frac{\partial \psi}{\partial t}$

This equation is known as the one-dimensional time-dependent Schrödinger equation. If the potential energy $E_p$ is known, this equation can be solved in principle, and the solution will yield the possible wavefunctions that we can associate with the particle.

First Principle

The Schrödinger equation cannot be derived from first principles; it is the first principle, which cannot be mathematically derived, just as Newton's laws of motion are not derivable. The justification lies in the fact that its predictions agree with the experiment.

The Schrödinger equation is to quantum machanics what Newton's second law is to classical physics.

## # Expectation Values

One of the consequences of the uncertainty principle is that the position, momentum, energy, and so forth of a particle cannot, in general, be precisely determined. We can, however, talk about the average value of these quantities, i.e., expectation values. Can we determine these expectation values without doing the experiment if we know the wavefunction $\psi (x,t)$ associated with the system? The answer is yes.

$P$, the probalility density of a particle's position (at a certain time, or assume $P$ independent of time). The probalility of finding the particle located within $dx$, or $\Delta x$, centered around $x$ is

$P(x) dx$

$\sum P(x) \Delta x = 1, \Delta x \to 0$

When $\Delta x \to 0$,

$\sum P(x) \Delta x = 1 \Rightarrow \int_{-\infty}^\infty P(x) dx = 1$

$\bar{x} = - \infty[P(-\infty) \Delta x] + \cdots + x_n[P(x_n) \Delta x] + \cdots + \infty[P(\infty) \Delta x]$

$\Rightarrow \bar{x} = \sum{x_i} [P(x_i) \Delta x]$

When $\Delta x \to 0$,

$\bar{x} = \int_{-\infty}^{\infty} x P(x) dx = \int_{-\infty}^{\infty} x |\psi|^2 dx = \int_{-\infty}^{\infty} x \psi^{*}\psi dx$

For this relation to be valid, $\psi$ must be normalized so that the sum of their probabilities for all $x$ is unity, that is,

$\int_{-\infty} ^{\infty} \psi^{*} \psi dx = 1$

It is customary to write the numerator with the $x$ between the two wavefunctions as

$\bar{x} = \int_{-\infty}^{\infty} \psi^{*}x\psi dx$

Similarly, the potential energy expectation can be obtained.

$\bar{E_p} = \int_{-\infty}^{\infty} \psi^{*} E_p(x) \psi dx$

To find $\bar{p}$, the operator $-i\hbar \frac{\partial}{\partial x}$ is used.

$\bar{p} = \int_{-\infty}^{\infty} \psi^{*} p \psi dx = \int_{-\infty}^{\infty} \psi^{*} \left(-i\hbar \frac{\partial \psi}{\partial x}\right)dx$

To find $\bar{E}$, the operator $i \hbar \frac{\partial}{\partial t}$ is used.

$\bar{E} = \int_{-\infty}^{\infty} \psi^{*} E \psi dx = \int_{-\infty}^{\infty} \psi^{*} \left( i \hbar \frac{\partial \psi}{\partial t} \right) dx$

## # Separation of variables

How to solve Schrödinger equantion. One standard technique for solving partial differential eqautions is the method of separation of variables. The method consists in trying a solution that is a product of two functions: one a function of $x$ alone and the oher a function of $t$ alone; that is, let

$\psi(x,t) = \chi(x) \Gamma(t)$

WARNING

Why we can separate $x$ and $t$ into two functions?

Let us substitute it into Schrödinger equantion,

$-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}+E_{p}(x) \psi=i \hbar \frac{\partial \psi}{\partial t}$

$-\frac{\hbar^{2}}{2 m} \Gamma(t) \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x) \chi(x) \Gamma(t)=i \hbar \chi(x) \frac{d \Gamma}{d t}$

If we devide through by $\chi(x) \Gamma(t)$, we get

$-\frac{\hbar^{2}}{2 m} \frac{1}{\chi(x)} \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x)=i \hbar \frac{1}{\Gamma(t)} \frac{d \Gamma(t)}{d t}$

The right side of the equation is a function of $t$ alone, and the left side is a function of $x$. Because $x$ and $t$ are independent variables, the equation will be correct if both sides are equal to the same constant $G$ which is called the separation constant, that is,

$-\frac{\hbar^{2}}{2 m} \frac{1}{\chi(x)} \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x) = G$

$i \hbar \frac{1}{\Gamma(t)} \frac{d \Gamma(t)}{d t} = G$

We now have two ordinary differential equations, one involving $x$ alone and the other involving $t$ alone.

