Quantum Mechanics 1 Lecture 2: Wave Mechanics

Quantum Mechanics 1

Lecture 2: Wave Mechanics

Massimiliano Grazzini

University of Zurich

Outline

The double slit experiment

The structure of the quantum amplitude

Classical limit

Sum over the paths

The uncertainty principle

The free-particle kernel

The wave function and the Schrödinger equation

Free particles

The double slit experiment

screen

detector

source

1 2

A source radiates electrons all of the same energy that come out in all directions The screen has two holes, 1 and 2 through which the electrons can pass

D x

Experimentally one finds that the probability as a function of is the complicated curve on the right with maxima and minima and there are regions of in which the electrons hardly arrive

x x

The detector counts how many electrons arrive at _x

A relatively recent experiment has verified this phenomenon with the largest and most classical objects, which are currently the fullerene molecules

(Nairz, Arndt, and Zeilinger (2003)

The fullerene molecule , consisting of 60 carbon atoms arranged in a truncated

icosahedral shape, is the smallest known natural soccer ball.

C

Given the large dimension one would expect that these molecules behave like particles under ordinary circumstances.

The multi-slit experiment by Nairz et al.

has instead shown their wave nature

The double slit experiment

The double slit experiment

With this experimental setup, it would seem reasonable to assume that:

Each electron must go through the slit 1 or through the slit 2

The chance to arrive to should be the sum of two parts: the chance to arrive through the slit 1 and the chance to arrive through the slit 2 ^x _P

^P

screen

detector

source

1 2

D

x

The double slit experiment

The sum of the two clearly does not give the previous curve so we conclude that

screen

detector

source

1 2

D x

We may easily check this by closing one of the two slits and doing again the experiment: the results are shown below

P ≠ P

+ P

The curve is familiar: it is exactly the

interference pattern expected if waves starting from the source pass through the two holes

P(x)

The simplest way to represent wave amplitudes is through complex numbers

We are led to introduce two complex amplitudes and such that _{P = | A |}

with _{A = A}

_{+ A}

and _P

_{= | A}

_|

_P

^A ₌

_{| A}

_|

^A

Logical difficulty: it is not true that _{P = P}

_{+ P}

It is not true that when both holes are open the particle goes from one hole or the other !

If this were the case we could classify all arrivals at in two disjoint classes:

those coming from the hole 1 and those from the hole 2 ^x

The effect of observation

We have concluded on logical grounds that since we cannot assume that the electron passes either through hole 1 or hole 2 ^{P ≠ P}

^{+ P}

But we can devise an experiment to check this directly

source

A source of light behind the screen allows us to watch and see through which hole the electron passes

Electron scatter light: if light is scattered behind hole 1, we may conclude that an electron passed through hole 1; if it is scattered close to hole 2, the electron

passed through hole 2

1

2

For each electron arriving at we can now experimentally say whether it came through hole 1 or 2

x

The result is that the probability distributions and are indeed given by the curves on the right, and that

P

P

P = P

+ P

When we watch the electrons to see from which hole they passed we find

When we do not watch we get a different result

The answer is that we have used light: the light in collision with the electron may alter its motion and its chance to arrive to a given point on the detector

We could use very low intensity, but weak light does not mean weak disturbance When we have low intensity, it means that less photons are used, but this way we may miss the electron

What happened ?

But when we do see an electron, it means that a photon was scattered, and a momentum of order _{h / λ} was transferred to the electron

How about taking very large wavelengths ? This way the photon momentum

would be smaller ^λ

But there is a limit to this: if too long a wavelength is used, we will not be able to say whether the scattering was behind hole 1 or hole 2

We thus see that any experiment designed to determine through which hole the electron passes must produce enough disturbance to alter the ensuing distribution It was first pointed out by Heisenberg in his uncertainty principle that the

consistency of the new mechanics implies a limitation to the precision to which an experiment could be performed

