today’s outline - august 22, 2012csrri.iit.edu/~segre/phys405/12f/lecture_02.pdftoday’s outline...
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Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
Today’s Outline - August 22, 2012
• 1D Schrodinger equation
• The wave function
• Probability review
• Probability density
• Normalization
• Position and momentum operators
Reading Assignment: Chapter 2.1–2.2
Homework Assignment #01:Chapter 1: 1, 3, 8, 11, 15, 17due Wednesday, August 29, 2012
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 1 / 13
1D Schrodinger equation
i~∂Ψ
∂t= − ~2
2m
∂2Ψ
∂x2+ VΨ
i~∂Ψ
∂t= Total Energy
− ~2
2m
∂2Ψ
∂x2= Kinetic Energy
VΨ = Potential Energy
where the wave function,Ψ(x , t) is a function of bothtime and space
this equation can be viewed asan expression of conservation ofenergy
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 2 / 13
1D Schrodinger equation
i~∂Ψ
∂t= − ~2
2m
∂2Ψ
∂x2+ VΨ
i~∂Ψ
∂t= Total Energy
− ~2
2m
∂2Ψ
∂x2= Kinetic Energy
VΨ = Potential Energy
where the wave function,Ψ(x , t) is a function of bothtime and space
this equation can be viewed asan expression of conservation ofenergy
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 2 / 13
1D Schrodinger equation
i~∂Ψ
∂t= − ~2
2m
∂2Ψ
∂x2+ VΨ
i~∂Ψ
∂t= Total Energy
− ~2
2m
∂2Ψ
∂x2= Kinetic Energy
VΨ = Potential Energy
where the wave function,Ψ(x , t) is a function of bothtime and space
this equation can be viewed asan expression of conservation ofenergy
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 2 / 13
1D Schrodinger equation
i~∂Ψ
∂t= − ~2
2m
∂2Ψ
∂x2+ VΨ
i~∂Ψ
∂t= Total Energy
− ~2
2m
∂2Ψ
∂x2= Kinetic Energy
VΨ = Potential Energy
where the wave function,Ψ(x , t) is a function of bothtime and space
this equation can be viewed asan expression of conservation ofenergy
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 2 / 13
1D Schrodinger equation
i~∂Ψ
∂t= − ~2
2m
∂2Ψ
∂x2+ VΨ
i~∂Ψ
∂t= Total Energy
− ~2
2m
∂2Ψ
∂x2= Kinetic Energy
VΨ = Potential Energy
where the wave function,Ψ(x , t) is a function of bothtime and space
this equation can be viewed asan expression of conservation ofenergy
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 2 / 13
1D Schrodinger equation
i~∂Ψ
∂t= − ~2
2m
∂2Ψ
∂x2+ VΨ
i~∂Ψ
∂t= Total Energy
− ~2
2m
∂2Ψ
∂x2= Kinetic Energy
VΨ = Potential Energy
where the wave function,Ψ(x , t) is a function of bothtime and space
this equation can be viewed asan expression of conservation ofenergy
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 2 / 13
Meaning of the wave function
The wave function, Ψ(x , t) de-scribes everything about a par-ticle (system)
a complex quantity but itsphase is meaningless
spatial integral gives probabilityof the particle being found inthe interval from a to b
Copenhagen interpretation hasproven to be correct one – col-lapse of the wave function aftermeasurement!
∫ b
a|Ψ(x , t)|2 dx
|Ψ|2
xa b
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 3 / 13
Meaning of the wave function
The wave function, Ψ(x , t) de-scribes everything about a par-ticle (system)
a complex quantity but itsphase is meaningless
spatial integral gives probabilityof the particle being found inthe interval from a to b
Copenhagen interpretation hasproven to be correct one – col-lapse of the wave function aftermeasurement!
∫ b
a|Ψ(x , t)|2 dx
|Ψ|2
xa b
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 3 / 13
Meaning of the wave function
The wave function, Ψ(x , t) de-scribes everything about a par-ticle (system)
a complex quantity but itsphase is meaningless
spatial integral gives probabilityof the particle being found inthe interval from a to b
Copenhagen interpretation hasproven to be correct one – col-lapse of the wave function aftermeasurement!
