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:

\(\langle y|x\rangle =\langle y|x\rangle_{\rm first~route}+\langle y|x\rangle_{\rm second~route}. \) <assumption>

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:

\(P(x\rightarrow y)=\langle y|x\rangle^2=\langle y|x\rangle\langle y|x\rangle^\star. \) <assumption>

Base states

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

\( |x\rangle=\sum\limits_i|i\rangle. \) <assumption>

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

\(\langle x|p\rangle=e^{ipx/\hbar}.\) <assumption>

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:

\(i\hbar\frac{\partial\phi}{\partial t}=\frac{-\hbar^2}{2m}\nabla^2\phi+V\phi,\) <assumption>

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&mdash;multiplying on the left by \(\phi^\star\), multiplying the equation's complex conjugate on the left by \(\phi\), and subtracting one from the other&mdash;can lead to the continuity equation:

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

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

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

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