In the previous example we have seen that quantum mechanics allows us to calculate the experimentally verifiable probabilities. In this post we will further build our theory by taking another simple example, and by considering the quantum mechanical description of that system.

Let’s consider a particle trapped in a box.

Its classical description comprises the position and momentum information of the particle at some instant of time. From this initial data and using the classical equations of motion we can calculate the future behavior of the particle for all times.

Now lets look at the quantum description of the system.

How to write the state vector for the particle.

From our previous discussion we know that quantum mechanics associates a vector with the state of the particle. So our first task is to work out how to write this vector. For this we consider some experimentally measurable observable — for example the energy of the particle.

Lets assume that the various energy measurements of the particle result in values $latex E_1$, $latex E_2$,… We can associate a vector with each of these energy values. Now, since the count of such vectors will be infinite ( for infinite possible values of the energy), we can not easily represent these vectors as column matrices as we did in our previous example. So we need to introduce a new compact notation.

A new notation

We adopt a new notation as follows. To each energy value $latex E_n$, we associate a vector denoted as $latex \left|\phi_n\right\rangle$, with $latex n=1,2,3,…$

The vectors $latex |\phi_n\rangle$ span a space called the state space of the particle, and all possible states of the particle are some linear combination of the vectors $latex \left|\phi_n\right\rangle$. We will come to this point in a short time. Lets first consider some properties of the vectors $latex \left|\phi_n\right\rangle$.

Scalar product

These vectors $latex \phi_n\rangle$ satisfy the following relation.
$latex \langle \phi_n|\phi_m\rangle=0 &s=2$

The left side of the above equation represents the scalar ( or inner) product of the two vector $latex \left|\phi_n\right\rangle $ and $latex \left|\phi_m\right\rangle$. The value 0 on the right side tells us that the two vectors are orthogonal to each other. This means that the two vectors are linearly independent, and that the one can not be written in terms of the other.

Now if we consider the scalar product of $latex \left|\phi_n\right\rangle$ with itself, we get
$latex \langle \phi_n|\phi_n\rangle=1 &s=2$

In this case the scalar product on left hand side is a measure of the norm of the vector (It is the square of the norm of the vector $latex |\phi_n\rangle$). From above equation we see that the vectors $latex |\phi_n\rangle$ are normalized.

The last two relations can be written in a more compact form as
$latex \langle \phi_n|\phi_n\rangle=\delta_{nm} &s=2$
where $latex \delta_{nm}$ is called Kronecker delta. Its value is 1 if the two indices in the subscript are the same; else the value is 0.

More generally, the scalar product of two arbitrary vectors $latex \langle \psi|\chi\rangle$ is a complex number. It quantifies how much the vector $latex |\psi\rangle$ resembles $latex |\chi\rangle$, and it satisfies the following properties.

$latex \langle\psi|\chi\rangle = \langle\chi|\psi\rangle^* &s=2 $

$latex \langle\psi| (a|\chi_1\rangle + b |\chi_2\rangle) =a \langle\psi|\chi_1\rangle + b \langle \psi|\chi_2\rangle &s=2 $

Here $latex a$ and $latex b$ are complex numbers.

Outer product

The outer product of the vectors $latex |\phi_n\rangle$ is an operator, that can transform a vector into another. For example consider

$latex (|\phi_n\rangle\langle\phi_n|)|\psi\rangle = \langle \phi_n|\psi\rangle |\phi_n\rangle &s=2$

The operator $latex |\phi_n\rangle\langle\phi_n|$ in this case is a projector that projects any arbitrary vector $latex |\psi\rangle$ to the vector $latex |\phi_n\rangle$. One interesting property of the vectors $latex |\phi_n\rangle$ is that the sum of all the projectors they give rise to equals identity operator. Mathematically it can be written as
$latex \sum_{n=1}^\infty |\phi_n\rangle\langle\phi_n| = \hat I &s=2$

This is called the completion relation. It allows us to treat the vectors $latex |\phi_n\rangle$ as base vectors, and use these base vectors to write any vector, operator, or a scalar product. As example

$latex |\psi\rangle= \hat I |\psi\rangle = \sum_{n=1}^\infty
|\phi_n\rangle\langle\phi_n|\psi\rangle = \sum_{n=1}^\infty c_n |\phi_n\rangle &s=2$

Here, $latex c_n=\langle\phi_n|\psi\rangle$ are called expansion coefficients for the expansion of $latex |\psi\rangle$ in the base vectors $latex |\phi_n\rangle$.