∫ b
a|Ψ(x , t)|2 dx
|Ψ|2
xa b
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 3 / 13
Meaning of the wave function
The wave function, Ψ(x , t) de-scribes everything about a par-ticle (system)
a complex quantity but itsphase is meaningless
spatial integral gives probabilityof the particle being found inthe interval from a to b
Copenhagen interpretation hasproven to be correct one – col-lapse of the wave function aftermeasurement!
∫ b
a|Ψ(x , t)|2 dx
|Ψ|2
xa b
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 3 / 13
Meaning of the wave function
The wave function, Ψ(x , t) de-scribes everything about a par-ticle (system)
a complex quantity but itsphase is meaningless
spatial integral gives probabilityof the particle being found inthe interval from a to b
Copenhagen interpretation hasproven to be correct one – col-lapse of the wave function aftermeasurement!
∫ b
a|Ψ(x , t)|2 dx
|Ψ|2
xa b
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 3 / 13
Meaning of the wave function
The wave function, Ψ(x , t) de-scribes everything about a par-ticle (system)
a complex quantity but itsphase is meaningless
spatial integral gives probabilityof the particle being found inthe interval from a to b
Copenhagen interpretation hasproven to be correct one – col-lapse of the wave function aftermeasurement!
∫ b
a|Ψ(x , t)|2 dx
|Ψ|2
xa b
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 3 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j .
Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Probability review
N =∞∑j=0
N(j)
P(j) =N(j)
N
1 =∞∑j=0
P(j)
〈j〉 =
∑jN(j)
N
=∞∑j=0
jP(j)
Suppose we have a distribution of discrete quantitiessuch as ages of people in a sports stadium, where N(j)is the number of individuals with the age j . Thetotal number of people, N, is
The probability of an individual chosen at randomfrom the crowd having the age j is
The sum of all the probabilities must be 1
The average value of the age (not the most probable)is given by
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 4 / 13
Expectation values
In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as
and given thename, expectation value
One particular quantity, the variance, de-scribes the “width” of the distributionand is given by
Where σ is called the standard deviationof the distribution
〈f (j)〉 =∞∑j=0
f (j)P(j)
σ2 ≡⟨(∆j)2
⟩σ =
√〈j2〉 − 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 5 / 13
Expectation values
In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as and given thename, expectation value
One particular quantity, the variance, de-scribes the “width” of the distributionand is given by
Where σ is called the standard deviationof the distribution
〈f (j)〉 =∞∑j=0
f (j)P(j)
σ2 ≡⟨(∆j)2
⟩σ =
√〈j2〉 − 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 5 / 13
Expectation values
In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as and given thename, expectation value
One particular quantity, the variance, de-scribes the “width” of the distributionand is given by
Where σ is called the standard deviationof the distribution
〈f (j)〉 =∞∑j=0
f (j)P(j)
σ2 ≡⟨(∆j)2
⟩σ =
√〈j2〉 − 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 5 / 13
Expectation values
In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as and given thename, expectation value
One particular quantity, the variance, de-scribes the “width” of the distributionand is given by
Where σ is called the standard deviationof the distribution
〈f (j)〉 =∞∑j=0
f (j)P(j)
σ2 ≡⟨(∆j)2
⟩
σ =
√〈j2〉 − 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 5 / 13
Expectation values
In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as and given thename, expectation value
One particular quantity, the variance, de-scribes the “width” of the distributionand is given by
Where σ is called the standard deviationof the distribution
〈f (j)〉 =∞∑j=0
f (j)P(j)
σ2 ≡⟨(∆j)2
⟩
σ =
√〈j2〉 − 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 5 / 13
Expectation values
In general, the average value of any quan-tity, f (j) which depends on this distribu-tion may be calculated as and given thename, expectation value
One particular quantity, the variance, de-scribes the “width” of the distributionand is given by
Where σ is called the standard deviationof the distribution
〈f (j)〉 =∞∑j=0
f (j)P(j)
σ2 ≡⟨(∆j)2
⟩σ =
√〈j2〉 − 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 5 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩
=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2
=⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Computing the variance
σ2 =⟨(∆j)2
⟩=∑
(∆j)2 P(j)
=∑
(j − 〈j〉)2 P(j)
=∑⟨
(j2 − 2j 〈j〉+ 〈j〉2)P(j)
=∑
j2P(j) +∑
2j 〈j〉P(j) +∑〈j〉2 P(j)
=⟨j2⟩− 2 〈j〉 〈j〉+ 〈j〉2 =
⟨j2⟩− 〈j〉2
∆j = (j − 〈j〉)
expanding the square
dividing into threesums
Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 6 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variables
with the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
N
ρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j)
1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j)
〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2
σ2 ≡⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Continuous variables
We can extend all of these quantities to a system of continuous variableswith the introduction of the probability density, ρ(x) = |Ψ|2
P(j) =N(j)
Nρ(x)
1 =∞∑j=0
P(j) 1 =
∫ +∞
−∞ρ(x)dx
〈f (j)〉 =∞∑j=0
f (j)P(j) 〈f (x)〉 =
∫ +∞
−∞f (x)ρ(x)dx
σ2 ≡⟨(∆j)2
⟩=⟨j2⟩− 〈j〉2 σ2 ≡
⟨(∆x)2
⟩=⟨x2⟩− 〈x〉2
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 7 / 13
Normalizing the wave function
Beyond satisfying the Schrodingerequation, the wave function must alsohave physical significance.
