Quantum Science Philippines
Quantum Science Philippines

AN UNDULATORY THEORY OF THE MECHANICS OF ATOMS AND MOLECULES by E. Schrodinger

EDMAR G. PANTOHAN

This report is based on the very interesting researches of L. de Broglie on what he called “phase waves”. The advantages of the wave theory,

  1. The laws of motion and quantum condition can be derived from Hamiltonian principle.
  2. The discrepancy between the frequency of motion and frequency of emission disappears when the latter frequencies coincide with the difference of the former.
  3. It possible to pursue the so called transitions.
  4. This wave theory is better supported by experiment

Consider a point mass m moving in a conservative field of force in q-space. Using the well known Hamiltonian principle and the kinetic energy of the particle, we can have the Hamiltonian partial differential equation. To solve this equation we put W=Et+S(x,y,z), geometrically we described W as a system of surfaces. By allowing this to vary with time, the phase velocity u of the wave is solve. But this velocity u is not the velocity of the particle which proportional to the square root of the difference between energy and potential.

Though in the above we are dealing with wave surfaces and calculating phase velocity, the whole established analogy is more on geometrical optics than real physical or undulatory optics. The fundamental mechanical conception is that of the path or the orbit of the particle and it corresponds to the conception of rays in optical analogy. But the concept of rays loses its significance in real physical optics as soon as the dimensions of the beam or of material become comparable with the wavelength . Considering this striking fact, the ordinary mechanics is really not applicable to mechanical system of a very small atomic dimensions. The same kind as the non-applicability of geometrical optics to the phenomena of diffraction or interference.

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Perturbation of a 3-dimensional infinite cubical well

Karl Patrick S. Casas

Consider a three-dimensional infinite cubical well

Cubical Well with perturation

[eq]V(x,y,z)=\left\{

\begin{array} {cccccc}

0, & if &0<x<a,& 0<y<a,& and& 0<z<a \\

\infty, &otherwise& & & & \\

\end{array}[/eq]

The stationary states are

[eq]\psi^{0}_{n_x,n_y,n_z}(x,y,z)=\left(2/a\right)^{3/2}\sin\left(\frac{n_x \pi}{a}x\right)\sin\left(\frac{n_y \pi}{a}y\right)\sin\left(\frac{n_z \pi}{a}z\right)[/eq]

and the allowed ground state energy is given by

[eq]E^0_0=3\frac{\pi^2\hbar^2}{2ma^2}[/eq]

.

The first excited state is triply degenerate,

[eq]E^0_1=3\frac{\pi^2\hbar^2}{ma^2}[/eq]

and we denote each degenerate state as

[eq]\psi_a=\psi_{112}, \psi_b=\psi_{121}, \psi_c=\psi_{211}[/eq]

Now, we introduce the perturbation  [eq]H'[/eq] (shown in the figure above),

[eq]H’=\left\{

\begin{array} {cccccc}

V_0, & if &0<x<a/2,& a/2<y<a,& and& 0<z<a \\

0, &otherwise& & & & \\

\end{array}[/eq]

We can get the first-order correction to the ground state energy by

[eq]E^1_0=\langle\psi_{111}|H’|\psi_{111}\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq]

Using the integral formula,

[eq]\int{\sin^2\alpha xdx}=\frac{x}{2}-\frac{\sin2\alpha x}{4\alpha}[/eq].

And so, solving each integral,

[eq]\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx=\left[\frac{x}{2}-\frac{\sin2\pi x/a}{4\pi/a}\right]_0^{a/2}=\frac{x}{2}\left|_0^{a/2}=\frac{a}{4}[/eq]

[eq]\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy=\left[\frac{y}{2}-\frac{\sin2\pi y/a}{4\pi/a}\right]_{a/2}^a=\frac{y}{2}\left|_{a/2}^a=\frac{a}{4}[/eq]

[eq]\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz=\left[\frac{z}{2}-\frac{\sin2\pi z/a}{4\pi/a}\right]_0^{a}=\frac{z}{2}\left|_0^{a}=\frac{a}{2}[/eq]

Thus,

[eq]E_0^1=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{2}\right)^3\left(\frac{1}{4}\right)=\frac{V_0}{4}[/eq].