Step 1: To solve $\Gamma(t)$,

$\frac{d \Gamma(t)}{dt} = \frac{G}{i\hbar} \Gamma(t)$

$\Gamma(t) = ke^{\frac{G}{i\hbar}t}= ke^{\frac{-iG}{\hbar}t} = e^{\frac{-iG}{\hbar}t} , k = 1$

Now, the question is, what is $G$? To find it out, let't operate on the wavefunction $\psi(x,t)$ with the energy operator $i\hbar \frac{\partial}{\partial t}$.

$i\hbar \frac{\partial \psi}{\partial t} = i \hbar \frac{\partial \left( \chi(x) \Gamma (t)\right)}{\partial t} = i\hbar\chi (x) \frac{\partial \Gamma (t)}{\partial t} = i \hbar \chi(x) \left( \frac{-iG}{\hbar} \right) \Gamma(t) = G\chi(x)\Gamma(t) = G\psi$

As for $i\hbar\frac{\partial \psi}{\partial t} = E \psi$, we see that the separation constant $G$ is the total energy $E$ of the sysytem. Thus the time-dependent part of the wavefunction $\psi(x,t)$ is,

$\Gamma(t) = e^{\frac{-iE}{\hbar}t}$

Step2: To solve $\chi(x)$,

$-\frac{\hbar^{2}}{2 m} \frac{1}{\chi(x)} \frac{d^{2} \chi(x)}{d x^{2}}+E_{p}(x)= G = E$

Then we get

$-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}+E_{p}(x) \chi = E \chi$

This equation is called time-independent Schrödinger equation. For a given $E_p(x)$, the equation has to be solved to find the possible $\chi(x)$'s that can be associated with the system.

So far, there is no restrictions on the values of the total energy $E$ of the system. Thus, for different $E$'s we expect to get different functions $\chi(x)$. We will see, however, that $\chi(x)$ must satisfy certain requirements. That requirements will, in some cases, restrict the number of physically acceptable $\chi(x)$'s and consequently the values of $E$ that the system may have.

The solutions $\chi$ are called the eigenfunctions or eighenstates; the corresponding values of $E$ are called the eigenvalues. (In German, "eigen" means "proper")

TIP

Different eigenvalues correspond to different eigenstates, different wavefunctions. That is amazing.

### # Required Properties of the Eighenfunction and its Derivative

$\chi(x)$ must be well-behaved. This means that the eighenfunciton $\chi(x)$ and it derivative $d\chi/dx$ must have the following properties:

1. They must be finite everywhere.
2. They must be single-valued everywhere.
3. They must be continuous everywhere.(In some special cases $d\chi/dx$ is not continuous.)

When these conditions are satisfied, the eigenfunction ($\chi(x)$) and the associated wavefunction ($\psi(x, t) = \chi(x) \Gamma(t)$) are said to be well-behaved. Why must these conditions be satisfied?

Recall that

$\bar{x} = \int_{-\infty}^{\infty} \psi^{*}x\psi dx$

$\bar{p} = \int_{-\infty}^{\infty} \psi^* ( -i \hbar \frac{\partial}{\partial x}) \psi dx$

If $\chi$ or $d\chi/dx$ are not finite, the average value of the momentum will be infinite; infinite momentum means infinite energy; this is not physically possible.

If $\chi$ or $d\chi/dx$ were not single-valued, the average position and the average momentum would not be defined.

The requirement of contunuity is closely tied to the requrement of finiteness. If $\chi$ is not continuous at a given point in space, then $d\chi/dx$ would be infinite at that point. This, as we have indicated, is not allowed. If $d\chi/dx$ is discontinuous at some point, then $d^2 \chi / d x^2$ would be infinite at that point. From the time-independent Schrödinger equation,

$-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}+E_{p}(x) \chi = E \chi$

this would imply ahat either $E$ or $E_p$ is infinite, which is not physically possible.

TIP

Let $A = a + bi$, $B= c + di$.

$A^* = a - bi, B^* = c - di$

$A^* B^* = (a - bi) (c - di) = ac -(bc + ad) i + bidi = ac - bd - (bc + ad) i$

Then

$AB = (a + bi) ( c + di) = ac + (bc + ad)i + bidi = ac - bd + (bc + ad) i$

$(AB)^* = ac - bd -(bc + ad)i$

Therefore

$(AB)^* = A^* B^*$