If we are tobelieve the statistical interpretation,the wavefunction must be normalized,that is the integral of the probabilitydensity over all space must be unity.
But if we normalize at t = 0 whatguarantees that the wave function willremain normalized over all times?
1 =
∫ +∞
−∞|Ψ(x , t)|2 dx
This can be proven by starting withthe time derivative of the normal-ization integral.
d
dt
∫ +∞
−∞|Ψ(x , t)|2 dx =
∫ +∞
−∞
∂
∂t|Ψ(x , t)|2 dx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 8 / 13
Normalizing the wave function
Beyond satisfying the Schrodingerequation, the wave function must alsohave physical significance. If we are tobelieve the statistical interpretation,the wavefunction must be normalized,that is the integral of the probabilitydensity over all space must be unity.
But if we normalize at t = 0 whatguarantees that the wave function willremain normalized over all times?
1 =
∫ +∞
−∞|Ψ(x , t)|2 dx
This can be proven by starting withthe time derivative of the normal-ization integral.
d
dt
∫ +∞
−∞|Ψ(x , t)|2 dx =
∫ +∞
−∞
∂
∂t|Ψ(x , t)|2 dx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 8 / 13
Normalizing the wave function
Beyond satisfying the Schrodingerequation, the wave function must alsohave physical significance. If we are tobelieve the statistical interpretation,the wavefunction must be normalized,that is the integral of the probabilitydensity over all space must be unity.
But if we normalize at t = 0 whatguarantees that the wave function willremain normalized over all times?
1 =
∫ +∞
−∞|Ψ(x , t)|2 dx
This can be proven by starting withthe time derivative of the normal-ization integral.
d
dt
∫ +∞
−∞|Ψ(x , t)|2 dx =
∫ +∞
−∞
∂
∂t|Ψ(x , t)|2 dx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 8 / 13
Normalizing the wave function
Beyond satisfying the Schrodingerequation, the wave function must alsohave physical significance. If we are tobelieve the statistical interpretation,the wavefunction must be normalized,that is the integral of the probabilitydensity over all space must be unity.
But if we normalize at t = 0 whatguarantees that the wave function willremain normalized over all times?
1 =
∫ +∞
−∞|Ψ(x , t)|2 dx
This can be proven by starting withthe time derivative of the normal-ization integral.
d
dt
∫ +∞
−∞|Ψ(x , t)|2 dx =
∫ +∞
−∞
∂
∂t|Ψ(x , t)|2 dx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 8 / 13
Normalizing the wave function
Beyond satisfying the Schrodingerequation, the wave function must alsohave physical significance. If we are tobelieve the statistical interpretation,the wavefunction must be normalized,that is the integral of the probabilitydensity over all space must be unity.
But if we normalize at t = 0 whatguarantees that the wave function willremain normalized over all times?
1 =
∫ +∞
−∞|Ψ(x , t)|2 dx
This can be proven by starting withthe time derivative of the normal-ization integral.
d
dt
∫ +∞
−∞|Ψ(x , t)|2 dx =
∫ +∞
−∞
∂
∂t|Ψ(x , t)|2 dx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 8 / 13
Normalizing the wave function
Beyond satisfying the Schrodingerequation, the wave function must alsohave physical significance. If we are tobelieve the statistical interpretation,the wavefunction must be normalized,that is the integral of the probabilitydensity over all space must be unity.