In our case the principle states that any attempt to design an apparatus to determine

through which hole the electron passed, and delicate enough so as not to deflect the

electron enough to destroy the interference pattern is bound to fail

The uncertainty principle

source

When an electron is deflected the vertical component of its

momentum is changed

Suppose that the screen is not fixed but it is free to move up and down and that we can measure its displacement

p p

l

a

δp

δp

For small deflection angles the momentum keeps the same

magnitude and we have

| p

− p

| / | p | = | δp | / p ∼ a / l p

p

The uncertainty principle

The interference pattern of the original experiment is the sign of a wave-like behavior of the electrons

The two beams of electrons will interfere constructively and give a maximum in the diffraction pattern at the center of the screen

As we move down from the center the next maximum will occur at a distance d such that the beam from the hole 1 has travelled exactly one wavelength more than the beam from 2

The following relation holds

First (central) maximum Second

maximum

λ

a ∼ d

l

The uncertainty principle

If we want to say which hole the electron passed through, the momentum imbalance must be measured with precision better than , that is _δp

If we want to see the interference pattern we must control the position of the screen with _{Δx < d}

Combining the two requirements and using ^λ and we obtain

a ∼ d

l δp / p ∼ a / l

We thus conclude that if we want to say which hole the electron passed

through and at the same time we want to observe the interference pattern we must have

Δp < δp = λ

d p < λ

Δx p = h Δx

Use De Broglie relation

which is (as order of magnitude) what the uncertainty principle forbids !

ΔxΔp < h

Δp < δp

screen

detector

source

D

Suppose now that we add more holes….

The structure of the amplitude

screen

detector

source

D

Suppose now that we add more holes….

The structure of the amplitude

screen

detector

source

D

Suppose now that we add more holes….

The structure of the amplitude

screen

detector

source

D

The amplitude will be the sum of the

amplitudes over the interfering alternatives Suppose now that we add more holes….

Eventually, if we remove the screen we obtain

The structure of the amplitude

A = ∑

i

A

A = ∑ A

detector

source

D

The structure of the amplitude

But what is the form of _A

?

Sum over all the possible paths connecting the source and a point on the screen

It can be obtained by considering the analogy with geometrical optics

A = ∑

paths

A

The structure of the amplitude

In geometrical optics the path of light rays is determined by the Fermat principle: the “optical path” of the ray, that is, the phase

difference in the initial and final points of the path must be minimum In classical mechanics the particle trajectory can be determined

according to the principle of minimal action: the action functional computed along the classical trajectory must be minimum

Starting from this analogy we can assume that the phase of the

amplitude on a given path must be proportional to the classical action computed along that path. The proportionality constant should be

identified with the Planck constant _ℏ

Geometrical optics is obtained from wave optics by taking the limit _{λ → 0}

A

∼ exp{iS[path]/ ℏ}

detector

source

D

Classical limit

Sum over all the possible paths connecting the source and a point on the screen

In the classical limit the action is large in absolute value and oscillates very rapidly In this limit ( ) only the classical trajectory contributes: the

contributions from all the other trajectories cancel out ^{ℏ → 0} A = ∑

paths

exp{iS[path]/ ℏ}

detector

source

D

Classical limit

Sum over all the possible paths connecting the source and a point on the screen

In this limit ( ) only the classical trajectory contributes: the contributions from all the other trajectories cancel out ^{ℏ → 0}

In the classical limit the action is large in absolute value and oscillates very rapidly A = ∑

paths

exp{iS[path]/ ℏ}

→ exp{iS

/ ℏ}

Classical limit

We can thus summarise this as

Quantum Mechanics Wave optics

Classical mechanics ~ Geometrical optics

ℏ → 0 λ → 0

~

Sum over the paths

We consider a particle moving in one dimension from the point at time to

the point at time _x

_t

^x

^t

The amplitude to go from to is obtained as _{a b K(b, a) = ∑}

x(t)

exp{iS(x(t))/ ℏ}

Although the qualitative idea of the sum over the paths is clear, we need a precise

mathematical definition

The Riemann integral of a function is

defined as ^{f (x)}

Δx→0

lim ∑

i

f (x

)Δx

We can follow an analogous procedure to define the sum over the paths

x f (x)

x

x

Δx

t_b

t_a

x_a x_b

x(t)

