State

A state is something that a system is in. The system is where I perform my measurement; the state is the result of that measurement.

When I perform one kind of a measurement, and then I perform another kind of a measurement, the two results are correlated. In particular, the first result may determine the probability that the second measurement will yield a specific value, i.e., that the system will be in a specific state with respect to the second measurement, as opposed to some other state.

The abstract symbol for a state $x$ is $|x\rangle$. This is just a label; it is not a number.

The symbol for the transition from state $x$ to state $y$ is $\langle y|x\rangle$. This, actually, is a number!

Amplitude

Specifically, it is a complex number called the amplitude. The reason why it is a complex number is experimental: we found that if there are two possible ways for a system to reach state $y$ starting from state $x$, it is not the probabilities, but these complex numbers that will need to be summed:

 $\langle y|x\rangle =\langle y|x\rangle_{\rm first~route}+\langle y|x\rangle_{\rm second~route}.$

Probability

The actual probability that the system in state $x$ will also be in state $y$ is computed as the square of the absolute value of the complex number:

 $P(x\rightarrow y)=\langle y|x\rangle^2=\langle y|x\rangle\langle y|x\rangle^\star.$

Base states

It is assumed as an axiom that any state can be expressed as a sum of base states:

 $|x\rangle=\sum\limits_i|i\rangle.$

What we know about the base states is that they are orthogonal:

$\langle i|j\rangle=\delta_{ij}.$

State vector

The contribution of each base state $|i\rangle$ to $|x\rangle$ is characterized by , which is just a complex number. The state $|x\rangle$ can, therefore, be viewed as a vector that is expressed in terms of base vectors $| i\rangle$ in some complex vector space.

The transition amplitude from state $x$ to state $y$ can be expressed through a set of base states as:

 $\langle y|x\rangle=\sum\limits_i\langle y|i\rangle\langle i|x\rangle.$ (1)

The probability that a system in state $x$ is in state $x$ is unity:

$\langle x|x\rangle=\sum\limits_i\langle x|i\rangle\langle i|x\rangle=1.$

The probability that a system in state $x$ is found in some base state is also unity:

$\sum\limits_i|\langle i|x\rangle|^2=\sum\limits_i\langle i|x\rangle\langle i|x\rangle^\star=1.$

From this one can see that

$\langle i|x\rangle=\langle x|i\rangle^\star.$

And since any state $y$ can be a base state in some set of base states, it is true in general that

$\langle y|x\rangle=\langle x|y\rangle^\star.$

Operators

When you do something to a system, you change its state. This is expressed by an operator acting on that state:

$|y\rangle=\hat A|x\rangle.$

This is defined to mean the following:

$\hat A|x\rangle=\sum\limits_{ij}|i\rangle\langle i|\hat A|j\rangle\langle j|x\rangle,$

which means that $\hat A$ is just a collection of matrix elements $A_{ij}$, expressed with respect to some set of base states.

Expectation value

A measurement may be expressed in the form of an operator. If this is the case, the average (expectation) value of that measurement can be expressed as:

$A_\mathrm{av}=\langle x|\hat A|x\rangle,$

which really is just shorthand for

$A_\mathrm{av}=\sum\limits_{ij}\langle x|i\rangle\langle i|\hat A|j\rangle\langle j|x\rangle.$

Wave function

What if there is an infinite number of states? For instance, a particle's position $l$ along a line may be expressed in terms of the set of individual positions as base states. But there is an infinite number of such positions possible. Thus, our sum

$|l\rangle=\sum\limits_x|x\rangle\langle x|l\rangle$

becomes instead the integral

$|l\rangle=\int|x\rangle\langle x|l\rangle dx.$

This equation has little practical meaning since $|x\rangle$ is just an abstract symbol. However, the probability that a system in state $l$ is later found in state $k$, previously expressed as the sum (1):

$\langle k|l\rangle=\sum\limits_x\langle k|x\rangle\langle x|l\rangle.$

is now the integral

$\langle k|l\rangle=\int\langle k|x\rangle\langle x|l\rangle dx.$

Both $\langle k|x\rangle$ and $\langle x|l\rangle$ are just complex numbers; complex-valued functions, in fact, of the continuous variable $x$:

$\langle k|x\rangle=\langle x|k\rangle^\star=\psi^\star(x),$

$\langle l|x\rangle=\psi(x).$

These functions are called wave functions mainly because they typically appear in the form of periodic complex-valued functions. With their help, the transitional probability can now be expressed as:

$P(\phi\rightarrow\psi)=\int\phi^\star(x)\psi(x)dx,$

and the expectation value of an operator can be written as

 $A_\mathrm{av}=\int\phi^\star(x)\hat A\phi(x)dx.$ (2)

Algebraic operator

In this context, $\hat A$ no longer works as a matrix operator converting a state vector into another state vector, but as an algebraic operator converting a wave function into another wave function. How the matrix operator, expressed in terms of states and amplitudes, and the algebraic operator, expressed usually as a differential operator, relate to each other is another question!

Position operator

If we know the probability $P(x)$ that a particle will be at position $x$, we can compute the average position of the particle after many measurements as follows:

$\bar x=\int xP(x)dx.$

But this is the same as

$\int x\phi^\star(x)\phi(x)dx=\int\phi^\star(x)x\phi(x)dx,$

which is formally identical to the expectation value (2) for a measurement that can be expressed in the form of an operator $\hat x$. In other words, $\hat x$ can be viewed as the position operator. When the base states are positions, the position operator is just a multiplication of the wave function by $x$.

Momentum operator

The same computation can be performed for the momentum, using as base states states of definite momentum:

$\int p\phi^\star(p)\phi(p)dp=\int\phi^\star(p)p\phi(p)dp.$

Question is, can the momentum operator be expressed in terms of base states of position?

The amplitude of a system, which is in state $\beta$, to be found in a state of definite momentum $p$, is just the definite integral

$\int\limits_{-\infty}^\infty\langle p|x\rangle\langle x|\beta\rangle dx.$

The relationship between position and momentum, specifically the amplitude for a particle to be found at position $x$ after it has been measured to have momentum $p$ is assumed to be

 $\langle x|p\rangle=e^{ipx/\hbar}.$

So our integral becomes

 $\langle x|p\rangle =\int\limits_{-\infty}^\infty e^{-ipx/\hbar}\langle x|\beta\rangle dx.$ (3)

Now let's use, as $|\beta\rangle$, the state $\hat p|k\rangle$ (it can be any state after all) where $\langle x|k\rangle=\phi(x)$, and the $k$'s are assumed to be states of definite momentum. This way, our earlier expression becomes the expectation value of the momentum:

$\int\limits_{-\infty}^\infty\langle p|x\rangle\langle x|\beta\rangle dx=\int\limits_{-\infty}^\infty\langle p|x\rangle\langle x|\hat p|k\rangle dx=\int\limits_{-\infty}^\infty\langle p|x\rangle p\langle x|k\rangle dx.$

Then $\langle x|\beta\rangle$ is just $p\langle x|k\rangle=p\phi(x)$, and:

$\langle p|\beta\rangle=\int\limits_{-\infty}^\infty e^{-ipx/\hbar}p\phi(x)dx.$

The integral can be computed by observing that $de^{-ipx/\hbar}/dx=(-i/\hbar)pe^{-ipx/hbar}$, integrating in parts, and assuming that $\phi(x)=0$ when $x=\pm\infty$:

$\langle p|\beta\rangle=\frac{\hbar}{i}\int\limits_{-\infty}^\infty e^{-ipx/\hbar}\frac{\partial\phi}{\partial x}dx,$

so, from (3):

$\langle x|\beta\rangle=\frac{\hbar}{i}\frac{\partial\phi}{\partial x},$

and we now have an expression for the momentum operator $\hat p$:

$\hat p=\frac{\hbar}{i}\frac{\partial}{\partial x}.$

Time displacement

How does a system evolve over time? Let's consider the time displacement operator $\hat U(t_1,t_2)$:

$\langle\chi|\hat U(t_1,t_2)|\phi\rangle.$

S-matrix

When $t_1\rightarrow-\infty$ and $t_2\rightarrow+\infty$, we call $\hat U(t_1,t_2)$ the S-matrix.