But if we normalize at t = 0 whatguarantees that the wave function willremain normalized over all times?
1 =
∫ +∞
−∞|Ψ(x , t)|2 dx
This can be proven by starting withthe time derivative of the normal-ization integral.
d
dt
∫ +∞
−∞|Ψ(x , t)|2 dx
=
∫ +∞
−∞
∂
∂t|Ψ(x , t)|2 dx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 8 / 13
Normalizing the wave function
Beyond satisfying the Schrodingerequation, the wave function must alsohave physical significance. If we are tobelieve the statistical interpretation,the wavefunction must be normalized,that is the integral of the probabilitydensity over all space must be unity.
But if we normalize at t = 0 whatguarantees that the wave function willremain normalized over all times?
1 =
∫ +∞
−∞|Ψ(x , t)|2 dx
This can be proven by starting withthe time derivative of the normal-ization integral.
d
dt
∫ +∞
−∞|Ψ(x , t)|2 dx =
∫ +∞
−∞
∂
∂t|Ψ(x , t)|2 dx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 8 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)
=∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Look more closely at the integrand andapply the product rule.
∂
∂t|Ψ|2 =
∂
∂t(Ψ∗Ψ)
= Ψ∗∂Ψ
∂t+∂Ψ∗
∂tΨ
= Ψ∗i~2m
∂2Ψ
∂x2− i~
2m
∂2Ψ∗
∂x2Ψ
=i~2m
(Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ
)
using the Schrodinger equation
∂Ψ
∂t=
i~2m
∂2Ψ
∂x2− i
~VΨ
and its complex conjugate
∂Ψ∗
∂t= − i~
2m
∂2Ψ∗
∂x2+
i
~VΨ∗
adding and subtracting ∂Ψ∂x
∂Ψ∗
∂xpermits factoring
∂
∂t|Ψ|2 =
i~2m
(∂Ψ
∂x
∂Ψ∗
∂x+ Ψ∗
∂2Ψ
∂x2− ∂2Ψ∗
∂x2Ψ− ∂Ψ
∂x
∂Ψ∗
∂x
)=
∂
∂x
[i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)]C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 9 / 13
Time independence of normalization
Putting the integrand back into the original expression reveals that wehave an exact differential which can be immediately integrated
d
dt
∫ +∞
−∞|Ψ|2 dx =
∫ +∞
−∞
∂
∂t|Ψ|2 dx
=
∫ +∞
−∞
∂
∂x
i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
= 0
This must vanish if the wave function is well-behaved and approaches 0 at±∞
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 10 / 13
Time independence of normalization
Putting the integrand back into the original expression reveals that wehave an exact differential which can be immediately integrated
d
dt
∫ +∞
−∞|Ψ|2 dx =
∫ +∞
−∞
∂
∂t|Ψ|2 dx
=
∫ +∞
−∞
∂
∂x
i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
= 0
This must vanish if the wave function is well-behaved and approaches 0 at±∞
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 10 / 13
Time independence of normalization
Putting the integrand back into the original expression reveals that wehave an exact differential which can be immediately integrated
d
dt
∫ +∞
−∞|Ψ|2 dx =
∫ +∞
−∞
∂
∂t|Ψ|2 dx
=
∫ +∞
−∞
∂
∂x
i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
= 0
This must vanish if the wave function is well-behaved and approaches 0 at±∞
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 10 / 13
Time independence of normalization
Putting the integrand back into the original expression reveals that wehave an exact differential which can be immediately integrated
d
dt
∫ +∞
−∞|Ψ|2 dx =
∫ +∞
−∞
∂
∂t|Ψ|2 dx
=
∫ +∞
−∞
∂
∂x
i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
= 0
This must vanish if the wave function is well-behaved and approaches 0 at±∞
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 10 / 13
Time independence of normalization
Putting the integrand back into the original expression reveals that wehave an exact differential which can be immediately integrated
d
dt
∫ +∞
−∞|Ψ|2 dx =
∫ +∞
−∞
∂
∂t|Ψ|2 dx
=
∫ +∞
−∞
∂
∂x
i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
= 0
This must vanish if the wave function is well-behaved and approaches 0 at±∞
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 10 / 13
Expectation value of position
Suppose we have many systems allprepared in the same way and all ina state described by the wave func-tion Ψ(x , t).