We divide the time interval in steps of width

t

− t

ϵ

t

t

x

x

x

x

t

t

ϵ

K(b, a) = lim

ϵ→0

1 A ∫

+∞

−∞

dx

A ∫

+∞

−∞

dx

A . . . . ∫

+∞

−∞

dx

A exp{iS(x(t))/ ℏ}

For the moment we limit ourselves to consider a free particle

L (x, · x, t) = 1

2 m x ·

∼ m

2 ( x

− x

ϵ )

2

For each of the values at the time we integrate over _t

^x

_x

S[x(t)] = ∫

t_b

t_a

L(x, · x, t)dt

We introduce a constant whose precise value will be fixed later

A

≡ ∫ 𝒟x(t)exp{iS(x(t))/ ℏ}

S[x(t)] = m ϵ 2

N−1

∑

( x

− x

ϵ )

2

K(b, a) = lim

ϵ→0

1 A ∫

+∞

−∞

dx

A ∫

+∞

−∞

dx

A . . . . ∫

+∞

−∞

dx

A exp{iS(x(t))/ ℏ}

We can then write the amplitude _{K(b, a)}

= lim

ϵ→0

1 A

∫

+∞

−∞

dx

. . . . dx

exp { im

2ℏ ϵ

∑

i=0

( x

− x

ϵ )

2

}

= lim

ϵ→0

1 A^N ∫

+∞

−∞ dx₂. . . . dx_N−1exp{ im 2ℏϵ

N−1

∑i=2 (x_i+1 − x_i)²}∫

+∞

−∞ dx₁exp{ im

2ℏϵ [(x₂ − x₁)² + (x₁ − x_a)²]} 2(x₁−(x₂+x_a)/2)²+(x₂−x_a)²/2

The integral over is now a Gaussian integral and it gives _x

∫

+∞

−∞

dx

exp { im

2ℏϵ [ (x

− x

)

+ (x

− x

)

] } = π iℏϵ

m exp { im

4ℏϵ (x

− x

)

}

(recall _{∫ dx exp{iax}

_{} = iπ / a} )

K(b, a) = lim

ϵ→0

1 A

∫

+∞

−∞

dx

. . . . dx

exp { im 2ℏϵ

N−1

∑

( x

− x

)

} π iℏϵ

m exp { im

4ℏϵ (x

− x

)

}

We can then proceed in a similar way with the integral over _x

= lim

ϵ→0

1

A^N iπℏϵ m ∫

+∞

−∞ dx₃. . . . dx_N−1exp{ im 2ℏϵ

N−1

∑i=3 (x_i+1 − x_i)²}∫

+∞

−∞ dx₂ exp{ im

2ℏϵ [(x₃ − x₂)² + (x₂ − x_a)²/2]} 3/2(x₂−(2x₃+x_a)/3)²+(x₃−x_a)²/3

= lim

ϵ→0

1

A

2iπ ℏϵ 2m ∫

+∞

−∞

dx

. . . . dx

exp { im 2ℏϵ

N−1

∑

( x

− x

)

} 4iπ ℏϵ

3m exp { im

6ℏϵ (x

− x

)

}

Factor

Factor

By repeating the procedure recursively we obtain

K(b, a) = lim

ϵ→0

1

A

( 2π iℏϵ

m )

(N−1)/2

1

2 ⋅ 2

3 ⋅ 3

4 . . . . N − 1

N exp { im

2ℏϵN (x

− x

)

}

= lim

ϵ→0

1

A ( 2π iℏϵ A

m )

