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Lecture 5
A Particle in a Box
Study Goal of This Lecture
- Solving a particle in a box
- Know the eignevalue an eigenfunction of particle in a box
- Calculate the expectation value of particle in a box
- Solving a particle in a multidimensional box
5.1 Solving a particle in a box
5.1.1 General Procedure for Solving the T.I.S.E.
Up to this point we have covered the basic ”rules” of quantum mechancis and in the following lectures we will basically repeatedly ”practicing” applications of these rules and approximated methods. It takes some work to familize/internalized quantum mechanics rules. We will start with simple/exactly solvable models. Note that we have gone over some mathematics without doing much practices/examples, particularly, we spent quite some time discuss about ”measurement”, and now we are ready to apply the formalism to real problems. After we have spent some time to practice some simple quantum mechanics problems, we will returm to the issue of fundamental rules of quantum mechanics and explicitly state the ”postulates” or ”ground rules” of quantum mechanic.
Now, lets focus on the applications of quantum mechanics to ”solving” simple problems. We spent some time talking about the important of ”eigenfunction” and ”eigenvalues” of quantum mechanical operators, indeed, we state that a single mea- Eigenfunctions and eigenvalues are impor- tant.
surement causes a wavefunction to collapse into a eigenfunction of the associated observable and results an outcome of a value associated with the eigenfunction. It can’t be emphasized enough that eigenfunctions of an quantum mechanical operator plays the central role in quantum mechanics. Actually, the Schr¨odinger equation is an eigenvalue problem.
Hψˆ = Eψ, Hˆ = pˆ
2 2 m + Vˆ (x). (5.1)
⇒ eigenfunctions of Hˆ, i.e. energy eigenstates, are important ⇒ they represent Hˆ also governs time evolution.
Some declaration or no- tations: Aφ^ ˆ 1 = a 1 φ 1 , this is a eigenproblem, a 1 is a eigenvalue and φ 1 is a eigenfunction. Eigen- function = wavefunc- tion = state.
the likely stable/stationary ”energy” states. (In other words, eigenstate of Hˆ are energetically stable states)
5.1.2 Particle in a 1-D box
The simplest quantum mechanical system, hence always the first problem to solve in a quantum mechanic class, is the particle in a one dimensional box. Consider a particle of mass m constrainted to move in a 1-D box of length a.
Figure 5.1: Superposition of several wave.
The potential energy is:
V (x) =
∞ if x < 0 0 if 0 ≤ x ≤ a ∞ if x > a
En = ℏ
(^2) k 2 2 m =^
ℏ^2
2 m ×^
n^2 π^2 a^2 =^
n^2 h^2 8 ma^2.^ (5.12) corresponding eigenfunction:
ψn(x) = B sin(kx) = B sin
( (^) nπx a
To determine B, we consider the normalization condition
1 =
∫ (^) a 0
ψ∗ n(x)ψn(x)dx = |B|^2 ·
∫ (^) a 0
sin^2 ( nπxa )dx (5.14)
= |B|^2 ·
∫ (^) a 0
[ 1
2 −^
2 cos
( (^2) nπx a
)]
dx
= |B|^2 ·
[ 1
2 a^ −^
a 2 nπ sin
( (^2) nπx a
∣∣a 0
]
=^12 |B|^2 a.
∴ B =
a ,^ (5.15) Any arbitrary φ(x) can be written as φ(x) = ∑ n Cnψn(x).
ψn(x) =
a sin^
nπx a.^ (5.16) It is easy to show that
ψn(x)ψm(x)dx = 0 for n 6 = m, that is {ψn} are orthornor- mal. From these eigenfunctions, we can calculate any properties of the particle.
Let’s summerize the systematic way of solving the Schr¨odinger equation :
- Write down Hˆ
- Write down Schr¨odinger equation and simplify.
- Find general solutions to the Schr¨odinger equation.
- Find boundary condtions.
- Plug in the boundary conditions and find quantization conditions.
- Find En and ψn(x).
- Normalize the eigenfunction. With the eigenfunctions, we can calculate experimental expectation values when the system is prepared in any of the states (Notice that one wavefunction is one quantum state).
5.2 Properties of a Particle in a Box
Let’s plot these eigenfunctions → stationary states.
Figure 5.2: Plot of the wavefunction of particle in a box.
Observation:
- Energy ∝ n^2 , not equally spaced
- As E increases, number of nodes increases too (Number of node = n − 1.)
- The probability |ψ(x)|^2 is more localized in the center at n = 1 and then spread out as n ↑
- The zero point energy is (^8) mnh^22
- Energy ∝ (^) a^12 , so when size of the box increases, the energy drops rapidlly
- Return to classical state at n → ∞ Note that the particle is not ”fixed” localized in space, instead, we can only calculate the ”probability” of finding the particle at a position. Now let’s calculated
5.3 Particle in a 3-D box
This can be genrerlized to higher dimensions, consider a box of dimension a, b, c along x, y, z axis. Thus the Schr¨odinger equation can be written as
− ℏ
2 2 m (^
∂^2
∂x^2 +^
∂^2
∂y^2 +^
∂^2
∂z^2 )ψ(x, y, z) =^ Eψ(x, y, z).^ (5.27)
Note that ψ is a function of three variables, and the variables in the Hamiltonian are independent. i.e. no terms such as (^) ∂x∂ ∂y∂ or xy · · ·. We call this kind of system a ”seperable” system. In general, the solution can be written as a ”product” of indepedent functions Independent degree of freedom → product!! To write in a product is a huge reduction of com- plexity.
ψ(x, y, z) = X(x)Y (y)Z(z). (5.28)
Plugging into the equation, we obtain:
− ℏ
2 2 m [Y^ (y)Z(z)^
∂^2
∂x^2 X(x) +^ X(x)Z(z)^
∂^2
∂y^2 Y^ (y) +^ X(x)Y^ (y)^
∂^2
∂z^2 Z(z)] = EX(x)Y (y)Z(z).
Divide by X(x)Y (y)Z(z) on both sides, we obtain
− ℏ
2 2 m [^
X(x)
∂^2
∂x^2 X(x) +^
Y (y)
∂^2
∂y^2 Y^ (y) +^
Z(z)
∂^2
∂z^2 Z(z)] =^ E^ (5.30) Why it can be expanded in this manner? ⇒
− 2 ℏm^2 ∂x∂^22 X(x) = ExX(x) − 2 ℏm^2 ∂y∂^22 Y (y) = EyY (y) − 2 ℏm^2 ∂z∂^22 Z(z) = Ez Z(z)
, Ex + Ey + Ez = E. (5.31)
The original equation is now separated into three independent equations, and we know the solutions ⇒ each one is a 1-D particle in a box.
The solution for 1-D particle in a box is:
X(x) =
a sin
( (^) nxπx a
Therefore, the wavefunctions of 3-D particle in a box is:
ψnx,ny ,nz (x, y, z) =^ A product state.
abc sin
( (^) nxπx a
sin
( (^) nyπy b
sin
( (^) nz πz c
energy is:
Enx,ny ,nz = h
2 8 m (^
n^2 x a^2 +^
n^2 y b^2 +^
n^2 z c^2 ).^ (5.34) Now the eigenstate are determined by three quantum numbers, i.e. we must assign quantum number for each coordinate to define a state. Observe from the result above, we could find that:
energy → sum, wave function → products Only for independent subsystems! For a cubic box, energy levels on all three directions are equal, so Degenerate: different wavefunctions (eigenfunctions) with the same energy (eigenvalue).
nx ny nz
The first column in the right hand side is the ground state. The second to forth column represent first excited states and it is three-fold degenerate.
