Quantum Mechanics 1 Lecture 4: One dimensional QM

Quantum Mechanics 1

Lecture 4: One dimensional QM

Massimiliano Grazzini

University of Zurich

Outline

The Potential Step

- Tunneling effect Potential Well

The Potential Barrier

General discussion

- structure of the spectrum

- the infinite potential well

One-dimensional problems

We now discuss the determination of the spectrum of the Hamiltonian and the solution of the Schrödinger equation for some simple forms of the potential _V(x)

In both nuclear and solid-state physics there are often situations in which we can identify regions where the potential is roughly constant but the transition from one region to another happens in a rather sharp way

This situations can be modelled by assuming that the potential has a step

The other important case is the one in which There is a potential well

V(x)

V(x)

− ℏ 2m

ψ ′ ′ (x) + V(x)ψ (x) = E ψ (x)

Potential Steps

V(x)

a x

We consider a one-dimensional potential with a discontinuity at _{x = a}

The time-independent Schrödinger equation can be written as

ψ ′ ′ (x) = − 2m

ℏ

(E − V(x))ψ (x)

We now ask ourselves how the wave function should behave at _{x = a} Suppose that _{ψ (x)} or _{ψ ′ (x)} are discontinuous at _a

If _{ψ (x) ∼ θ(x − a)} then _{ψ ′ ′ (x) ∼ δ′ (x − a)} (recall MMP2 !)

But according to the Schrödinger equation should have at most a finite

step at _{x = a} ^{ψ ′ ′ (x)}

Analogously if _{ψ ′ (x) ∼ θ(x − a)} then _{ψ ′ ′ (x) ∼ δ(x − a)}

We must require continuity for and

at even if the potential has a step ^{ψ (x)}

ψ ′ (x) x = a

As an example of a problem of the type we have just discussed, we consider the potential step

V(x) V

x V(x) = V

θ(x) V

≥ 0

The Schrödinger equation in the two regions reads:

ψ ′ ′ (x) = − 2mE ℏ

ψ

ψ ′ ′ (x) = − 2m(E − V

) ℏ

ψ

x < 0

x > 0

and we have to exploit the continuity conditions on _{ψ (x)} and _{ψ ′ (x)}

We distinguish two cases: and since they correspond to different physical situations ^{E > V}

^{E < V}

E I II

Since the corresponding wave functions are not normalisable we cannot

directly interpret the solution as describing the behaviour of a single particle Note:

We are looking for stationary states with definite energy _E

The connection with realistic particle states can be done as in Lecture 3

through the use of wave packets

As an example of a problem of the type we have just discussed, we consider the potential step

V(x) V

x V(x) = V

θ(x) V

≥ 0

The Schrödinger equation in the two regions reads:

ψ ′ ′ (x) = − 2mE ℏ

ψ

ψ ′ ′ (x) = − 2m(E − V

) ℏ

ψ

x < 0

x > 0

and we have to exploit the continuity conditions on _{ψ (x)} and _{ψ ′ (x)}

We distinguish two cases: and since they correspond to different physical situations ^{E > V}

^{E < V}

- The case _{E > V}

In this case the equation in both regions reads _{ψ ′ ′ (x) + k}

_{ψ = 0}

k = 2 mE

ℏ

= k

k = 2m(E − V

)

ℏ

= k

x < 0 x > 0

E I II

The energy eigenfunctions in the regions I ( _{x < 0} ) and II ( _{x > 0} ) are

ψ

(x) = Ae

+ Be

ψ

(x) = Ce

+ De

With the conditions _ψ

_{(0) = ψ}

₍₀₎ and _ψ

_{′ (0) = ψ}

_{′ (0)} which imply the constraints and

A + B = C + D ik

A − ik

B = ik

C − ik

D

The solutions are plane waves in both regions with momenta and _k

_k

These are two constraints over the four complex coefficients _{A, B, C, D} The overall normalisation

and phase are irrelevant We end up with three independent parameters We can determine one solution by fixing one of the four coefficients If we set _{D = 0} we can express and in terms of _{B C A}

B = k

− k

k

+ k

A C = 2k

k

+ k

A

A + B = C

ik

A − ik

B = ik

C

These conditions completely determine the wave function up to the overall normalisation _A

j = − iℏ

2m ( ψ *ψ ′ − ψ ψ ′ * )

To interpret the solution we can use the probability current we have introduced in Lecture 3