(N−1)/2

ϵ

ϵN exp { im

2ℏ(t − t ) (x

− x

)

}

If we now set _{A = 2} ^{π iℏϵ} we find

m

K(b, a) = m

2π iℏ(t

− t

) exp { im

2ℏ(t

− t

) (x

− x

)

}

By setting _t

_{= 0, x}

_{= 0, t}

_{= t, x}

_{= x} we can simply write

K(b, a) → K(x, t) = m

2π iℏt exp { im

2ℏt x

}

It is interesting to observe that the factor ¹

2 m (x

− x

)

t

− t

corresponds to the classical action along the path from to so we can

write ^S

^{a b}

K(b, a) = m

2π iℏ(t

− t

) exp { i

ℏ S

} K(b, a) = lim

ϵ→0

1

A ( 2π iℏϵ A

m )

(N−1)/2

ϵ

ϵN exp { im

2ℏ(t

− t

) (x

− x

)

}

t_b − t_a

The free-particle kernel

K(x, t) = m

2π iℏt exp { im

2ℏt x

}

We now want to discuss some

implications of the free-particle kernel

At fixed time the real part of the amplitude varies with as _{t x}

Re(K)

x

We see that as we get farther from the origin the oscillations become more and more rapid

What is the wavelength at large ? We can find it by solving _x

2π = m(x + λ)²

2ℏt − mx²

2ℏt = mxλ

ℏt + mλ² 2ℏt

The free-particle kernel

K(x, t) = m

2π iℏt exp { im

2ℏt x

}

We now want to discuss some

implications of the free-particle kernel

Re(K)

x

We see that as we get farther from the origin the oscillations become more and more rapid

2π = m(x + λ)²

2ℏt − mx²

2ℏt = mxλ

ℏt + mλ² 2ℏt

If we assume _{λ ≪ x} we can neglect the second term and we find

λ = 2 π ℏ m(x / t)

At fixed time the real part of the amplitude varies with as _{t x}

A classical free particle which moves from the origin in the time interval has

velocity and momentum _{x / t mx / t} ^t

We can interpret the previous result by saying that when the motion can be described by assigning a classical momentum

to the particle the amplitude varies with a wavelength

p = mx / t

λ = h / p

We can repeat this discussion for the time dependence Suppose we keep the distance fixed and vary _t

Re(K)

t

The oscillation period is defined as the time required to increase the phase by ^T _2π

2π = mx

2ℏt − mx

2ℏ(t + T )

= mx²

2ℏt (1 − 1

1 + T/t) = mx²

2ℏt² T 1 + T/t

If we assume _{T ≪ t} we find _{ω = 2} ^π

T ∼ m

2ℏ ( x t )

2

But is nothing but the classical energy of the particle, so we conclude that

E = 1 2 m ( x t )

2

E = ℏω = hν

The above conclusions can also be obtained by recalling that in classical mechanics the momentum and energy at the time and position are _t

_x

p = ∂S

∂x

E = − ∂S

∂t

and that in the classical limit the kernel is

K(b, a) ∼ exp{ i

ℏS_cl(b,a)}

Events occurring in succession

The amplitude to go from to is obtained as _{a b K(b, a) = ∑}

x(t)

exp{iS(x(t))/ ℏ}

Suppose that is some time between and : The action along any path between and can be written as _a ^t

_b ^t

^t

S[b, a] = S[b, c] + S[c, a]

This follows from the definition of the action as an integral over time (and on the fact that it depends only on position and velocity)

It is possible to split any path in two

parts: the first part will have end points and and the second part

would have end points and

x

x

= x(t

)

x

x

We thus write _{K(b, a) = ∫} 𝒟x(t)exp{iS(x(t))/ ℏ} = ∫ 𝒟x(t)exp{i / ℏS[b, c] + i / ℏS[c, a]}

x

t

c

x a

t b

Events occurring in succession

x

t

c

x a

t b

We then have to integrate over all paths from to , over all paths from to and finally to integrate the result over all possible values of ^{a c} _x

