Staff Site Universitas Negeri Yogyakarta
量子力学
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
Chapter 6
Time-independent
Perturbation Theory
What
&
Why
Infinite square well with small perturbation
Problem
Suppose we have solved the (time - independen t)
Schr o dinger equation for some potential, now we
perturb the potential sligntly.
Difficulty & why
What
For this problem, we' d like to solve for
Perturbation theory is a sys -
the new eigenfunct ions and eigenvalues :
HΨn = En Ψn ,
tematic procedure for obtai ning approximate solutions
But unless we are very lucky, we' re
unlikely to be able to solve the Shrondi -
to the perturbed problem by
buiding on the known exact
nger equation exactly, for this more co mplicated potential.
solutions to the unperturbed
case.
OUTLINE
•
Non-degenerate perturbation theory
• Degenerate perturbation theory
• Some examples for application
Non-degenerate Perturbation Theory
• General formulation
• First-order theory
• Second-order energies
General formulation
The unperturbed case
Time - independent Shrodinger equation :
0
0
n
0
n
0
n
H Ψ =E Ψ
Eigenvalues:
Eigenfunctions:
E
0
n
Requirement:
0
n
Ψ
Ψ
0
n
Ψ
0
m
= δ nm
Considering the perturbed case, we should solve
for the new eigenvalues and eigenfunctions.
We can write the Hamiltonian
of the new system into the
following two parts
0
H = H + H′
Hamiltonian of the
unperturbed system
Additional
Hamiltonian
of the
perturbed
system
Rewrite the new Hamiltonian as the sum of two terms
A small number ;later
we will crank it up to
1,and H will be the true,
exact Hamiltonian
H = H 0 + λH ′
Write the nth eigenfunction and eigenvalue as
power series in λ as follows
0
n
1
n
2
2
n
Ψn = Ψ + λΨ + λ Ψ +
0
n
1
n
2
2
n
En = E + λE + λ E +
The first-order
correction
The second-order
correction
Plugging the above
H, n and En into the
time-independent
equation and
Collecting like
powers of λ
HΨn = EnΨn
(
)
(
)
H 0 Ψn0 + λ H 0 Ψn1 + H ′Ψ + λ2 H 0 Ψn2 + H ′Ψn1 +
(
)
(
)
= En0 Ψn0 + λ En0 Ψn1 + En1 Ψn0 + λ2 En0 Ψn2 + En1 Ψn1 + En2 Ψn0 +
λ is a device to keep track of the different orders
To lowest order:
0
0
n
0
n
0
n
H Ψ =E Ψ
To first order:
0
1
n
0
n
0
n
1
n
1
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ
To second order：
0
2
n
1
n
0
n
2
n
1
n
1
n
2
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ + E Ψ
And so on
First-Order Theory
• The first-order correction to the
energy
• The first-order correction to the
wave function
The first-order
correction to the energy
0
1
n
0
n
0
n
1
n
1
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ
0
n
0
1
n
Multiplying by (Ψn0 )∗
and integrating
0
n
0
0
0
1
1
0
0
′
Ψ H Ψ + Ψ H Ψn = En Ψn Ψn + En Ψn Ψn
And,
So,
Ψn0 Ψn0 = 1
0
n
0
1
n
0
n
0
n
1
n
Ψ H Ψ =E Ψ Ψ
1
n
0
n
0
n
E = Ψ H′ Ψ
Results: The first-order correction to the energy is the
expectation value of the perturbation in the unperturbed state.
Example
The first-order correction
to the wave function
Rewrite
equation
0
1
n
0
0 1
1
0
′
H Ψ + H Ψn = En Ψn + En Ψn
H 0 − En0）Ψn1 = −
（H ′ − En1）Ψn0
The unperturbed wave functi-
1
n
Ψ =
ons constitute a complete set
Additional,
H 0 Ψn0 = En0 Ψn0
(n)
m
0
m
C Ψ
m≠n
( Em0 − En0 )Cm( n ) Ψm0 = −
（H ′ − En1）Ψn0
m≠n
0
l
Take the inner product with Ψ
( Em0 − En0 )Cm( n ) Ψl0 Ψm0 = − Ψl0 H ′ Ψn0 + En1 Ψl0 Ψn0
m≠ n
1
n
If l = n, we get E ;
C
(n)
m
If l ≠ n, we get C .