Finally the vectors $latex |\phi_n\rangle$ satisfy the equation
$latex \hat H |\phi_n\rangle = E_n |\phi_n\rangle &s=2 $

In this case we see that the vector is not changed after the action of the operator $latex \hat H$, and is merely multiplied by a number $latex E_n$. This is called an eigenvalue equation. The operator on the left hand side is called the Hamiltonian. It represents the total energy of the system. The above equation tells us that our base vectors $latex |\phi_n\rangle$ are the eigenvectors of the Hamiltonian, and the eigenvalues associated with these vectors are the experimentally observable energies of the system ($latex E_1, E_2, …$ from where we started our discussion).

Lets go back to our particle in a box, and to our question of writing the state vector for the particle. The state of the particle in the box at some time $latex t=0$ is given by $latex |\psi(0)\rangle$. We can expand it in the eigenstates of the Hamiltonian as

$latex |\psi(0)\rangle = \sum_{n=1}^\infty c_n |\phi_n\rangle &s=2$
where $latex c_n=\langle\phi_n|\psi(0)\rangle$ is the overlap or the component of $latex |\psi(0)\rangle$ along the base vector $latex |\phi_n\rangle$.

The time evolution of the system is given by the famous Schrodinger equation

$latex i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle =\hat H |\psi(t)\rangle &s=2$
Where $latex \hat H$ is again the Hamiltonian of the system representing the total energy of the system.
The solution of this equation (in our case) is
$latex |\psi(t)\rangle = e^{-i \hat H t/\hbar } |\psi(0)\rangle &s=2$
Using the eigenvalue equation for the Hamiltonian, and the expansion of the state vector in the eigenvectors of the Hamiltonian we can write the solution as

$latex |\psi(t)\rangle = \sum_{n=1}^\infty c_n e^{-iE_n t/\hbar}|\phi_n\rangle &s=2$

This is the complete description of the particle in the box for all time.
Lets now see what information we can extract from this description. First lets ask the question what is the energy of the particle. For this we will have to perform an experiment where we measure this energy. If we perform such experiment and we keep in mind that quantum mechanics allows us to calculate the probability of random events happening in the nature, we get the following answer

  • The result of the energy measurement will be a random value among all the eigenvlaue of the  Hamiltonian.
  • The probability of obtaining a given eigenvalue  (say $latex E_n$ ) is given by $latex |c_n|^2=|\langle\phi_n|\psi(t)\rangle|^2$. This is the modulus square of the amplitude associated with the path $latex |\phi_n\rangle$.

In our next post we will see how to extract more information from the state vector of the particle.

4 responses to “A Particle in a Box- 1”

  1. […] After this short summary you can look again at A Particle in a Box- 1. […]

  2. […] A Particle in a Box- 1 we have seen that we can write the quantum state of a particle in a box in the eigen-basis of […]

  3. Asad Rana Avatar
    Asad Rana

    Aslam o Alaikum sir
    Can you please elaborate complteness relation? As you mentioned in your belog. “Some of all projectors they give rise to the equals identity Operator”.How it is possible?

    1. Faheel Hashmi Avatar

      It is not some. It is sum of projectors.
      We know that the observables have eigenvectors, and we can make the outer product of these eigenvectors whose job is to project a ket to the corresponding eigenvector.
      Consider a two level system with basis vectors $latex |g\rangle$ and $latex |e\rangle$. The corresponding projector operators will be
      $latex |g\rangle\langle g|$ and $latex |e\rangle\langle e|$. These two operators will act on any arbitrary vector in this two level system state space and will project those vectors to $latex |g\rangle$ and $latex |e\rangle$, respectively. You can convince yourself by carrying out this operation.

      Now the completeness relation says that the sum of these operators is identity operator. For the two level system just discussed it will mean
      $latex |g\rangle\langle g| + |e\rangle\langle e| = \hat I$

      For a particle in a box with (Hamiltonian) eigenvectors $latex |\phi_n\rangle$ the completeness relation will be
      $latex \sum_{n=0}^{\infty} |\phi_n\rangle\langle \phi_n|= \hat I$
      This is what is written in the blog.

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