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Lecture Notes on Quantum Theory I | PHYS 321, Study notes of Physics

Material Type: Notes; Class: Quantum Theory I; Subject: Physics; University: Mesa State College; Term: Fall 2008;

Typology: Study notes

Pre 2010

Uploaded on 08/16/2009

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Exercise: Show that
Ψp(x) = Aeαx
satisfies
i~Ψp(x)
∂x =pΨp(x)
provided that α=ip/~.
The wavefunction for a state with definite momentum is identical to the complex repre-
sentation of a wave, eikx where k= 2π/λ as the wavenumber. From this it follows that the
wavefunction for a particle is exactly that of a wave where the momentum and wavelength
are related by
λ=h
p.(IV.1.59)
This is the deBroglie relationship.
It is often necessary to consider function of momentum such as the kinetic energy of a
particle of mass m.
K=p2
2m
In general these will be constructed in terms of a series such as
A(p) = a0+a1p+a2p2+...=X
n
anpn.
The observable associated with this is defined to be
ˆ
A=a0+a1ˆp+a2ˆp2+... =X
n
anˆxpn.(IV.1.60)
The following theorem is useful.
Theorem: Consider the operator ˆxnwhere nis any integer. Then
ˆpn|Ψi (i~)nnΨ
∂xn.(IV.1.61)
Proof: To follow.
Thus
ˆ
A|Ψi X
n
an(i~)nnΨ
∂xn.
1.4 Determining Energy Eigenstates
It is often useful to provide a quantum mechanical analog of a classical system. Examples
include quantum mechanical descriptions of free particles and harmonic oscillators. The
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Exercise: Show that Ψp(x) = Aeαx satisfies −iℏ

∂Ψp(x) ∂x = pΨp(x)

provided that α = ip/ℏ.

The wavefunction for a state with definite momentum is identical to the complex repre- sentation of a wave, eikx^ where k = 2π/λ as the wavenumber. From this it follows that the wavefunction for a particle is exactly that of a wave where the momentum and wavelength are related by

λ = h p

. (IV.1.59)

This is the deBroglie relationship. It is often necessary to consider function of momentum such as the kinetic energy of a particle of mass m.

K = p^2 2 m In general these will be constructed in terms of a series such as

A(p) = a 0 + a 1 p + a 2 p^2 +... =

n

anpn.

The observable associated with this is defined to be

Aˆ = a 0 + a 1 pˆ + a 2 pˆ^2 +... =

n

an xpˆ n. (IV.1.60)

The following theorem is useful.

Theorem: Consider the operator ˆxn^ where n is any integer. Then

pˆn^ |Ψ〉 ↔ (−iℏ)n^ ∂nΨ ∂xn^

. (IV.1.61)

Proof: To follow. •

Thus Aˆ |Ψ〉 ↔

n

an (−iℏ)n^ ∂nΨ ∂xn^

1.4 Determining Energy Eigenstates

It is often useful to provide a quantum mechanical analog of a classical system. Examples include quantum mechanical descriptions of free particles and harmonic oscillators. The

task will be to use the classical description as a guide for determining the states and dynamics of the analogous quantum system. In general, any quantum mechanical system can be described in terms of energy eigen- states; these are the states that return specific energy measurement outcomes with certainty. Energy eigenstates satisfy Hˆ |φE 〉 = E |φE 〉 (IV.1.62)

where where Hˆ is the Hamiltonian operator, E denotes a possible energy measurement outcome and |φE 〉 is the state that gives outcome E with certainty when the energy of the system is measured. The importance of these lies in the fact that a general state can always be expressed as a superposition of energy eigenstates, i.e.

|Ψ〉 =

E

cE |φE 〉

where cE are complex numbers. Additionally the time evolution of a system initially in an energy eigenstate, |Ψ(0)〉 = |φE 〉 proceeds via,

|Ψ(t)〉 = Uˆ (t) |Ψ(0)〉 = Uˆ (t) |φE 〉 = e−i^ Ht/ˆ ℏ^ |φE 〉 = e−iEt/ℏ^ |φE 〉.

This can be extended to superpositions in the general fashion,

|Ψ(0)〉 =

E

cE |φE 〉 =

E

cE e−iEt/ℏ^ |φE 〉.

Thus the key step in describing quantum systems in one dimension is to determine the Hamiltonian and to find its eigenstates. This is informed by the energy for the corresponding classical system. The total energy of a particle of mass m is

E =

p^2 2 m

  • V (x)

where p is the particle’s momentum and V (x) is the potential energy. As for spin-1/ systems, the transition to the Hamiltonian is accomplished by replacing the physically measurable quantities x and p by observables, i.e. Hermitian operators, ˆx and ˆp. Thus

Hˆ = pˆ

2 2 m

  • V (ˆx). (IV.1.63)

Eq. (IV.1.62) gives an operator equation

[ pˆ^2 2 m

  • V (ˆx)

]

|φE 〉 = E |φE 〉.

In some circumstances, such as the quantum harmonic oscillator, it is possible to proceed directly from this algebraic equation and via a series of operator manipulations arrive at

Inside the well (0 6 x 6 L), the time-independent Schr¨odinger equation gives

ℏ^2

2 m

∂^2 φE (x) ∂x^2 = EφE (x).

and the corresponding boundary conditions are

φE (0) = φE (L) = 0.

Thus the task is to solve the differential equation

∂^2 φ(x) ∂x^2

2 mE ℏ^2 φ(x)

subject to the given boundary conditions. The general solution is

φE (x) = A sin kx + B cos kx

where A and B are complex constants and k satisfies

k^2 = 2 mE ℏ^2

. (IV.1.66)

The first boundary condition implies

0 = φE (0) = A sin k0 + B cos k0 = B

while the second gives

0 = φE (L) = A sin kL + B cos kL = A sin kL.

the only non-trivial solutions are that

k =

nπ L

where n = 1, 2 , 3 ,.... Thus the possible energy values and states can be labeled by a discrete quantity, namely n. Substituting into Eq. (IV.1.66) gives the possible energy values

En =

n^2 π^2 ℏ^2 2 mL^2

(IV.1.67)

where the index n labels the possibilities. This can be used to label the corresponding eigenstates, i.e. |φn〉 which corresponds to wavefunction

φn(x) = A sin

( (^) nπx L

. (IV.1.68)

The constant A may be determined by applying the normalization condition 〈φn|φn〉 = 1.

Exercise: Show that the normalization condition for any infinite well energy eigen- state implies that |A|^2 =

L