(n)
m
=
0
′
Ψ H Ψn
Ψ =
m≠ n
0
n
0
n
Ψ H′ Ψ
0
m
1
n
0
m
E −E
0
m
0
n
E −E
0
m
0
m
Ψ
Notice: although perturbation theory often
yields surprising by accurate energies, the
wave functions are notoriously poor.
Second-order energies
0
2
n
1
0
2
1
1
2
0
′
H Ψ + H Ψn = E n Ψn + E n Ψn + E n Ψn
Take the inner product with Ψn0
Ψn0 H 0 Ψn2 + Ψn0 H ′Ψn1 = En0 Ψn0 Ψn2 + En1 Ψn0 Ψn1 + En2 Ψn0 Ψn0
Ψn0 H 0 Ψn2 = H 0 Ψn0 Ψn2 = En0 Ψn0 Ψn2
0
n
0
n
Meanwhile，Ψ Ψ
0
n
1
n
Ψ Ψ =
C
(n)
m
= 1；
0
n
0
m
Ψ Ψ
m≠n
2
Ψm0 H ′ Ψn0
2
n
E =
m≠ n
En0 − Em0
The Hermiticity of H0
We could proceed to
calculate the second-order
correction to the function,
the third-order correction
to the energy, and so on,
but in practice, the above
results is ordinarily as high
as it is useful to pursue this
method.
Results
Correction to the eigenvalue:
2
0
m
0
n
Ψ H′ Ψ
0
n
′ +
En = E + H nn
m ≠n
0
n
E −E
0
m
+
(1)
Correction to the eigenfunction:
Ψm0 H ′ Ψn0
0
n
Ψn = Ψ +
m≠ n
0
n
E −E
0
m
0
m
Ψ +
(2)
Degenerate Perturbation Theory
• Two-ford degeneracy
• Higher-order degeneracy
Non-degenerate perturbation theory fail?
• In many cases, where two (or more) distinct
states share the same energy, non-degenerate
perturbation theory fail !
• Sometimes, two energy levels exists so near
that non-degenerate perturbation theory
cannot give us a satisfied answer.
So, we must look for some other way to
handle the problem.
Two-ford degeneracy
In order to see how the method generalizes,
we begin with the twofold degeneracy
Suppose ,
0
0
a
0
0
a
0
0
b
0
0
b
H Ψ =E Ψ ,
E0
H Ψ =E Ψ ,
λ
“Lifting of a degeneracy by a
paerturbation
0
a
0
b
Ψ Ψ
=0 .
The perturbation will
“break” the degeneracy
Any linear combination of the above states
0
0
a
0
b
Ψ = αΨ + β Ψ
is still an eigenstate of H0,with the same eigenvalue E0
0
0
0
H Ψ =E Ψ
Essential problem:
0
When we turn off the
perturbation, the “upper” state reduces down to one
0
0
linear combination of Ψa and Ψb , and the “lower”
state reduce to some other linear combination, but
we don’t know a priori what these good linear
combination will be. For this reason we can’t even
calculate the first-order energy (equation 1) because
we don’t know what unperturbed states to use.
As before, we write H, E,
H = H
0
E = E
0
Ψ = Ψ
in the following form
+ λH ′ ,
+ λE
0
1
+ λΨ
2
+ λ E
1
2
2
+ λ Ψ
+
2
,
+
Plug into the stationary equation
To lowest order, H 0 Ψn0 = En0 Ψn0
To first order, H 0 Ψ1 + H ′Ψ 0 = E 0 Ψ1 + E 1Ψ 0
0
1
0
0 1
1 0
′
H Ψ +HΨ =E Ψ +E Ψ
Take the inner
product wi th Ψa0
Ψa0 H 0 Ψ1 + Ψa0 H ′Ψ 0 = E 0 Ψa0 Ψ1 + E 1 Ψa0 Ψ 0
0
0
a
0
0
a
0
0
b
0
0
b
H Ψ =E Ψ ,
H Ψ =E Ψ ,
0
a
0
b
Ψ Ψ
0
=0 ,
0
a
0
b
Ψ = αΨ + β Ψ
0
a
0
a
Meanwhile, H 0 is
Hermitian, the first
term on the left
cancels the first
term on the right
0
a
0
b
α Ψ H′ Ψ + β Ψ H′ Ψ
= αE
1
Results
αWaa + βWab = αE
0
i
1
Where, Wij ≡ Ψ H ′ Ψ
0
j
The “matrix
elements” of H'
, (i, j = a, b)
0
b
Similarly, the inner product with Ψ yields
αWba + βWbb = βE
1
Eliminate βWab
α [WabWba − ( E −Waa )( E −Wbb )] = 0
1
1
If α is not zero, we can get the following equation
1 2
1
( E ) − E (Waa + Wbb ) + (WaaWbb − WabWba ) = 0
1
2
2
E = Waa + Wbb ± (Waa − Wbb ) + 4 Wab
2
1
±
The two roots
correspond to the
two perturbed
energies
(3)
But what if
α is zero?