In the region I ( _{x < 0} ) we have _ψ

_{(x) = Ae}

_{+ Be}

and thus

j

(x) = − iℏ

2m ( ( A*e

+ B*e

) ik

( Ae

− Be

) − cc )

= − iℏ

2m ( ik

( | A |

− | B |

) + ik

( − A*Be

+ B* Ae

) − cc )

This term is purely real:

cancels out with cc

= ℏk

m ( | A |

− | B |

)

This term is purely imaginary:

picks up a factor of 2 from cc

j = − iℏ

2m ( ψ * ∇ψ − ψ ∇ ψ * )

which in one dimension reads

j

(x) = ℏk

m ( | A |

− | B |

) j

(x) = ℏk

m | C |

is the incident probability current on the step

j

= ℏk

m | A |

Looking at these results we can give the following interpretation

is the reflected probability current

j

= ℏk

m | B |

j

= ℏk

m | C |

The reflection coefficient is _{R ≡} ^j

j

= | B |

| A |

= (k

− k

)

(k

+ k

)

The transmission coefficient is _{T ≡} ^j

j

= k

k

| C |

| A |

= 4k

k

(k

+ k

)

In the region II ( _{x > 0} ) we have _ψ

_{(x) = Ce}

and thus

j

(x) = − iℏ

2m ( C*e

(ik

)Ce

− cc ) = ℏk

m | C |

is the transmitted probability current from region I to region II

Note: they are all

constant !

V(x) V

x E

Comments:

Classically, a particle arriving on the

potential barrier with energy and velocity would be transmitted with unit probability, just with smaller velocity

E > V

v = 2E / m

v = 2(E − V

)/ m

In quantum mechanics the particle is reflected with probability : this

phenomenon is a wave phenomenon analogous to the reflection of light at the boundary surface of media with different refraction indices

R

In the limit _{E ≫ V}

we have _k

_{∼ k}

and _{R → 0} with _{T → 1} This means that in this limit there is no reflection

Conservation of particle number: from the previous results we can easily check that _j

_{= j}

or _j

_{= j}

_{+ j}

Regressive solution:

Up to now we have assumed (particle coming from the left): an

independent solution can be found setting ^{D = 0} _{A = 0} (particle coming from the right)

E

- The case _{E < V}

In this case the Schrödinger equation in region I is unchanged _{ψ ′ ′ (x) + k}

_{ψ = 0}

k

= 2 mE

ℏ

x < 0

But in region II it becomes

ψ

(x) = Ce

+ De

ψ ′ ′ (x) = χ

ψ

whose solution are exponential functions

χ

= 2m(V

− E ) ℏ

In order to obtain a physically acceptable solution we have to set _{D = 0}

I II

ψ

(x) = Ce

ψ

(x) = Ae

+ Be

ψ

(x) = Ae

+ Be

ψ

(x) = Ce

E > V

E < V

We can repeat the same steps as for the case

setting _k

_{= iχ}

^{E > V}

R = k

− iχ

k

+ iχ

2

= 1

We find

V(x) V

x

Comments:

Classically, a particle arriving on the

potential barrier with energy cannot penetrate the barrier and is reflected ^{E < V}

In quantum mechanics the probability to pass through the barrier is non vanishing, although exponentially suppressed (tunneling effect)

The penetration depth is of order _χ

No particle net flux takes place because one can check that _j

_{(x) = 0} The limit of infinitely high potential step ( _V

_{→ ∞} )

In this limit we have and there is no penetration: the wave is fully reflected ( _ψ

_{(x) = 0} and ^χ

^{→ ∞} _{B = − A} )

We have _ψ

_{(x) = A(e}

_{− e}

₎ and thus _ψ

_{(0) = 0}

E

V(x) V

x

Potential barrier

V(x) V

x E

We now investigate what happens in the presence of the square potential barrier

− a a

V(x) = V

θ(a − | x | )

We consider only the case _{E < V}

As in the case of the potential step, a classical particle would be completely reflected by the barrier

However, we have seen in that case that in quantum mechanics the particle can have a non vanishing probability of crossing the step, and thus, we expect this to happen also for the potential barrier

The general solution of the Schrödinger equation is

ψ (x) =

Ae

+ Be

Ce

+ De

Fe

+ Ge

k = 2mE / ℏ χ = 2m(V

− E )/ ℏ

x < − a

− a < x < a

x > a

We have to impose the matching conditions at and on the wave function and its derivative ^{x = − a x = a}

