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)

14. Example #14: Free Particle

We now turn to a case of free particle. In this case, the potential V (x) = 0 everywhere and the time-independent Schrödinger equation is:

$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=E\psi $                                                      (14.1)

Or, with $k\equiv {\sqrt{2mE}}/{\hbar }\;$ :

$\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=-{{k}^{2}}\psi $                                                      (14.2)

The general solution to this Schrödinger equation is:

$\psi (x)=A{{e}^{ikx}}+B{{e}^{-ikx}}$                                                (14.3)

It is the same as the case of the inside of the infinite square well. However, unlike the infinite square well, there are no boundary conditions to restrict the possible value of E, therefore, the free particle can carry any energy. The standard time-dependent solution is given by:

$\Psi (x,t)=A{{e}^{ik\left( x-\frac{\hbar k}{2m}t \right)}}+B{{e}^{-ik\left( x+\frac{\hbar k}{2m}t \right)}}$                                    (14.4)

This equation also represents wave of fixed profile, traveling with constant velocity in the positive or negative x-direction. Since every point on the waveform is moving along with constant velocity, its shape doesn’t change as it propagates. The first term in equation (14.4) represents a wave traveling to the right, and the second term represents a wave of the same energy traveling to the left. Let us say that k is negative for waves traveling to the left and k is positive for waves traveling to the right:

$k\equiv \pm \frac{\sqrt{2mE}}{\hbar }$                                                               (14.5)

And equation (14.4) can be rewritten:

${{\Psi }_{k}}(x,t)=A{{e}^{i\left( kx-\frac{\hbar {{k}^{2}}}{2m}t \right)}}$                                                     (14.6)

The velocity of the waves, i.e. the coefficient of t over the coefficient of x, is:

$v=\frac{\hbar |k|}{2m}=\sqrt{\frac{E}{2m}}$                                                       (14.7)

It is half of the velocity of a classical particle: ${{v}_{classical}}=\sqrt{\frac{2E}{m}}$ with energy $E=({1}/{2}\;)mv_{classical}^{2}$. This ‘classical velocity’ of the particle is in fact the group velocity of the particle, i.e. the velocity of the envelope, in contrast to the phase velocity of the individual ripples. Generally, the group velocity is given by: ${{v}_{group}}=\frac{d\omega }{dk}=\frac{\hbar k}{m}$ with $\omega ={{{\hbar }^{2}}k}/{2m}\;$ in our case. And the phase velocity is given by: ${{v}_{phase}}=\frac{\omega }{k}=\frac{\hbar k}{2m}$ which is half of group velocity, thus ${{v}_{classical}}={{v}_{group}}=2{{v}_{phase}}$ .

However, wave function (14.6) is non-normalizable:

$\int\limits_{-\infty }^{\infty }{\Psi _{k}^{*}{{\Psi }_{k}}dx=|A{{|}^{2}}\int\limits_{-\infty }^{\infty }{1}dx=|A{{|}^{2}}\infty }$                                  (14.8)

Physical interpretation of this characteristic is that: A free particle cannot exist in a stationary state, so there is no such thing as a free particle with a definite energy. Nevertheless, the separable solutions still play a mathematical role that is completely independent of their physical interpretation. The general solution to time-dependent Schrödinger equation is still a linear combination of separable solutions (in analogy to equation (10.3) in section 10):

$\Psi (x,t)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\phi (k){{e}^{i(kx-\frac{\hbar {{k}^{2}}}{2m}t)}}dk}$                                     (14.9)

The quantity ${1}/{\sqrt{2\pi }}\;$ usually appears in Fourier analysis and factored out for convenience. While the combination $\left( {1}/{\sqrt{2\pi }}\; \right)\phi (k)dk$ plays the ‘role’ of the coefficient c_n in equation (10.3). Now, the wave function (14.9) can be normalized for appropriate $\phi (k)$. The problem is now to determine $\phi (k)$ so as to fit the initial wave function which is given by:

$\Psi (x,0)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\phi (k){{e}^{ikx}}dk}$                                         (14.10)

The answer of this classical problem is provided by means of Fourier transform:

$f(x)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{F(k){{e}^{ikx}}dk}$ ↔ $F(k)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{f(x){{e}^{-ikx}}dx}$                  (14.11) 

Here, F (k) is the Fourier transform of f (x), and f (x) is the inverse Fourier transform of F (k). These integrals have to exist: the necessary and sufficient condition on f (x) is that $\int\limits_{-\infty }^{\infty }{|f(x){{|}^{2}}dx}$ be finite, in this case $\int\limits_{-\infty }^{\infty }{|F(k){{|}^{2}}dk}$ is also finite. For our purposes, the physical requirement that $\Psi (x,0)$ is normalizable, guarantees that the integrals exist. Therefore, the general solution of the Schrödinger equation for free particle is equation (14.9) with:

$\phi (k)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\Psi (x,0){{e}^{-ikx}}dx}$                                       (14.12)