How should we do in practice
Idea: It would be greatly to our advantage if
we could somehow guess the “good” states
right from the start.
M
O
R
A
L
1. Look around for some Hermitian operator
A that commutes with H' ;
2. Pick as your unperturbed states ones that
are simultaneously eigenfunctions of H0
and A ;
3. Use ordinary first-order perturbation theory.
If you can’t find such an operator, you’ll have to resort to
Equation 3, but in practice this is seldom necessary
Theorem
Proof
Let A be a Hermitian operator
Byassumption, [A, H′] = 0, so
that commutes with H ′.If Ψa0
and Ψb0 are eigenfunctions of A
0
′
Ψ [A, H ]Ψb = 0
0
a
with distinct eigenvalues,
= Ψa0 AH′Ψb0 − Ψa0 H′AΨb0
AΨa0 = µΨa0 , AΨb0 = νΨb0 , µ ≠ ν ,
= AΨa0 H′Ψb0 − Ψa0 H′νΨb0
Then Wab = 0 (and hence Ψa0 and
Ψb0 are the " good" states to use in
perturbation theory).
= (µ −ν ) Ψa0 H′Ψb0 = (µ −ν )Wab.
But µ ≠ν , soWab = 0. QED
Higher-order degeneracy
Higher-order degeneracy
0
E =E
0
1
0
n
0
，Ψ =
0
Matrix equation
0
nk
akΨ ,
0
1
1
H Ψ + H ′Ψ = E Ψ + E Ψ
( H k′′k − E 1δ k ′k )ak = 0
0
k
Secular equation
1
det H k′′k − E δ k ′k = 0
Get E'
Get ak ,
then
get 0
Diagonalizes the perturbation H′
If you can think of an operator A that commutes with H',
and use the simultaneous eigenfunctions of A and H0, then
the W matrix will automatically be diagonal, and you
won’t have to fuss with solving the characteristic equation.
Example
Some Examples For Application
• The fine structure of hydrogen
• The zeeman effect
• Hyperfine splitting
The Fine Structure of Hydrogen
• The relativistic correction
• Spin-orbit coupling
2
2
The simplest
Hamiltonian of
the hydrogen
1
H=−
∇ −
2m
4πε 0 r
Replace m by the
reduced mass
Correct for the motion
of the nucleus
2
e
Relativistic correction
Spin-orbit coupling
Quantization of the Coulomb field
The magnetic interaction
between the dipole moments of
the electron and the proton
Fine structure
Lamb shift
Hyperfine structure
Hierarchy of correction to the Bohr
energies of hydrogen
Bohr energies:
Fine structure:
Lamb shift :
Hyperfine splitting:
2
2
of order
4
2
α mc
of order
5
2
α mc
of order
m
of order
α mc
4
2
α
mc
mp
An application of time-independent
perturbation theory
The relativistic correction
General discussion
2
2
1
2
∇ −
H=−
2m
4πε 0 r
e
Classical formula
Represent
kinetic energy
T
=
2 m
The relativistic formula
T=
The total
relativistic
energy
p
2
mc
2
( c)
1− υ
2
− mc
The rest energy
2
Express T in terms of the relativistic momentum
mυ
The relativistic momentum：p =
( c)
1− υ
( )
m υ c + m c 1− υ
c
2 2
2 4
p c +m c =
2
υ
1−
2
2 2
2 4
( c)
2
2
(
= T + mc
)
2 2
The nonrelativistic
T=
p 2 c 2 + m 2 c 4 − mc 2
1 p
2
T = mc 1 +
2 mc
2
1 p
−
8 mc
limit p
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
Chapter 6
Time-independent
Perturbation Theory
What
&
Why
Infinite square well with small perturbation
Problem
Suppose we have solved the (time - independen t)
Schr o dinger equation for some potential, now we
perturb the potential sligntly.