^{c b} In doing the first step the factor _{S[b, c]} is constant, so we can write

K(b, a) = ∫

dx

∫ 𝒟x(t)exp{i / ℏS[b, c]}K(c, a)

In the next step we integrate over the paths at fixed and we find _x

K(b, a) = ∫

dx

K(b, c)K(c, a)

The kernel to go from to is the sum over all the possible values of of amplitudes for the particle to go from to and then to

a b

x

a c b

Amplitudes for events occurring in

succession in time multiply

Wave function

The kernel _{K(b, a)} describes the propagation of a particle from to _{a b}

We want to generalise this result by considering a particle for which we do not know the initial position

By using the result on amplitudes occurring in succession we can write

Such amplitude has the same probability characteristics as those we already encountered: its square gives the probability of finding the particle at the point and at the time ^{| ψ (x, t) |}

2

x t

We will call this amplitude wave function

The wave function _{ψ (x, t)} describes the state of the system at the time _t

ψ (x

, t

) = ∫

+∞

−∞

dx

K(x

, t

; x

, t

)ψ (x

, t

)

We call the total amplitude to arrive at _{(x, t)} from the past _{ψ (x, t)}

Towards new principles

We now want to summarise what we have learned so far

Quantum systems have an intrinsically casual behavior. We cannot say what

happens until we observe it. Only the probability distribution of a measurement is a physical property of the system

Probabilities should be computed by squaring transition amplitudes The amplitudes for different alternatives sum up like waves

(superposition principle) and interference phenomena occur The amplitude for a particle to go from to is obtained by summing the amplitude on each possible path (path integral)

The contribution of the amplitude from a single path is obtained by

where is the classical action evaluated on that path

In the classical limit

only the classical path contributes

a

b

The state of a system is described by the wave function

The Schrödinger Equation

In Quantum Mechanics the wave function determines in a complete way the

state of a physical system ^ψ

This implies that such function, at a given time, defines also the behaviour of the system in the future

Mathematically this means that the derivative of the wave function must be

determined from at the same time _ψ ^{∂ψ / ∂t}

According to the superposition principle, such dependence must be linear We thus write in general

i ℏ ∂ ψ

∂t = H ψ

Where is a linear operator and the factor has been added for convenience _{H iℏ}

Now, given the function its modulus squared gives the probability to find the particle at the time in the position ^{ψ (x, t) | ψ (x, t) |}

2

t x

If we integrate over the total probability must be independent on time _x

We then have 0 = d

dt ∫ | ψ |

dx = ∫ ∂ψ *

∂t ψ dx + ∫ ψ * ∂ ψ

∂t dx

Multiplying by and using the definition of we find _{iℏ H}

0 = ∫ ( ψ *(H ψ) − (Hψ)*ψ ) dx

Using the definition of adjoint _{∫ (Hψ)*ψ = ∫ ψ *(H}

_ψ) (remember MMP1 !) we find

0 = ∫ ψ * ( H

− H ) ψ dx

Given the arbitrarity of we conclude that must be self-adjoint _{ψ H}

so its eigenvalues are real. Do they correspond to a physical quantity ?

H = H

To answer this question we can check what happens in the classical limit In this limit we have

ψ = a exp{iS / ℏ}

Where a can be treated as a constant

If we act on this form of the wave function with _{iℏ ∂} we obtain

∂t iℏ ∂ ψ

∂t = iℏa exp{iS / ℏ} i

ℏ ∂S

∂t = − ∂ S

∂t ψ

Comparing with _{i ℏ ∂} ^ψ

∂t = H ψ

we conclude that in the classical limit the eigenvalues of the operator correspond to _{H −∂S / ∂t}

But this quantity is nothing but the energy of the system

The operator is the Hamilton operator for the system _H

We now want to see that the path integral formulation that we have outlined naturally leads to the Schrödinger equation

