The confinement of a particle in a three-dimensional potential is discussed in this section [4, 6]. The potential is defined as (Figure 6)
V¼ 0, 0≤x<a; 0≤y<b; 0≤z<c
∞, Otherwise
The three dimensional time-independent Schrödinger equation is given as
∇2Ψx, y, z
�2m
ℏ2VΨx, y, z
¼ �EΨx, y, z
(65)
Let the eigen functionΨx, y, z
is taken as the product ofΨxð Þ,x Ψyð Þy andΨzð Þz according to the technique of separation of variables. i.e.,Ψx, y, z
¼ Ψxð ÞΨx yð ÞΨy zð Þ.z
Ψyð ÞΨy zð Þz d2Ψxð Þx
dx2 þΨxð ÞΨx zð Þz d2Ψyð Þy
dy2 þΨxð ÞΨx yð Þy d2Ψzð Þz dz2 �2m
ℏ2 VΨx, y, z
¼ �2m
ℏ2EΨx, y, z
Divide the above equation byΨx, y, z
gives us 1
Ψxð Þx
d2Ψxð Þx dx2 þ 1
Ψyð Þy
d2Ψyð Þy dy2 þ 1
Ψzð Þz
d2Ψzð Þz
dz2 ¼ �2m
ℏ2E (66) Now we can boldly write E as Exð Þ þx Eyð Þ þy Ezð Þz
1 Ψxð Þx
d2Ψxð Þx dx2 þ 1
Ψyð Þy
d2Ψyð Þy dy2 þ 1
Ψzð Þz
d2Ψzð Þz
dz2 ¼ �2m
ℏ2 Exð Þ þx Eyð Þ þy Ezð Þz (67)
Figure 6.
Three-dimensional potential box.
Quantum Mechanics
Now the equation can be separated as follows:
d2Ψxð Þx dx2 þ2m
ℏ2 Exð ÞΨx xð Þ ¼x 0 d2Ψyð Þy
dy2 þ2m
ℏ2Eyð ÞΨy yð Þ ¼y 0 d2Ψzð Þz
dz2 þ2m
ℏ2 Ezð ÞΨz zð Þ ¼z 0 The normalized eigen functionΨxð Þx is given as
Ψxð Þ ¼x 2 a
� �1=2
sin nxπx a
� �
In the same way,Ψyð Þy andΨzð Þz are given as Ψyð Þ ¼y 2
b
� �1=2
sin nyπy b
� �
Ψzð Þ ¼z 2 c
� �1=2
sin nzπz c
� �
Hence, the eigen function�x, y, z�
is given as Ψ�x, y, z�
¼Ψxð ÞΨx yð ÞΨy zð Þ ¼z 8 abc
� �1=2
sin nxπx a
� �
sin nyπy b
� �
sin nzπz c
� �
(68) The energy given values are given as
Exð Þ ¼x n2xπ2ℏ2 2ma2 Ey� �y
¼n2yπ2ℏ2 2mb2 Ezð Þ ¼z n2zπ2ℏ2 2mc2 The total energy E is
E¼Exð Þ þx Eyð Þ þy Ezð Þ ¼z π2ℏ2 2m
n2x a2þn2y
b2þn2z c2
!
(69)
Some of the results are summarized here:
• In a cubical potential box, a¼b¼c, then the energy eigen value becomes, E¼ π2ℏ2
2ma2�n2xþn2yþn2z� : Exactly Solvable Problems in Quantum Mechanics
DOI: http://dx.doi.org/10.5772/intechopen.93317
7. Particle in a 3D box
The confinement of a particle in a three-dimensional potential is discussed in this section [4, 6]. The potential is defined as (Figure 6)
V¼ 0, 0≤x<a; 0≤y<b; 0≤z<c
∞, Otherwise
The three dimensional time-independent Schrödinger equation is given as
∇2Ψx, y, z
�2m
ℏ2 VΨx, y, z
¼ �EΨx, y, z
(65)
Let the eigen functionΨx, y, z
is taken as the product ofΨxð Þ,x Ψyð Þy andΨzð Þz according to the technique of separation of variables. i.e.,Ψx, y, z
¼ Ψxð ÞΨx yð ÞΨy zð Þ.z
Ψyð ÞΨy zð Þz d2Ψxð Þx
dx2 þΨxð ÞΨx zð Þz d2Ψyð Þy
dy2 þΨxð ÞΨx yð Þy d2Ψzð Þz dz2 �2m
ℏ2VΨx, y, z
¼ �2m
ℏ2 EΨx, y, z
Divide the above equation byΨx, y, z
gives us 1
Ψxð Þx
d2Ψxð Þx dx2 þ 1
Ψyð Þy
d2Ψyð Þy dy2 þ 1
Ψzð Þz
d2Ψzð Þz
dz2 ¼ �2m
ℏ2 E (66) Now we can boldly write E as Exð Þ þx Eyð Þ þy Ezð Þz
1 Ψxð Þx
d2Ψxð Þx dx2 þ 1
Ψyð Þy d2Ψyð Þy
dy2 þ 1 Ψzð Þz
d2Ψzð Þz
dz2 ¼ �2m
ℏ2 Exð Þ þx Eyð Þ þy Ezð Þz (67)
Figure 6.
Three-dimensional potential box.