To solve for the eigenvalue [eq]E_1^1[/eq], we consider the components of

[eq]W=\left(

\begin{array} {ccc}

W_{aa} & W_{ab} & W_{ac}\\

W_{ba} & W_{bb} & W_{bc}\\

W_{ca} & W_{cb} & W_{cc}

\end{array} \right)[/eq]

where we can then solve the corresponding eigenvalue equation,

[eq]\left(

\begin{array} {ccc}

W_{aa} & W_{ab} & W_{ac}\\

W_{ba} & W_{bb} & W_{bc}\\

W_{ca} & W_{cb} & W_{cc}\\

\end{array} \right)[/eq]

[eq]\times\left(

\begin{array} {c}

\alpha \\

\beta\\

\gamma

\end{array}

\right)

= E’ \left(

\begin{array} {c}

\alpha \\

\beta\\

\gamma

\end{array}

\right)[/eq]

Solving for the components,

[eq]W_{aa}=\langle\psi_a|H’|\psi_a\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{2\pi}{a}z\right)[/eq]

we already have the first and second integral. Now, consider the z-part

[eq]\int_0^a\sin^2(2\pi z/a )dz=\left[\frac{z}{2}-\frac{\sin2\left(2\pi/a\right)z}{4\left(2\pi/a\right)}\right]_0^a=\frac{z}{2}\left|_0^a=\frac{a}{2}[/eq]

hence,

[eq]W_{aa}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}[/eq]

Now,

[eq]W_{bb}=\langle\psi_b|H’|\psi_b\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq]

solving the y-integral

[eq]\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}y\big)dy[/eq]

[eq]=\left[\frac{y}{2}-\frac{\sin2\left(2\pi/a\right)y}{4\left(2\pi/a\right)}\right]_0^a=\frac{y}{2}\left|_{a/2}^a =\frac{a}{2}-\frac{a}{4}=\frac{a}{4}[/eq].

So,

[eq]W_{bb}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}[/eq].

And then,

[eq]W_{cc}=\langle\psi_c|H’|\psi_c\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{2\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq].

Considering the x-part,

[eq]\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}x\right)dx=\frac{x}{2}-\frac{\sin2\left(2\pi/a\right)x}{4\left(2\pi/a\right)}\left]_0^a=\frac{x}{2}\left|_{a/2}^a =\frac{a}{2}-\frac{a}{4}=\frac{a}{4}[/eq]

Thus,

[eq]W_{cc}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}[/eq]

Next,

[eq]W_{ab}=\langle\psi_a|H’|\psi_b\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx[/eq]

[eq]\times\int_{a/2}^{a}\sin\left(\frac{\pi}{a}y\right)\sin\left(\frac{2\pi}{a}y\right)dy\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz[/eq]

Here, it’s different from the previous. So we use the following integral,

[eq]\int\sin px\sin qxdx = \frac{\sin(p-q)x}{2(p-q)}-\frac{\sin(p+q)x}{2(p+q)}[/eq]

Solving for the y- and z- integrals,

[eq]\int_{a/2}^{a}\sin\left(\frac{\pi}{a}y\right)\sin\left(\frac{2\pi}{a}y\right)dy=\frac{\sin(1-2)(\pi/a)y}{2(1-2)(\pi/a)}\left|_{a/2}^a-\frac{\sin(1+2)(\pi/a)y}{2(1+2)(\pi/a)}\right|_{a/2}^a[/eq]

[eq]=-\frac{2}{2\pi}-\frac{a}{6\pi}=-\frac{4a}{6\pi}=-\frac{2a}{3\pi}[/eq]

[eq]\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz[/eq]