We start from the form of the wave function in terms of the kernel _{K(b, a)}

We focus on a single particle in one dimension subject to a potential with

Lagrangian ^{V(x, t)}

ℒ(x, · x, t) = 1

2 m x ·

− V(x, t)

Choosing _t

_{= t + ϵ} , _x

_{= x} and _x

_{= y t} ,

_{= t} we can write

ψ (x, t + ϵ) = ∫ dy

A exp { i

ℏ ϵℒ ( x + y

2 , x − y

ϵ , t + ϵ

2 ) } ψ ( y, t)

= ∫ dy

A exp { im

2ℏϵ (x − y)

} exp { − iϵ

ℏ V ( x + y

2 , t + ϵ

2 ) } ψ ( y, t)

ψ (x

, t

) = ∫

+∞

−∞

dx

K(x

, t

; x

, t

)ψ (x

, t

)

If is small the difference must also be small, or the velocity should be very

large ^{ϵ x − y}

If is small the factor oscillates very rapidly and damps the integration to the region

ϵ exp { im

2ℏϵ (x − y)

} y

x − y ∼ ϵℏ / m

It makes sense to expand the second factor in powers of and _{ϵ (x − y)}

We start from

ψ (x, t) + ϵ∂

ψ (x, t) ≃ ∫ dy

A exp { im(x − y)

2ℏϵ } ( 1 − iϵ

ℏ V(x, t) )

× [ ψ (x, t) + (x − y)∂

ψ (x, t) + ( x − y)

2 ∂

ψ (x, t) ]

: comparing the left and right handed side we find

𝒪(ϵ

)

ψ (x, t) = ∫ dy

A exp { im(x − y)

2ℏϵ } ψ (x, t) = 2 π iℏϵ

m 1

A ψ (x, t) ψ (x, t + ϵ) = ∫ dy

A exp { im

2ℏϵ (x − y)

} exp { − iϵ

ℏ V ( x + y

2 , t + ϵ

2 ) } ψ (y, t)

and we find

So we can set consistently with what we have found for the free-particle kernel

A = 2 π iℏϵ m

Multiplying by we find _iℏ

iℏ∂

ψ (x, t) = − ℏ

2m ∂

ψ + V(x, t)ψ ψ (x, t) + ϵ∂

ψ (x, t) ≃ ∫ dy

A exp { im(x − y)

2ℏϵ } ( 1 − iϵ

ℏ V(x, t) )

× [ ψ (x, t) + (x − y)∂

ψ (x, t) + ( x − y)

2 ∂

ψ (x, t) ]

The term of on the right-hand side vanishes because the integral is odd, consistently with the absence of ^{𝒪(x − y) ∼ 𝒪( ϵ )} _{𝒪( ϵ )} on the left hand side

: let’s look again at our equation

𝒪( ϵ )

: at this order we have to keep also the term on the right

𝒪(ϵ) 𝒪 ( (x − y)

)

ϵ∂

ψ (x, t) = ∫ dy

2 A exp { im(x − y)

2ℏϵ } (x − y)

∂

ψ + ϵ

iℏ V(x, t)ψ (x, t)

iℏϵ/(2m)

*

* Use _∫r²_dre^−ar2 ₌ ^π

2 a^−3/2

This equation can be written in the form i ℏ ∂ ψ

∂t = H ψ If we identify

H = − ℏ 2m

∂

+ V(x, t)

Comments:

Hamilton operator

The classical Hamiltonian of the system is

H = p

2m + V(x, t)

The Hamilton operator is obtained from the classical Hamiltonian with the replacement

p → − iℏ∂

Schrödinger equation

The Schrödinger equation is a first-order differential equation in the time variable : the form of the wave function is sufficient as initial condition ^{t ψ (x, t = t}

⁾

The _K(x

_{, t}

_{, x}

_{, t}

_{) ≡ K(2,1)} is the Green function of the operator

iℏ∂

+ ℏ

2m ∂

− V(x, t)

