Today’s Outline - August 22, 2012csrri.iit.edu/~segre/phys405/12F/lecture_02.pdfToday’s Outline - August 22, 2012 1D Schr odinger equation The wave function Probability review

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






• Normalization










• Normalization










• Normalization










• Normalization










• Normalization










• Normalization










• Normalization










• Normalization





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



i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ







i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ







i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ







i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ







i~∂Ψ

∂t= − ~2

2m

∂2Ψ

∂x2+ VΨ

i~∂Ψ

∂t= Total Energy

− ~2

2m

∂2Ψ






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







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b







∫ b

a|Ψ(x , t)|2 dx

|Ψ|2

xa b


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


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





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)






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)






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)






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)






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)






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)






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)






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)






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


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



〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩σ =

√〈j2〉 − 〈j〉2


Expectation values




〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩σ =

√〈j2〉 − 〈j〉2


Expectation values




〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩

σ =

√〈j2〉 − 〈j〉2


Expectation values




〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩

σ =

√〈j2〉 − 〈j〉2


Expectation values




〈f (j)〉 =∞∑j=0

f (j)P(j)

σ2 ≡⟨(∆j)2

⟩σ =

√〈j2〉 − 〈j〉2


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



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2



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〉)



Since σ2 ≥ 0,⟨j2⟩≥ 〈j〉2


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



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




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




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




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




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




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




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




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




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


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



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.


1 =

∫ +∞

−∞|Ψ(x , t)|2 dx


d

dt

∫ +∞

−∞|Ψ(x , t)|2 dx =

∫ +∞

−∞

∂

∂t|Ψ(x , t)|2 dx





1 =

∫ +∞

−∞|Ψ(x , t)|2 dx


d

dt

∫ +∞

−∞|Ψ(x , t)|2 dx =

∫ +∞

−∞

∂

∂t|Ψ(x , t)|2 dx





1 =

∫ +∞

−∞|Ψ(x , t)|2 dx


d

dt

∫ +∞

−∞|Ψ(x , t)|2 dx =

∫ +∞

−∞

∂

∂t|Ψ(x , t)|2 dx





1 =

∫ +∞

−∞|Ψ(x , t)|2 dx


d

dt

∫ +∞

−∞|Ψ(x , t)|2 dx =

∫ +∞

−∞

∂

∂t|Ψ(x , t)|2 dx





1 =

∫ +∞

−∞|Ψ(x , t)|2 dx


d

dt

∫ +∞

−∞|Ψ(x , t)|2 dx

=

∫ +∞

−∞

∂

∂t|Ψ(x , t)|2 dx





1 =

∫ +∞

−∞|Ψ(x , t)|2 dx


d

dt

∫ +∞

−∞|Ψ(x , t)|2 dx =

∫ +∞

−∞

∂

∂t|Ψ(x , t)|2 dx


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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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

)]




∂

∂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

)


∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ


∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+

i

~VΨ∗


∂Ψ∗


∂

∂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


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±∞




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





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





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





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



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.






〈x〉 =

∫x |Ψ(x , t)|2 dx

=

∫Ψ∗xΨdx








〈x〉 =

∫x |Ψ(x , t)|2 dx

=

∫Ψ∗xΨdx








〈x〉 =

∫x |Ψ(x , t)|2 dx

=

∫Ψ∗xΨdx








〈x〉 =

∫x |Ψ(x , t)|2 dx

=

∫Ψ∗xΨdx








〈x〉 =

∫x |Ψ(x , t)|2 dx

=

∫Ψ∗xΨdx




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








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








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








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








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








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








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








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








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








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


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


Momentum operator




d 〈x〉dx

= 〈v〉

〈p〉 = −i~∫

xΨ∗∂Ψ

∂xdx

=

∫Ψ∗(−i~ ∂

∂x

)Ψdx

xop = x

pop = −i~ ∂∂x


Momentum operator




d 〈x〉dx

= 〈v〉

〈p〉 = −i~∫

xΨ∗∂Ψ

∂xdx

=

∫Ψ∗(−i~ ∂

∂x

)Ψdx

xop = x

pop = −i~ ∂∂x


Momentum operator




d 〈x〉dx

= 〈v〉

〈p〉 = −i~∫

xΨ∗∂Ψ

∂xdx

=

∫Ψ∗(−i~ ∂

∂x

)Ψdx

xop = x

pop = −i~ ∂∂x


Momentum operator



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


Momentum operator



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


Momentum operator




d 〈x〉dx

= 〈v〉

〈p〉 = −i~∫

xΨ∗∂Ψ

∂xdx

=

∫Ψ∗(−i~ ∂

∂x

)Ψdx

xop = x

pop = −i~ ∂∂x


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