Making $t_1=t$ and $t_2=t+\Delta t$, observing that when $\Delta t=0$, $U_{ij}$ (in some coordinate representation) must be $\delta_{ij}$, and assuming that for small $\Delta t$, the change in $\phi$ will be linear, we get:

$U_{ij}=\delta_{ij}-\frac{i}{\hbar}H_{ij}(t)\Delta t.$

(the factor $-i/\hbar$ is introduced for reasons of convenience.)

In other words, the difference between the wave function of the two states can be expressed as:

$\phi'-\phi=-\frac{i}{\hbar}\Delta t\bar H\phi,$

or, dividing by $\Delta t$ and recognizing the left-hand side as a time differential:

$i\hbar\frac{\partial\phi}{\partial t}=\bar H\phi.$

Schrödinger equation

Schrödinger, that kind chap, then just decided to use in place of $\hat H$ an operator that he concocted up on the basis of the classical expression for energy:

$E=\frac{p^2}{2m}+V.$

His equation:

 $i\hbar\frac{\partial\phi}{\partial t}=\frac{-\hbar^2}{2m}\nabla^2\phi+V\phi,$

describes the wave function of a particle moving in a potential field $V$.

A crucial thought is that the Schrödinger equation is not as fundamental as you might have been led to believe. Indeed, there's no single Schrödinger equation; the actual equation of a system depends on the characteristics of that system, and is often derived heuristically, through the process of operator substitution.

Operator substitutions

One result is a "rule of thumb": substitution rules that are used to derive quantum operators from the classical quantities of momentum, energy, and position:

$\hat p\rightarrow\frac{\hbar}{i}\frac{\partial}{\partial x},$

$\hat H\rightarrow i\hbar\frac{\partial}{\partial t},$

$\hat x\rightarrow x.$

Commutativity

The operators $\hat x$ and $\hat p$ do not commute:

$(\hat x\circ\hat p)\phi=x\frac{\hbar}{i}\frac{\partial\phi}{\partial x},$

$(\hat p\circ\hat x)\phi=\frac{\hbar}{i}\frac{\partial(x\phi)}{\partial x}=\frac{\hbar}{i}\frac{\partial x}{\partial x}\phi+\frac{\hbar}{i}x\frac{\partial\phi}{\partial x},$

$(\hat p\circ\hat x-\hat x\circ\hat p)\phi=\frac{\hbar}{i}\phi.$

Probability Current

A simple manipulation of the Schrödinger equation—multiplying on the left by $\phi^\star$, multiplying the equation's complex conjugate on the left by $\phi$, and subtracting one from the other—can lead to the continuity equation:

$\phi^\star\left(\frac{\hbar^2}{2m}\nabla^2\phi+V\phi-i\hbar\frac{\partial\phi}{\partial t}\right)-\phi\left(\frac{-\hbar^2}{2m}\nabla^2\phi^\star+V\phi^\star+i\hbar\frac{\partial\phi^\star}{\partial t}\right)$

$=\frac{-\hbar^2}{2m}(\phi^\star\nabla^2\phi+\nabla\phi^\star\nabla\phi-\nabla\phi\nabla\phi^\star-\phi\nabla^2\phi^\star)-i\hbar\left(\phi^\star\frac{\partial\phi}{\partial t}+\phi\frac{\partial\phi^\star}{\partial t}\right)$

$=\frac{-\hbar^2}{2m}\nabla(\phi^\star\nabla\phi-\phi\nabla\phi^\star)-i\hbar\frac{\partial\phi^\star\phi}{\partial t},$

or, substituting

${\bf\mathrm{j}}=\frac{-i\hbar^2}{2m}(\phi^\star\nabla\phi-\phi\nabla\phi^\star),$

$\rho=\hbar\phi^\star\phi,$

we get

$-i\left(\nabla{\bf\mathrm{j}}+\frac{\partial\rho}{\partial t}\right)=0,$

$\nabla{\bf\mathrm{j}}+\frac{\partial\rho}{\partial t}=0.$

References

Feynman, Richard P., The Feynman Lectures on Physics III., Addison-Wesley, 1977
Aitchison, I. J. R. & Hey, A. J. G., Gauge Theories in Particle Physics, Institute of Physics Publishing, 1996