[eq]=\frac{\sin(2-1)(\pi/a)y}{2(2-1)(\pi/a)}\left|_{0}^a-\frac{\sin(2+1)(\pi/a)y}{2(2+1)(\pi/a)}\right|_{0}^a=0[/eq]

Hence,

[eq]W_{ab}=W_{ba}=0[/eq]

Moving on,

[eq]W_{ac}=\langle\psi_a|H’|\psi_c\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx[/eq]

[eq]\times\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz[/eq]

We have seen in the previous section that the z-part is zero, and so

[eq]W_{ac}=W_{ca}=0[/eq]

Lastly,

[eq]W_{bc}=\langle\psi_b|H’|\psi_c\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx[/eq]

[eq]\int_{a/2}^{a}\sin\left(\frac{2\pi}{a}y\right)\sin\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq]

Solving for the x-integral,

[eq]\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx=\frac{\sin(1-2)(\pi/a)y}{2(1-2)(\pi/a)}\left|_{0}^a-\frac{\sin(1+2)(\pi/a)y}{2(1+2)(\pi/a)}\right|_{0}^a[/eq]

[eq]=\frac{a}{2\pi}+\frac{a}{6\pi}=\frac{4a}{6\pi}=\frac{2a}{3\pi}[/eq]

Therefore,

[eq]W_{bc}=\left(\frac{2}{a}\right)^3V_0\left(\frac{2a}{3\pi}\right)\left(\frac{2a}{3\pi}\right)\left(\frac{a}{2}\right)=-\frac{16}{9\pi^2}V_0[/eq]

[eq]=\left(-\frac{V_0}{4}\right)\left(\frac{8}{3\pi}\right)^2=-\frac{V_0}{4}\kappa=W_{cb}[/eq],

where [eq]\kappa=(8/3\pi)^2[/eq]

Finally,

[eq]W=\frac{V_0}{4}\left(

\begin{array} {ccc}

1 & 0 & 0 \\

0 & 1 & -\kappa \\

0 & -\kappa & 1

\end{array} \right)[/eq]

We can now solve for the characteristic equation given by

[eq]\left|

\begin{array} {ccc}

\frac{V_0}{4}-\lambda & 0 & 0 \\

0 & \frac{V_0}{4}-\lambda & -\kappa\frac{V_0}{4} \\

0 & -\kappa\frac{V_0}{4} & \frac{V_0}{4}-\lambda

\end{array} \right|[/eq]

[eq]= \left(\frac{V_0}{4}-\lambda\right)\left[\left(\frac{V_0}{4}-\lambda\right)^2-\left(-\kappa\frac{V_0}{4}\right)^2\right]=0[/eq]

[eq]\Rightarrow\frac{V_0}{4}-\lambda=0\Rightarrow \lambda=\frac{V_0}{4}[/eq]

[eq]\Rightarrow\left(\frac{V_0}{4}-\lambda\right)=\pm\left(\kappa\frac{V_0}{4}\right)[/eq]

[eq]\Rightarrow\lambda=\frac{V_0}{4}\left(1\mp\kappa\right)[/eq]

Therefore,

[eq]E’_{\lambda}=\left\{

\begin{array} {c}

E_1^0 + \lambda(V_0/4) \\

E_1^0 + \lambda(1+\kappa)(V_0/4) \\

E_1^0 + \lambda(1-\kappa)(V_0/4)

\end{array}[/eq]

This result is exactly the same as in the sample problem related to this given in Grifftihs “Introduction to quantum mechanics”.

About the author: Karl Patrick S. Casas is a masters student of Mindanao State University-Iligan Institute of Technology. He hopes to finish his degree as soon as possible.