Quantum Mechanics
Now the equation can be separated as follows:
d2Ψxð Þx dx2 þ2m
ℏ2 Exð ÞΨx xð Þ ¼x 0 d2Ψyð Þy
dy2 þ2m
ℏ2 Eyð ÞΨy yð Þ ¼y 0 d2Ψzð Þz
dz2 þ2m
ℏ2 Ezð ÞΨz zð Þ ¼z 0 The normalized eigen functionΨxð Þx is given as
Ψxð Þ ¼x 2 a
� �1=2
sin nxπx a
� �
In the same way,Ψyð Þy andΨzð Þz are given as Ψyð Þ ¼y 2
b
� �1=2
sin nyπy b
� �
Ψzð Þ ¼z 2 c
� �1=2
sin nzπz c
� �
Hence, the eigen function�x, y, z�
is given as Ψ�x, y, z�
¼Ψxð ÞΨx yð ÞΨy zð Þ ¼z 8 abc
� �1=2
sin nxπx a
� �
sin nyπy b
� �
sin nzπz c
� �
(68) The energy given values are given as
Exð Þ ¼x n2xπ2ℏ2 2ma2 Ey� �y
¼n2yπ2ℏ2 2mb2 Ezð Þ ¼z n2zπ2ℏ2 2mc2 The total energy E is
E¼Exð Þ þx Eyð Þ þy Ezð Þ ¼z π2ℏ2 2m
n2x a2þn2y
b2þn2z c2
!
(69)
Some of the results are summarized here:
• In a cubical potential box, a¼b¼c, then the energy eigen value becomes, E¼ π2ℏ2
2ma2�n2xþn2yþn2z� : Exactly Solvable Problems in Quantum Mechanics
DOI: http://dx.doi.org/10.5772/intechopen.93317
• The minimum energy that corresponds to the ground state is E1¼32maπ2ℏ22. Here nx¼ny¼nz¼1.
• Different states with different quantum numbers may have the same energy.
This phenomenon is known as degeneracy. For example, the states (i) nx¼ 2; ny¼nz¼1, (ii) ny¼2; nx¼nz¼1; and (iii) nz¼2; nx¼ny¼1 have the same energy of E¼6πma2ℏ22. So we can say that the energy6πma2ℏ22has a 3-fold degenerate.
• The states (111), (222), (333), (444), … . has no degeneracy.
• In this problem, the state may have zero-fold degeneracy, 3-fold degeneracy or 6-fold degeneracy.
Author details
Lourdhu Bruno Chandrasekar1*, Kanagasabapathi Gnanasekar2and Marimuthu Karunakaran3
1 Department of Physics, Periyar Maniammai Institute of Science and Technology, Vallam, India
2 Department of Physics, The American College, Madurai, India
3 Department of Physics, Alagappa Government Arts College, Karaikudi, India
*Address all correspondence to: [email protected]
© 2020 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/
by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Quantum Mechanics
References
[1]Griffiths DJ. Introduction to Quantum Mechanics. 2nd ed. India:
Pearson
[2]Singh K, Singh SP. Elements of Quantum Mechanics. 1st ed. India: S.
Chand & Company Ltd
[3]Gasiorowicz S. Quantum Mechanics.
3rd ed. India: Wiley
[4]Schiff LI. Quantum Mechanics. 4th ed. India: McGraw Hill International Editions
[5]Peleg Y, Pnini R, Zaarur E, Hecht E.
Quantum Mechanics. 2nd ed. India:
McGraw Hill Editions
[6]Aruldhas G. Quantum Mechanics.
2nd ed. India: Prentice-Hall
Exactly Solvable Problems in Quantum Mechanics DOI: http://dx.doi.org/10.5772/intechopen.93317
• The minimum energy that corresponds to the ground state is E1¼32maπ2ℏ22. Here nx¼ny¼nz¼1.
• Different states with different quantum numbers may have the same energy.
This phenomenon is known as degeneracy. For example, the states (i) nx¼ 2; ny¼nz¼1, (ii) ny¼2; nx¼nz¼1; and (iii) nz¼2; nx¼ny¼1 have the same energy of E¼6πma2ℏ22. So we can say that the energy6πma2ℏ22has a 3-fold degenerate.
• The states (111), (222), (333), (444), … . has no degeneracy.
• In this problem, the state may have zero-fold degeneracy, 3-fold degeneracy or 6-fold degeneracy.
Author details
Lourdhu Bruno Chandrasekar1*, Kanagasabapathi Gnanasekar2and Marimuthu Karunakaran3
1 Department of Physics, Periyar Maniammai Institute of Science and Technology, Vallam, India
2 Department of Physics, The American College, Madurai, India
3 Department of Physics, Alagappa Government Arts College, Karaikudi, India
*Address all correspondence to: [email protected]
© 2020 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/
by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Quantum Mechanics
References
[1]Griffiths DJ. Introduction to Quantum Mechanics. 2nd ed. India:
Pearson
[2]Singh K, Singh SP. Elements of Quantum Mechanics. 1st ed. India: S.
Chand & Company Ltd
[3]Gasiorowicz S. Quantum Mechanics.
3rd ed. India: Wiley
[4]Schiff LI. Quantum Mechanics. 4th ed. India: McGraw Hill International Editions
[5]Peleg Y, Pnini R, Zaarur E, Hecht E.
Quantum Mechanics. 2nd ed. India:
McGraw Hill Editions
[6]Aruldhas G. Quantum Mechanics.
2nd ed. India: Prentice-Hall
Exactly Solvable Problems in Quantum Mechanics DOI: http://dx.doi.org/10.5772/intechopen.93317
Chapter 3
Transitions between Stationary States and the Measurement Problem
María Esther Burgos
Abstract
Accounting for projections during measurements is the traditional measurement problem. Transitions between stationary states require measurements, posing a different measurement problem. Both are compared. Several interpretations of quantum mechanics attempting to solve the traditional measurement problem are summarized. A highly desirable aim is to account for both problems. Not every interpretation of quantum mechanics achieves this goal.
Keywords: quantum measurement problem, transitions between stationary states, interpretations of quantum theory