- Condition at _{x = − a Ae}

_{+ Be}

_{= Ce}

_{+ De}

ik (Ae^−ika − Be^ika) = − χ(Ce^χa − De^−χa)

( e

e

e

− e

) ( A

B ) ⁼ ₍ ^e

χa

e

iχ

k

e

−

e

)

In matrix notation

( C D ) ( A

B ) ^{= 1} _{2 ( e} ^e

− ^e e

) (

e

e

iχ

k

e

−

e

) ( C D ) M(a)

We now observe that the symmetry of the problem requires ( F

G ) = M(− a) ( C

D ) and thus ₍ ^A _{B ) = M(a)M(− a)}

_{( G} ^F _{) = 𝒯 ( G} ^F ₎

The general solution is thus expressed in terms of the so called transfer matrix , which provides the coefficients in the region in terms of the coefficients in the region connecting the solutions where the particles propagates freely

x > a 𝒯 x < − a

- Condition at _{x = a}

For _M(a) and _{M(− a)}

we have

M(a) = 1 2

( 1 +

) e

( 1 −

) e

( 1 −

) e

( 1 +

) e

M(− a)

= 1 2

( 1 −

) e

( 1 +

) e

( 1 +

) e

( 1 −

) e

which leads to _{𝒯 =} ^{(cosh 2χa +}

iϵ2

sinh 2χa)e

sinh 2χa

−

sinh 2χa (cosh 2χa −

sinh 2χa)e

where and are defined as _{ϵ η}

ϵ = χ

k − k

χ η = χ

k + k χ

We can now specialise to the case of a particle incident from the left: we set _{G = 0} In this case we obtain from ₍ ^A

B ) = 𝒯 ( F G ) A = F(cosh 2χa + iϵ

2 sinh 2χa)e

B = F ( − iη

2 ) sinh 2χa

And we can compute the reflection and transmission coefficients as

R ≡ j

j

= | B |

| A |

= 1 − T T ≡ j

j

= | F |

| A |

and _F

A = (cosh 2χa + iϵ

2 sinh 2χa)

e

we thus find _{T =} ¹

1 + ( 1 +

) sinh

(2χa)

A classical particle with would be reflected by the barrier, but in the quantum theory we find a finite transmission probability across the barrier ^{E < V}

This purely quantum-mechanical effect is known as tunnelling effect

Important examples of this effect are alpha decay of nuclei and the cold emission

of electron from metals

We can study the expression for in the limit of high or wide barrier _T

T = 1

1 + ( 1 +

) sinh

(2χa)

Since _{sinh 2χa ∼ (1/2)e}

_{≫ 1} we find

T ∼

( 1 + ϵ

4 )

−1

4e

∼ exp { − 4 2m(V

− E )a / ℏ }

We see that in the limit of large or large width of the barrier the

probability to cross the barrier remains finite but is exponentially suppressed ^V

The Potential Well

We now want to study the bound states of the potential well

V(x) = − V

θ(a − | x | )

V(x)

− V

x

− a a

The potential well serves as a model for short-range forces such as those found in nuclear physics or at screened defects in solids

The tendency to bind a particle will grow with the depth and width of the well

For the solutions of the Schrödinger equation for are those of a free particle, so to find bound states we have to choose ^{E > 0} ₋ ^{| x} _V ^|

^> _≤ ^a _{E ≤ 0}

We have

with

ψ ′ ′ = χ

ψ χ = 2m(− E )/ ℏ _{| x | > a} with

ψ ′ ′ = − q

ψ q = 2m(E + V

)/ ℏ _{| x | < a}

A dimensionless parameter characterising this strength is _{ζ = 2mV}

_{a / ℏ}

To obtain physically acceptable solutions, we have to drop the exponentially increasing terms and set _{A = 0} and _{G = 0}

The solutions are

ψ (x) = Ae

+ Be

ψ (x) = C sin(qx) + D cos(qx)

ψ (x) = Fe

+ Ge

_{x > a}

x < − a

− a < x < a

The reflection symmetry of the potential strongly suggests to look for solution with definite parity ^{V(x) = V(−} _{ψ (−} ^x) _{x) = ψ (x)} (even) or _{ψ (− x) = − ψ (x)} (odd)

Even symmetry: we have _{C = 0} and _{F = B} The continuity conditions give:

Be

= D cos(qa) Bχe

= Dq sin(qa) _{tan(qa) =} ^χ

q

Since and depend on this is a transcendental equation that constrains the

possible values of the energy of the bound state, which give the discrete spectrum ^{χ q E}