Difficulty & why
What
For this problem, we' d like to solve for
Perturbation theory is a sys -
the new eigenfunct ions and eigenvalues :
HΨn = En Ψn ,
tematic procedure for obtai ning approximate solutions
But unless we are very lucky, we' re
unlikely to be able to solve the Shrondi -
to the perturbed problem by
buiding on the known exact
nger equation exactly, for this more co mplicated potential.
solutions to the unperturbed
case.
OUTLINE
•
Non-degenerate perturbation theory
• Degenerate perturbation theory
• Some examples for application
Non-degenerate Perturbation Theory
• General formulation
• First-order theory
• Second-order energies
General formulation
The unperturbed case
Time - independent Shrodinger equation :
0
0
n
0
n
0
n
H Ψ =E Ψ
Eigenvalues:
Eigenfunctions:
E
0
n
Requirement:
0
n
Ψ
Ψ
0
n
Ψ
0
m
= δ nm
Considering the perturbed case, we should solve
for the new eigenvalues and eigenfunctions.
We can write the Hamiltonian
of the new system into the
following two parts
0
H = H + H′
Hamiltonian of the
unperturbed system
Additional
Hamiltonian
of the
perturbed
system
Rewrite the new Hamiltonian as the sum of two terms
A small number ;later
we will crank it up to
1,and H will be the true,
exact Hamiltonian
H = H 0 + λH ′
Write the nth eigenfunction and eigenvalue as
power series in λ as follows
0
n
1
n
2
2
n
Ψn = Ψ + λΨ + λ Ψ +
0
n
1
n
2
2
n
En = E + λE + λ E +
The first-order
correction
The second-order
correction
Plugging the above
H, n and En into the
time-independent
equation and
Collecting like
powers of λ
HΨn = EnΨn
(
)
(
)
H 0 Ψn0 + λ H 0 Ψn1 + H ′Ψ + λ2 H 0 Ψn2 + H ′Ψn1 +
(
)
(
)
= En0 Ψn0 + λ En0 Ψn1 + En1 Ψn0 + λ2 En0 Ψn2 + En1 Ψn1 + En2 Ψn0 +
λ is a device to keep track of the different orders
To lowest order:
0
0
n
0
n
0
n
H Ψ =E Ψ
To first order:
0
1
n
0
n
0
n
1
n
1
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ
To second order：
0
2
n
1
n
0
n
2
n
1
n
1
n
2
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ + E Ψ
And so on
First-Order Theory
• The first-order correction to the
energy
• The first-order correction to the
wave function
The first-order
correction to the energy
0
1
n
0
n
0
n
1
n
1
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ
0
n
0
1
n
Multiplying by (Ψn0 )∗
and integrating
0
n
0
0
0
1
1
0
0
′
Ψ H Ψ + Ψ H Ψn = En Ψn Ψn + En Ψn Ψn
And,
So,
Ψn0 Ψn0 = 1
0
n
0
1
n
0
n
0
n
1
n
Ψ H Ψ =E Ψ Ψ
1
n
0
n
0
n
E = Ψ H′ Ψ
Results: The first-order correction to the energy is the
expectation value of the perturbation in the unperturbed state.
Example
The first-order correction
to the wave function
Rewrite
equation
0
1
n
0
0 1
1
0
′
H Ψ + H Ψn = En Ψn + En Ψn
H 0 − En0）Ψn1 = −
（H ′ − En1）Ψn0
The unperturbed wave functi-
1
n
Ψ =
ons constitute a complete set
Additional,
H 0 Ψn0 = En0 Ψn0
(n)
m
0
m
C Ψ
m≠n
( Em0 − En0 )Cm( n ) Ψm0 = −
（H ′ − En1）Ψn0
m≠n
0
l
Take the inner product with Ψ
( Em0 − En0 )Cm( n ) Ψl0 Ψm0 = − Ψl0 H ′ Ψn0 + En1 Ψl0 Ψn0
m≠ n
1
n
If l = n, we get E ;
C
(n)
m
If l ≠ n, we get C .
