15. Formalism of Quantum Mechanics: Vector Space

In quantum mechanics, the state of a system is described by an element of abstract vector space, the so-called state space. In Dirac notation, an element of this space is called a ket and is denoted by the symbol $\left| {} \right\rangle$

A vector space consists of a set of vectors | α >, | β >, | γ > ... together with a set of scalars (a, b, c …). In fact, we will be working with vectors that live in spaces of infinite dimension and the scalars will be ordinary complex numbers. Two important operations are: vector addition and scalar multiplication.

Vector addition:

The ‘sum’ of any two vectors is another vector: | α > + | β > = | γ >

Commutative: | α > + | β > = | β > + | α >         

Associative: | α > + (| β > + | γ >) = (| α > + | β >) + | γ >                    

There is a zero vector | 0 > with the property: | α > + | 0 > = | α >     

The associated inverse vector | -α > has the property: | α > + | -α >  = | 0 >

Scalar multiplication:
The ‘product’ of a scalar with a vector is another vector: a | α > = | γ >

Distributive:     a (| α > + | β >) = a | α > + a | β >

                           (a + b) | α > = a | α > + b | β >

Associative: a (b | α >) = (ab) | α >
Multiplication with scalars 0 and 1 has the properties: 0 | α > = 0; 1 | α > = | α > and |-α> = (-1)|α> 

Two important definitions in linear algebra are: linear combinations and linear dependence, these are closely related to systems of linear equations. A vector | v >  is a linear combination of vectors | α >, | β >, | γ > ... if there exist scalars (k₁, k₂, k₃ …) such that:

| v > = k₁ | α > + k₂ | β > + k₃ | γ > + ... + k_n | ω >                                         (15.1)

that is, if the vector equation:

| v > = x₁ | α > + x₂ | β > + x₃ | γ > + ...+ x_n | ω >                                        (15.2)

Has a solution where x_i are the unknown scalars.

Vectors | α >, | β >, | γ > ... are linearly dependent  if there exist scalars (k₁, k₂, k₃ …), not all zero, such that:

k₁ | α > + k₂ | β > + k₃ | γ > + ... + k_n | ω > = 0                                      (15.3)

that is, if the vector equation:

x₁ | α > + x₂ | β > + x₃ | γ > + ... + x_n | ω > = 0                                   (15.4)

has a nonzero solution where x_i are unknown scalars. Otherwise, the vectors are said to be linearly independent. For instance, in three dimensions the unit vector $\hat{k}$ is linearly independent of $\hat{i}$ and $\hat{j}$, but any vector in the xy-plane is linearly dependent on $\hat{i}$ and $\hat{j}$. A set of vectors is linearly independent if each one is linearly independent of all the rest. A collection of vectors span the space if every vector can be written as a linear combination of the members of this set. A set of linearly independent vectors that spans the space is called a basis. The number of vectors in any basis is the dimension of the space. Any given vector with n-tuple of its components | α > ↔ (a₁, a₂, a₃ ... a_n), can be represented with respect to a prescribed basis | e₁ >, | e₂ >, | e₃ > ... | e_n >:

| α > = a₁ | e₁ > + a₂ | e₂ > + a₃ | e₃ > + ... + a_n | e_n >                                 (15.5)

It is often more convenient to work with the components than with the ‘abstract’ vectors. For instance, addition of two vectors can be done by adding the corresponding components:

| α > + | β > ↔ (a₁ + b₁, a₂ + b₂, a₃ + b₃ ...  a_n + b_n)                                 (15.6)

Multiplication by a scalar can be simply done by multiplying each component:

c | α > ↔ (ca₁, ca₂, ca₃ ... ca_n)                                               (15.7)

Component of zero vector is represented by a string of zeroes:

| 0 > ↔ (0, 0, 0 ... 0)                                                     (15.8)

And the components of the inverse vector have their signs reversed:

| -α > ↔ (-a₁, -a₂, -a₃ ... -a_n)                                                (15.9)

There are two kinds of vector products in three dimensional space: the inner/dot product and cross product.

Vector space that is formed by inner product is called an inner product space. The dot product of two vectors | α > and | β > which is written as: < α | β >,  is a complex number with the following properties:

< β | α > = < α | β >^*                                                   (15.10)

< α | α >  ≥  0, and < α | α > = 0 ↔ | α > = | 0 >                                 (15.11)

< α | (b | β > + c | γ >) = b < α | β > + c < α | γ >                                (15.12)

The inner product of any vector is a non-negative number therefore, its square root is real. We call this the norm or the “length” of the vector:

|| α || ≡ (< α | α >)^1/2                                                   (15.13)

A unit vector with norm is 1, is said to be normalized and two vectors whose inner product is equal to zero are called orthogonal. An orthonormal set is a collection of mutually orthogonal normalized vectors, which can be defined as follows,

< α_i | α_j > ≡ δ_ij                                                    (15.14)

It is always possible and convenient to choose and orthonormal basis, so that the dot product can be written in terms of their components:

< α | β > = $a_{1}^{*}{{b}_{1}}+a_{2}^{*}{{b}_{2}}+...+a_{n}^{*}{{b}_{n}}$                          (15.15)

And the squared norm of the vector becomes:

< α | β > = | a₁ |² + | a₂ |² + ... + | a_n |²                                   (15.16)

With the components themselves expressed in term of basis < e_i | :

a_i = < e_i | α >                                                      (15.17)

The question then arises as to what is the angle between two vectors? In ordinary vector analysis the angle between two vectors is given by:

$\cos \theta =\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}$                                                    (15.18)

And by means of Schwarz inequality:

|< α | β >|²  ≤  < α | α > < β | β >                                                 (15.19)

The angle between | α > and | β > can be generalized by the formula:

$\cos \theta =\sqrt{\frac{<\alpha |\beta ><\beta |\alpha >}{<\alpha |\alpha ><\beta |\beta >}}$                                                 (15.20)

Solved Example #1: The Infinite Square Well

Solved Example #1: The Infinite Square Well is attached.

11. Example #1 (Application of Schrödinger equation): The Infinite Square Well

Consider a particle with mass m, subjected in a potential:

V (x) = 0 for 0 ≤ x ≤ a, and V (x) → ∞ elsewhere.

The particle is completely free, except at the two ends x = 0 and x = a where an infinite force prevents the particle from escaping. In this case, we would like to find the general solution to the Schrödinger equation for this particular potential.

 Figure 11.1. Infinite square well potential.

The probability of finding the particle outside the well is zero, i.e. ψ (x) = 0. While inside the well, the time-independent Schrödinger equation is:

$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=E\psi $                                                    (11.1)

Or:

$\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=-{{k}^{2}}\psi $, with $k=\frac{\sqrt{2mE}}{\hbar }$                                        (11.2)

Equation (11.2) is the simple harmonic oscillator equation, and the general solution is given by:

$\psi (x)=A\sin kx+B\cos kx$                                           (11.3)

With A and B are arbitrary constants. These constants are fixed by boundary conditions of the problem. ψ is continuous, when the potential V (x) goes to infinity, i.e. ψ (a) =  ψ (0) = 0.

Substituting the boundary condition ψ (0) = 0 into equation (11.3), we obtain: B = 0, and hence:

$\psi (x)=A\sin kx$                                                (11.4)

The boundary condition ψ (a) = A sin ka = 0, this condition can be satisfied for A = 0, then ψ (x) = 0. However, if A = 0, we get a trivial, “non-normalizable” solution. Therefore, sin ka = 0 which means that:

ka = 0, ±π, ±2π, ±3π, …                                           (11.5)

The negative solution gives nothing new and the minus sign can be absorbed into A. So the distinct solution is given by:

${{k}_{n}}=\frac{n\pi }{a}$  with n = 1, 2, 3, …                                         (11.6)

With the help of boundary condition at x = a, the possible allowed values of energy E_n can be determined:

${{E}_{n}}=\frac{{{\hbar }^{2}}k_{n}^{2}}{2m}=\frac{{{n}^{2}}{{\pi }^{2}}{{\hbar }^{2}}}{2m{{a}^{2}}}$                                                   (11.7)

In order to fix the constant A, we have to normalize ψ:

$\int\limits_{0}^{a}{{}}$|A|² sin² (kx) dx = 1, so that: |A|² = ${}^{2}/{}_{a}$                                (11.8)

The simplest is to pick the positive real root for constant A: A = $\sqrt{{2}/{a}\;}$then, the solutions inside the potential well are:

${{\psi }_{n}}(x)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right)$                                              (11.9)

ψ₁ is the state with the lowest energy, the so-called ground state, and the other states with higher energies are the excited states. Notice that the energy increases in proportion to n² and each successive state has one more node as we go up in energy (see figure 11.2 below where first three initial states are depicted). And this is a general property, that applies for any shape of the potential.

 Figure 11.2. The first three states of infinite square well.

As can be seen in figure 11.2, the states are alternately even and odd, with respect to the center of the potential well. This property applies whenever the potential is an even function.

Another important property of the states ψ_n is that, they are mutually orthogonal. And this property can be shown:

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx}=0$                                        (11.10)

For m ≠ n. And for m = n, the integral in equation (11.10) is equal to 1. We can also expressed equation (11.10) in terms of Kronecker delta function which is defined by: δ_mn = 0 if m ≠ n and if m = n, δ_mn = 1. Therefore,

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx={{\delta }_{mn}}}$                                   (11.11)

We say that the ψ’s are orthonormal.

Please note that any other function f (x), can be expressed as linear combination of state in equation (11.9), this is called Dirichlet’s theorem:

$f(x)=\sqrt{\frac{2}{a}}\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sin \left( \frac{n\pi }{a}x \right)}$                                       (11.12)

This expression for f (x), is in fact the Fourier expansion. The expansion coefficients c_n can be evaluated for a given f (x) and by applying the orthonormality of ψ_n. Thus, the m-th coefficient in the expansion of f(x) is given by:

${{c}_{m}}=\int{{{\psi }_{m}}}{{(x)}^{*}}f(x)dx$                                      (11.13)

The stationary states for the infinite square well are:

${{\Psi }_{n}}(x,t)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}$                              (11.14)

And the most general solution to the time-dependent Schrödinger equation is a linear combination of stationary states:

$\Psi (x,t)=\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}}$                           (11.15)