By using the characteristic length defined before we can also rewrite the previous equation as ^{ζ = 2mV}

^{a / ℏ}

tan(qa) = χ

q = 1 qa

2m(−E )

ℏ a = 1 qa

2mV

( 1 − (E + V

)/ V

) )

ℏ a = ζ

− (qa)

qa

This means that must fulfil _{q 0 ≤ qa ≤ ζ} The transcendental equation can be solved graphically by plotting the two functions as a function of _{z = qa}

0 1 2 3 4 5 6

2 4 6 8 10

tan(z)

ζ² − z² z

Example: _{ζ = 5} Since the function vanishes for

the number of intersections is given by the nearest natural number greater than

ζ

− z

z z = ζ

ζ/ π

The corresponding energy eigenvalues are

E = − ℏ

χ

2m = − V

( 1 − (qa)

ζ

) π /2 3π /2 z = qa

There is always an even bound state for _{ζ > 0} and their number increases with _ζ

n

= [ζ/ π]

The continuity conditions give:

Fe

= C sin(qa)

− Fχe

= qC cos(qa)

− cot(qa) = χ

q ₌ ^ζ

2

− (qa)

qa

0 1 2 3 4 5 6

2 4 6 8 10

π /2 π 3π /2 2π

−cot(z)

Example: _{ζ = 1} vs _{ζ = 5} As in the case of even symmetry the transcendental equation can be solved graphically by plotting the two functions as a function of _{z = qa}

In this case there are solutions only if

ζ = 2mV

a

ℏ > π 2

The number of odd states is determined by the condition ⁿ

π

2 (2n

− 1) < ζ < π

2 (2n

+ 1)

ζ²− z² z

ψ (x) = Be

ψ (x) = C sin(qx) + D cos(qx)

ψ (x) = Fe

_{x > a}

x < − a

− a < x < a

Odd symmetry: we have _{D = 0} and _{F = − B}

We can summarise the results we have obtained by saying that

E = − V

( 1 − (qa)

ζ

)

The energy levels are given in terms of the allowed wave number by

the formula ^qa

There are even and odd states that alternate with increasing energy and number of nodes ⁿ

ⁿ

Example: _{ζ = 5}

The Infinite Potential Well

We now consider the limit of infinitely deep well _V

_{→ ∞}

V(x)

−V₀ = − ∞

In this limit we have

ζ = 2mV

a / ℏ → ∞ − a a

and the solutions of the equations x

tan(qa) = ζ

− (qa)

qa _{− cot(qa) =} ^ζ

2

− (qa)

qa

move to the asymptotes of _tan(qa) and _cot(qa)

ψ (x) ∼ θ(a − | x | )cos(qx) qa = π

2 (2n + 1)

The even solutions are

The odd solutions are

ψ (x) ∼ θ(a − | x | )sin(qx) qa = n π

Odd multiples of

Even multiples of

Even and odd solutions can be normalised and written in the common form

ψ

(x) = 1

a sin ( k

(x + a) ) k

a = (n + 1) π

2 , n = 0,1,2...

And the energy levels are obtained by taking the limit of the

formula for the finite well ^V

^{→ ∞}

The above formula agrees with what we found in Lecture 1 with the Bohr- Sommerfeld rule

E

= − V

( 1 − (k

a)

ζ

)

ζ = 2mV₀a/ℏ

Overall infinite constant

to be dropped

Even+Odd multiples of

→ ℏ

k

2m = ℏ

2m ( π 2a )

2

(n + 1)

Symmetry properties

Parity

The parity operator is defined by _{P P ψ (x) = ψ (− x)}

Even functions are eigenvectors of with eigenvalue while odd functions are eigenvectors of with eigenvalue _{P −} ^P ₁ ⁺¹

Let us assume that the potential is even

PV(x) = V(− x) = V(x)

Since the Hamilton operator contains only a second derivative with respect to

we have ^x

PHψ (x) = Hψ (− x) = HP ψ (x)

Which means that _{[P, H ] = 0}

If we now consider a solution of the time independent Schrödinger equation

with energy _E ^{ψ (x)}

Hψ (x) = E ψ

and we apply we find _{P Hψ (− x) = E ψ (− x)}

which means that _{ψ (x)} and _{ψ (− x)} are solutions with the same eigenvalue _E

ψ

(x) ∼ ψ (x) + ψ (− x)

Considering even and odd combinations

ψ

(x) ∼ ψ (x) − ψ (− x)