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2nd-Order Correction

Rommel J. Jagus

Find the 2nd-order correction to the energies [eq](E_n^{2})[/eq] for the potential [eq]H=\alpha \delta (x-\frac{a}{2})[/eq]

Solution:

[eq] <\Psi_{m}^{0} | H | \Psi_{n}^{0}> = \frac{2} {a} \alpha \int_0^a sin(\frac{m x \pi }{a} ) \delta (x-\frac{a}{2}) sin(\frac{n x \pi }{a})[/eq]

[eq] <\Psi_{m}^{0} | H | \Psi_{n}^{0}> =\frac{2} {a} \alpha [a sin(\frac{m \pi }{a} ) sin(\frac{n \pi }{a})][/eq]

This is zero unless both m and n are odd in which case it is [eq]\pm\frac{2\alpha}{a} [/eq]

[eq] E_{n}^{2}=\sum_{m \ne n, odd} (\frac{2\alpha}{a})^2[/eq]

But

[eq] E_{n}^{o}=\frac{\pi^2 \hbar^2 n^2}{2ma^2}[/eq]

So,

[eq] E_{n}^{2}=2m(\frac{2\alpha}{\pi m})^2\sum_{m \ne n, odd} \frac{1}{(n^2-m^2)}[/eq]

Since [eq]\frac{1}{n^2-m^2}=\frac{1}{2n}(\frac{1}{m+n}-\frac{1}{m-n})[/eq]

For n=1

[eq] \sum= \frac{1}{2} \sum_{3,5,7} (\frac{1}{m+1}-\frac{1}{m-1}) [/eq]

[eq]\sum=\frac{1}{2}(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+[/eq]…[eq] -\frac{1}{2}-\frac{1}{4}-\frac{1}{6} – \frac{1}{8})[/eq]

[eq]\sum= \frac{1}{2}(\frac{-1}{2}) = – \frac{1}{4}[/eq]

For n=3

[eq] \sum= \frac{1}{6} \sum_{1,5,7} (\frac{1}{m+3}-\frac{1}{m-3}) [/eq]

[eq]\sum=\frac{1}{6}(\frac{1}{4}+\frac{1}{8}+\frac{1}{10}+[/eq]…[eq]-\frac{1}{2}-\frac{1}{4}-\frac{1}{8} – \frac{1}{10})[/eq]

[eq]\sum= \frac{1}{6}(\frac{-1}{6}) = – \frac{1}{36}[/eq]

Therefore

[eq] E_{n}^{2}=0 [/eq] if n is even

[eq] E_{n}^{2}=-2m (\frac{\alpha}{\pi\hbar^2\n})^2[/eq] if n is odd

Source: Problem 6.4 A in “Introduction to Quantum Mechanics” by David J. Griffiths

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Quantum Information Processing with Atoms and Photons

Michael J. Jabines

Quantum information science(QIS) is a new field of science and technology, combining and drawing on the disciplines of physical science, mathematics, computer science and engineering. Here, Quantum information processors exploit the quantum features of superposition and entanglement for application not possible in classical devices. It offers the potential for significant improvements in the communication and processing of information.

For an isolated quantum system, the fundamental unit of information is the quantum bit or qubit. Qubits are just quantum two-level system such as the spin of an electron or the polarization of a photon and can be prepared in a coherent superposition state. For a quantum information processor, there are three (3) requirements to have a good quantum hardware:

1. The quantum system must be initialized in a well-defined state.

2. Arbitrary unitary operators must be available and controlled to launch the initial state to an arbitrary entangled state.

3. Measurements of the qubits must be performed with high quantum efficiency.

The first requirements demand that the qubits are well isolated from the environment to ensure pure initial quantum states and to preserve their superposition character, but they must also interact strongly between one another to become entangled. Atomic gases and single photon are among promising candidates to implement quantum information technology because they can be well isolated from their environment. Despite this advantage it is challenging to design controllable interaction between these particles and to store or manipulate quantum information in a reliable way. Quantum information processing requires  qubits to behave as quantum memories for long-storage and for many applications to behave as quantum transmitters for long-distance communication. Cold and localized individual atoms are the natural choice for qubit memories and sources of local entanglement for quantum information processing. Photons, on the other hand are the natural source for the communication of quantum information, as they can traverse large distance through the atmosphere or optical fibers with minimal distance.