(n)
m
=
0
′
Ψ H Ψn
Ψ =
m≠ n
0
n
0
n
Ψ H′ Ψ
0
m
1
n
0
m
E −E
0
m
0
n
E −E
0
m
0
m
Ψ
Notice: although perturbation theory often
yields surprising by accurate energies, the
wave functions are notoriously poor.
Second-order energies
0
2
n
1
0
2
1
1
2
0
′
H Ψ + H Ψn = E n Ψn + E n Ψn + E n Ψn
Take the inner product with Ψn0
Ψn0 H 0 Ψn2 + Ψn0 H ′Ψn1 = En0 Ψn0 Ψn2 + En1 Ψn0 Ψn1 + En2 Ψn0 Ψn0
Ψn0 H 0 Ψn2 = H 0 Ψn0 Ψn2 = En0 Ψn0 Ψn2
0
n
0
n
Meanwhile，Ψ Ψ
0
n
1
n
Ψ Ψ =
C
(n)
m
= 1；
0
n
0
m
Ψ Ψ
m≠n
2
Ψm0 H ′ Ψn0
2
n
E =
m≠ n
En0 − Em0
The Hermiticity of H0
We could proceed to
calculate the second-order
correction to the function,
the third-order correction
to the energy, and so on,
but in practice, the above
results is ordinarily as high
as it is useful to pursue this
method.
Results
Correction to the eigenvalue:
2
0
m
0
n
Ψ H′ Ψ
0
n
′ +
En = E + H nn
m ≠n
0
n
E −E
0
m
+
(1)
Correction to the eigenfunction:
Ψm0 H ′ Ψn0
0
n
Ψn = Ψ +
m≠ n
0
n
E −E
0
m
0
m
Ψ +
(2)
Degenerate Perturbation Theory
• Two-ford degeneracy
• Higher-order degeneracy
Non-degenerate perturbation theory fail?
• In many cases, where two (or more) distinct
states share the same energy, non-degenerate
perturbation theory fail !
• Sometimes, two energy levels exists so near
that non-degenerate perturbation theory
cannot give us a satisfied answer.
So, we must look for some other way to
handle the problem.
Two-ford degeneracy
In order to see how the method generalizes,
we begin with the twofold degeneracy
Suppose ,
0
0
a
0
0
a
0
0
b
0
0
b
H Ψ =E Ψ ,
E0
H Ψ =E Ψ ,
λ
“Lifting of a degeneracy by a
paerturbation
0
a
0
b
Ψ Ψ
=0 .
The perturbation will
“break” the degeneracy
Any linear combination of the above states
0
0
a
0
b
Ψ = αΨ + β Ψ
is still an eigenstate of H0,with the same eigenvalue E0
0
0
0
H Ψ =E Ψ
Essential problem:
0
When we turn off the
perturbation, the “upper” state reduces down to one
0
0
linear combination of Ψa and Ψb , and the “lower”
state reduce to some other linear combination, but
we don’t know a priori what these good linear
combination will be. For this reason we can’t even
calculate the first-order energy (equation 1) because
we don’t know what unperturbed states to use.
As before, we write H, E,
H = H
0
E = E
0
Ψ = Ψ
in the following form
+ λH ′ ,
+ λE
0
1
+ λΨ
2
+ λ E
1
2
2
+ λ Ψ
+
2
,
+
Plug into the stationary equation
To lowest order, H 0 Ψn0 = En0 Ψn0
To first order, H 0 Ψ1 + H ′Ψ 0 = E 0 Ψ1 + E 1Ψ 0
0
1
0
0 1
1 0
′
H Ψ +HΨ =E Ψ +E Ψ
Take the inner
product wi th Ψa0
Ψa0 H 0 Ψ1 + Ψa0 H ′Ψ 0 = E 0 Ψa0 Ψ1 + E 1 Ψa0 Ψ 0
0
0
a
0
0
a
0
0
b
0
0
b
H Ψ =E Ψ ,
H Ψ =E Ψ ,
0
a
0
b
Ψ Ψ
0
=0 ,
0
a
0
b
Ψ = αΨ + β Ψ
0
a
0
a
Meanwhile, H 0 is
Hermitian, the first
term on the left
cancels the first
term on the right
0
a
0
b
α Ψ H′ Ψ + β Ψ H′ Ψ
= αE
1
Results
αWaa + βWab = αE
0
i
1
Where, Wij ≡ Ψ H ′ Ψ
0
j
The “matrix
elements” of H'
, (i, j = a, b)
0
b
Similarly, the inner product with Ψ yields
αWba + βWbb = βE
1
Eliminate βWab
α [WabWba − ( E −Waa )( E −Wbb )] = 0
1
1
If α is not zero, we can get the following equation
1 2
1
( E ) − E (Waa + Wbb ) + (WaaWbb − WabWba ) = 0
1
2
2
E = Waa + Wbb ± (Waa − Wbb ) + 4 Wab
2
1
±
The two roots
correspond to the
two perturbed
energies
(3)
But what if
α is zero?