If we measure the position of theparticle in each of these identicalsystems, we should obtain a re-sult consistent with the expectationvalue of the position x , which iscomputed as
We can expand this and write it ina slightly different way, with the xjust to left of Ψ.
〈x〉 =
∫x |Ψ(x , t)|2 dx
=
∫Ψ∗xΨdx
In this arrangement, x is said tobe an “operator” which acts on thewave function to its right
This will eventually lead directly towhat is called the “bra-ket” no-tation commonly used in quantummechanics.
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 11 / 13
Expectation value of position
Suppose we have many systems allprepared in the same way and all ina state described by the wave func-tion Ψ(x , t).
If we measure the position of theparticle in each of these identicalsystems, we should obtain a re-sult consistent with the expectationvalue of the position x , which iscomputed as
We can expand this and write it ina slightly different way, with the xjust to left of Ψ.
〈x〉 =
∫x |Ψ(x , t)|2 dx
=
∫Ψ∗xΨdx
In this arrangement, x is said tobe an “operator” which acts on thewave function to its right
This will eventually lead directly towhat is called the “bra-ket” no-tation commonly used in quantummechanics.
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 11 / 13
Expectation value of position
Suppose we have many systems allprepared in the same way and all ina state described by the wave func-tion Ψ(x , t).
If we measure the position of theparticle in each of these identicalsystems, we should obtain a re-sult consistent with the expectationvalue of the position x , which iscomputed as
We can expand this and write it ina slightly different way, with the xjust to left of Ψ.
〈x〉 =
∫x |Ψ(x , t)|2 dx
=
∫Ψ∗xΨdx
In this arrangement, x is said tobe an “operator” which acts on thewave function to its right
This will eventually lead directly towhat is called the “bra-ket” no-tation commonly used in quantummechanics.
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 11 / 13
Expectation value of position
Suppose we have many systems allprepared in the same way and all ina state described by the wave func-tion Ψ(x , t).
If we measure the position of theparticle in each of these identicalsystems, we should obtain a re-sult consistent with the expectationvalue of the position x , which iscomputed as
We can expand this and write it ina slightly different way, with the xjust to left of Ψ.
〈x〉 =
∫x |Ψ(x , t)|2 dx
=
∫Ψ∗xΨdx
In this arrangement, x is said tobe an “operator” which acts on thewave function to its right
This will eventually lead directly towhat is called the “bra-ket” no-tation commonly used in quantummechanics.
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 11 / 13
Expectation value of position
Suppose we have many systems allprepared in the same way and all ina state described by the wave func-tion Ψ(x , t).
If we measure the position of theparticle in each of these identicalsystems, we should obtain a re-sult consistent with the expectationvalue of the position x , which iscomputed as
We can expand this and write it ina slightly different way, with the xjust to left of Ψ.
〈x〉 =
∫x |Ψ(x , t)|2 dx
=
∫Ψ∗xΨdx
In this arrangement, x is said tobe an “operator” which acts on thewave function to its right
This will eventually lead directly towhat is called the “bra-ket” no-tation commonly used in quantummechanics.
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 11 / 13
Expectation value of position
Suppose we have many systems allprepared in the same way and all ina state described by the wave func-tion Ψ(x , t).
If we measure the position of theparticle in each of these identicalsystems, we should obtain a re-sult consistent with the expectationvalue of the position x , which iscomputed as
We can expand this and write it ina slightly different way, with the xjust to left of Ψ.
〈x〉 =
∫x |Ψ(x , t)|2 dx
=
∫Ψ∗xΨdx
In this arrangement, x is said tobe an “operator” which acts on thewave function to its right
This will eventually lead directly towhat is called the “bra-ket” no-tation commonly used in quantummechanics.