Indeed we have for _t

_{> t}

_iℏ∂

_{K(2,1) = − ℏ}

2m ∂

K(2,1) + V(2)K(2,1)

But from we have

as

ψ (x

, t

) = ∫

+∞

−∞

dx

K(x

, t

; x

, t

)ψ (x

, t

) K(2,1) → δ(x

− x

) t

→ t

If we define _{K(2,1) = 0} for _t

_{< t}

we have

[ iℏ∂

+ ℏ

2m ∂

− V(x

, t) ] K(2,1) = iℏδ(x

− x

)δ(t

− t

)

The superposition principle

We have already pointed out that the superposition principle requires that the differential laws of the new mechanics must be linear

The Schrödinger equation is a linear first-order differential equation in the time variable _t

This means that if and are solution of the Schrödinger equation _ψ

_ψ

ψ = a

ψ

+ a

ψ

Is also a solution

a

, a

∈ ℂ

Moreover, if in the state characterised by the function a certain observable takes the value and in the state characterised by the function it takes the value

ψ

O O

ψ

O

In the state characterised by the wave function the

observable can take either the value or the value _O

^{ψ = a} _O

^ψ

^{+ a}

^ψ

In the free case _{V(x, t) = 0} the Schrödinger equation reads

i ℏ ∂

ψ (x, t) = − ℏ 2m

∂

ψ (x, t)

i ℏ ∂

χ

(t)

χ

(t) = − ℏ 2m

∂

ϕ

(x)

ϕ

(x) ≡ E

The free-particle

We can look for a solution in factorised form _ψ

_{(x, t) = χ}

_(t)ϕ

_(x) which gives

with _E

_{= ℏω}

₌ ^ℏ

^k

and real

2m k

Left and right-handed side are now

function of different variables: they must be equal to a constant: let’s call it

and

χ

(t) = exp{− i / ℏE

t} ≡ exp{− iω

t} ϕ

(x) = exp{i 2mE

ℏ

x} ≡ exp{ikx}

By using and we obtain and , that is the right dispersion relation for a classical massive particle ^{k = 2π / λ λ = h / p p = ℏk E = p}

2

/2m

Recall MMP1: The functions _ϕ

(x) = exp{ikx} form a basis of L

[ℝ]

The functions _ψ

_{(x, t) = χ}

_(t)ϕ

(x) = exp{i(kx − ω

t)} = exp { i

ℏ ( px − E

t )}

Describe particles with momentum and energy but the wave function is a pure phase ! ^{p = ℏk E = p}

2

/2m

The probability is independent on

This means that all values of are equally likely and that the position is completely undetermined

| ψ

(x, t) |

= 1 x

x

This is the uncertainty principle !

Knowing _{ψ (x, t)} at _{t = 0} we can find a general solution of the Schrödinger equation

ψ (x, t) = ∫ dk

2π a(k) ψ

(x, t) = ∫ dk

2π a(k) exp{i(kx − ω

t)}

ψ (x,0) = ∫ dk

2π a(k) exp{ikx} a(k) = ∫ dy ψ ( y,0) exp{− iky} = ∫ dy ψ ( y,0) ψ

* ( y,0)

ψ (x, t) = ∫ dy ∫ dk

2π ψ

* (y,0) ψ

(x, t) ψ ( y,0)

Thus we can write

Comparing with the definition of the wave-function in terms of the kernel

ψ (x, t) = ∫ dy K(x, t, y,0) ψ (y,0)

we find

K(x, t, y,0) = ∫ dk

2π ψ

* ( y,0) ψ

(x, t)

= ∫ dk

2π e

e

= ∫ dk

2π exp { − iℏt

2m ( k − (x − y)m ℏt )

2

+ im(x − y)

2ℏt }

= m

2πiℏt exp{im(x − y)²

2ℏt }

Consistently with the direct

computation through the path integral