Quantum information technology is likely to have an important role in information processing after the demise of Moore’s Law. Current devices come primarily from the areas of quantum optics and atomic physics, usually involving laser-cooled and trapped atoms. But perhaps the most exciting feature of this field is that the first-large scale quantum computer will probably be built from a physical system that is not currently known. Current experiments that control the individual atoms and photons will continue to lead the bizzare features of quantum-mechanical foundations to the forefront. With this new language of information, hope we can gain more insight in the underlying quantum-physical principle, exactly as Shannon’s theory of classical information ushered advance physics responsible for the current digital age.

*A summary of Christopher Monroe’s “Quantum Information Processing with Atoms and Photons”

Michael Jabines is a graduate student in Physics of Mindanao State University-Iligan Institute of technology (MSU-IIT) Iligan City Philippines.

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ATOMIC STRUCTURE OF AN ATOM

Eric M. Alcantara

The present state of atomic theory is characterized by the fact that we not only believe that existence of atoms to be proved, but also we even believe that we have an intimate knowledge of the constituents of the individual atoms. It was in the early decades of the 19th century when the structure of the atom was coming in focus. We know that atoms consist of electrons, protons, neutrons and the nucleus. The discovery of the electron and the euclidation of its properties was the result of the work of J. J. Thompson. It was known that the protons and neutrons could group into the center region called nucleus, which was discovered by Ernest Rutherford in his gold foil experiment. But scientists could not think of any stable arrangements of particles. In 1902, Gilbert Lewis proposed a cubical model for atoms in which the electrons were positioned at the eight corners of the cube in a molecule. J. J. Thompson also suggests the plum pudding model in which electrons were jest embedded in the atom. But, both were disproved by Ernest Rutherford after his discovery of nucleus, he made a model that has a resemblance to a planetary system in which electrons orbit around the nucleus. But, Rutherford’s model violates classical electromagnetic theory because if the electrons were stationary, it would fall into the nucleus since the opposite charges on the particles would cause them to attract one another. It cannot be in an orbit circulating the nucleus either because circular motion requires consistent acceleration of circling body to keep it from flying away. Classical electromagnetic theory predicts that charged particles radiates light when it was accelerated. And this radiation will change in frequency as the electron loses energy and eventually it will spiral into the nucleus.
But by intuition, we know that the prediction is fake. Some atoms glow in the dark by emitting visible radiation but they do not change color which indicate frequency change neither they blow up or disappear.
Working under Rutherford was a brilliant young scientist who came to the rescue of Rutherford’s model. It was Neils Hendrik David Bohr who passed on to a study of the structure of the atom on the basis of Rutherford’s discovery of the nucleus. By borrowing the concept from the quantum theory of Max Planck, states that matter emitting or absorbing radiation is not continuous but in discrete bundles of energy called quantum. He suggested that the laws and behavior of the large particles are inadequate to explain all the motions and behavior of all the atoms and electrons. He proposed that an electron despite the fact that it is accelerating does not necessarily radiates energy. Bohr knew that electrons travel close to the nucleus in small orbit possess less energy than those moving in large orbits. This is because of Coulomb’s law that the electron near the nucleus is attracted more strongly by the nucleus than one that is farther away. Energy is required and work must be done in order to move it from the small orbit to a large orbit. Bohr reasoned that the energy difference between smaller and larger orbits must somehow be related to Plank’s quanta of energy. This leads Bohr to conclude that around the nucleus of an atom may occupy only certain precise orbits or energy levels, which leads him into his two postulates:
1. Every atom consists of nucleus and suitable number of electrons revolved around the nucleus in circular orbits. Electrons revolved only in certain non-radiating orbits called stationary orbits for which the total angular momentum is an integral multiple of h/2p where h is plank’s constant.
2. Radiation occurs when an electron jumps from one permitted orbit to another. It is emitted when electron jumps from higher orbit to a lower orbit.

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