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 11 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Computing the velocity
As time passes, the expecta-tion value of the position 〈x〉will change. This is a sortof velocity and may be cal-culated as
where we can use our previ-ous result to yield
Choosing u and dv and inte-grating by parts
Since the wave function goesto 0 at ±∞
The two terms can be shownto be identical with anotherintegration by parts and thus
d 〈x〉dt
=
∫x∂
∂tΨ∗Ψdx
=i~2m
∫x∂
∂x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
=i~2m
x
(Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)∣∣∣∣+∞−∞
− i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
= − i~2m
∫ (Ψ∗
∂Ψ
∂x− ∂Ψ∗
∂xΨ
)dx
d 〈x〉dt
= − i~m
∫Ψ∗
∂Ψ
∂xdx
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 12 / 13
Momentum operator
By assuming that this result is re-alted to the expectation value ofthe velocity, we can see how tocompute the expectation value ofthe momentum, p = mv .
If we cast this into the form for anoperator acting on Ψ, we obtain
We now have operators for positionand momentum.
d 〈x〉dx
= 〈v〉
〈p〉 = −i~∫
xΨ∗∂Ψ
∂xdx
=
∫Ψ∗(−i~ ∂
∂x
)Ψdx
xop = x
pop = −i~ ∂∂x
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 13 / 13
Momentum operator
By assuming that this result is re-alted to the expectation value ofthe velocity, we can see how tocompute the expectation value ofthe momentum, p = mv .
If we cast this into the form for anoperator acting on Ψ, we obtain
We now have operators for positionand momentum.
d 〈x〉dx
= 〈v〉
〈p〉 = −i~∫
xΨ∗∂Ψ
∂xdx
=
∫Ψ∗(−i~ ∂
∂x
)Ψdx
xop = x
pop = −i~ ∂∂x
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 13 / 13
Momentum operator
By assuming that this result is re-alted to the expectation value ofthe velocity, we can see how tocompute the expectation value ofthe momentum, p = mv .
If we cast this into the form for anoperator acting on Ψ, we obtain
We now have operators for positionand momentum.
d 〈x〉dx
= 〈v〉
〈p〉 = −i~∫
xΨ∗∂Ψ
∂xdx
=
∫Ψ∗(−i~ ∂
∂x
)Ψdx
xop = x
pop = −i~ ∂∂x
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 13 / 13
Momentum operator
By assuming that this result is re-alted to the expectation value ofthe velocity, we can see how tocompute the expectation value ofthe momentum, p = mv .
If we cast this into the form for anoperator acting on Ψ, we obtain
We now have operators for positionand momentum.
d 〈x〉dx
= 〈v〉
〈p〉 = −i~∫
xΨ∗∂Ψ
∂xdx
=
∫Ψ∗(−i~ ∂
∂x
)Ψdx
xop = x
pop = −i~ ∂∂x
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 13 / 13
Momentum operator
By assuming that this result is re-alted to the expectation value ofthe velocity, we can see how tocompute the expectation value ofthe momentum, p = mv .
If we cast this into the form for anoperator acting on Ψ, we obtain
We now have operators for position
and momentum.
d 〈x〉dx
= 〈v〉
〈p〉 = −i~∫
xΨ∗∂Ψ
∂xdx
=
∫Ψ∗(−i~ ∂
∂x
)Ψdx
xop = x
pop = −i~ ∂∂x
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 13 / 13
Momentum operator
By assuming that this result is re-alted to the expectation value ofthe velocity, we can see how tocompute the expectation value ofthe momentum, p = mv .
If we cast this into the form for anoperator acting on Ψ, we obtain
We now have operators for position
and momentum.
d 〈x〉dx
= 〈v〉
〈p〉 = −i~∫
xΨ∗∂Ψ
∂xdx
=
∫Ψ∗(−i~ ∂
∂x
)Ψdx
xop = x
pop = −i~ ∂∂x
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 13 / 13
Momentum operator
By assuming that this result is re-alted to the expectation value ofthe velocity, we can see how tocompute the expectation value ofthe momentum, p = mv .
If we cast this into the form for anoperator acting on Ψ, we obtain
We now have operators for positionand momentum.
d 〈x〉dx
= 〈v〉
〈p〉 = −i~∫
xΨ∗∂Ψ
∂xdx
=
∫Ψ∗(−i~ ∂
∂x
)Ψdx
xop = x
pop = −i~ ∂∂x
C. Segre (IIT) PHYS 405 - Fall 2012 August 22, 2012 13 / 13