How should we do in practice
Idea: It would be greatly to our advantage if
we could somehow guess the “good” states
right from the start.
M
O
R
A
L
1. Look around for some Hermitian operator
A that commutes with H' ;
2. Pick as your unperturbed states ones that
are simultaneously eigenfunctions of H0
and A ;
3. Use ordinary first-order perturbation theory.
If you can’t find such an operator, you’ll have to resort to
Equation 3, but in practice this is seldom necessary
Theorem
Proof
Let A be a Hermitian operator
Byassumption, [A, H′] = 0, so
that commutes with H ′.If Ψa0
and Ψb0 are eigenfunctions of A
0
′
Ψ [A, H ]Ψb = 0
0
a
with distinct eigenvalues,
= Ψa0 AH′Ψb0 − Ψa0 H′AΨb0
AΨa0 = µΨa0 , AΨb0 = νΨb0 , µ ≠ ν ,
= AΨa0 H′Ψb0 − Ψa0 H′νΨb0
Then Wab = 0 (and hence Ψa0 and
Ψb0 are the " good" states to use in
perturbation theory).
= (µ −ν ) Ψa0 H′Ψb0 = (µ −ν )Wab.
But µ ≠ν , soWab = 0. QED
Higher-order degeneracy
Higher-order degeneracy
0
E =E
0
1
0
n
0
，Ψ =
0
Matrix equation
0
nk
akΨ ,
0
1
1
H Ψ + H ′Ψ = E Ψ + E Ψ
( H k′′k − E 1δ k ′k )ak = 0
0
k
Secular equation
1
det H k′′k − E δ k ′k = 0
Get E'
Get ak ,
then
get 0
Diagonalizes the perturbation H′
If you can think of an operator A that commutes with H',
and use the simultaneous eigenfunctions of A and H0, then
the W matrix will automatically be diagonal, and you
won’t have to fuss with solving the characteristic equation.
Example
Some Examples For Application
• The fine structure of hydrogen
• The zeeman effect
• Hyperfine splitting
The Fine Structure of Hydrogen
• The relativistic correction
• Spin-orbit coupling
2
2
The simplest
Hamiltonian of
the hydrogen
1
H=−
∇ −
2m
4πε 0 r
Replace m by the
reduced mass
Correct for the motion
of the nucleus
2
e
Relativistic correction
Spin-orbit coupling
Quantization of the Coulomb field
The magnetic interaction
between the dipole moments of
the electron and the proton
Fine structure
Lamb shift
Hyperfine structure
Hierarchy of correction to the Bohr
energies of hydrogen
Bohr energies:
Fine structure:
Lamb shift :
Hyperfine splitting:
2
2
of order
4
2
α mc
of order
5
2
α mc
of order
m
of order
α mc
4
2
α
mc
mp
An application of time-independent
perturbation theory
The relativistic correction
General discussion
2
2
1
2
∇ −
H=−
2m
4πε 0 r
e
Classical formula
Represent
kinetic energy
T
=
2 m
The relativistic formula
T=
The total
relativistic
energy
p
2
mc
2
( c)
1− υ
2
− mc
The rest energy
2
Express T in terms of the relativistic momentum
mυ
The relativistic momentum：p =
( c)
1− υ
( )
m υ c + m c 1− υ
c
2 2
2 4
p c +m c =
2
υ
1−
2
2 2
2 4
( c)
2
2
(
= T + mc
)
2 2
The nonrelativistic
T=
p 2 c 2 + m 2 c 4 − mc 2
1 p
2
T = mc 1 +
2 mc
2
1 p
−
8 mc
limit p