We obtain eigenfunctions of with the same eigenvalue but with definite

parity _{P ψ} ^{H E}

e

(x) = ψ

(x) P ψ

(x) = − ψ

(x)

We conclude that for symmetric potentials we can choose a basis of stationary states with definite parity

Conjugation

Hψ *(x) = E ψ *(x)

Consider now the Schrödinger equation and its complex conjugate

Hψ (x) = E ψ (x)

Combining the two one obtains that _{ψ (x) + ψ *(x)} and _{(ψ (x) − ψ *(x))/ i} are real eigenfunctions with eigenvalue _E

If an energy eigenvalue is non-degenerate, its eigenfunction must be even or odd

Non degenerate eigenfunctions can thus be chosen to be real

General discussion

We now ask ourselves what are the qualitative features of the spectrum for a Hamiltonian of the general form

H = p

2m + V(x)

The Schrödinger equation reads

ψ ′ ′ (x) = 2 m

ℏ

( V(x) − E ) ψ (x)

and thus relates in each point the second derivative of the wave function to the function itself

This means that the qualitative behaviour of the solutions depends on the sign of

V(x) − E

We can first try to understand when the solutions are normalisable Indeed if for _{x → + ∞} or _{x → − ∞} we have _{V(x) > E}

The solutions in those regions will have an exponential behaviour

The solutions in those regions will have an oscillating behaviour We conclude that we can have normalisable solutions (and thus bound states) only if _{V(x) > E} for both _{x → + ∞} and _{x → − ∞}

On the contrary if for _{x → + ∞} or _{x → − ∞} we have _{V(x) < E}

This is in turn possible only if _V(x) has a minimum for finite values of _x

V(x) E

V(x)

E

bound states no bound states

We now want to show that the energy levels of the discrete spectrum are non degenerate

Indeed suppose that there are two eigenfunctions and corresponding to the

same energy _E ^ψ

^ψ

ψ

′ ′

ψ

= 2 m

ℏ

( V(x) − E ) = ψ

′ ′ ψ

We have

Which implies

ψ

′ ′ ψ

− ψ

ψ

′ ′ = 0

Integrating with respect to we obtain _x

ψ

′ ψ

− ψ

ψ

′ = const

But since the functions describe bound states they vanish at infinity and thus the constant must vanish, which means

ψ

′ / ψ

= ψ

′ / ψ

Integrating further we obtain , and, therefore, the functions are not

linearly independent ^ψ

^{= c ψ}

x V(x)

We now consider for simplicity the case of a symmetric potential with a unique minimum and with

and V(x) < 0

x→

lim

V(x) = const

The potential is defined up to a constant that we can set to zero

Then we can have regions of larger or smaller than _{E V(x)}

V(x) − E > 0

Here has the same sign as _{ψ ′ ′ ψ}

ψ > 0

ψ < 0

V

x V(x)

We now consider for simplicity the case of a symmetric potential with a unique minimum and with

and V(x) < 0

x→

lim

V(x) = const

The potential is defined up to a constant that we can set to zero

Then we can have regions of larger or smaller than _{E V(x)}

V(x) − E < 0

Here has the opposite sign with respect to _{ψ ′ ′ ψ}

ψ > 0

ψ < 0

V

The solutions _{ψ (x)} are smoothly constructed from the elements discussed above

1) _{E < V}

In this case _{V(x) − E > 0} everywhere and the solution would diverge to infinity In this case no solution is acceptable, consistently with the intuitive expectation that no states can have energy smaller than _V

2) _V

_{< E < 0}

This is the case of bound states. We can have regions of larger or smaller than and there are two turning points in which vanishes ^E

V(x) ± x

ψ ′ ′

x V(x)

V

E < 0 x

−x

ψ (x)

Outside the turning points the solution is roughly _e

with

χ = 2m(V(x) − E / ℏ

Inside the turning points the solution is oscillatory

with ^{− cos(qx)}

q = 2m(E − V(x)/ ℏ

1) _{E < V}

In this case _{V(x) − E > 0} everywhere and the solution would diverge to infinity In this case no solution is acceptable, consistently with the intuitive expectation that no states can have energy smaller than _V

2) _V

_{< E < 0}

This is the case of bound states. We can have regions of larger or smaller than and there are two turning points in which vanishes ^E

V(x) ± x

ψ ′ ′

x V(x)

V

E < 0 x

− x

ψ (x)

If the energy is sufficiently small the turning points get close, becomes small and the wave function does not develop zeros

q

1) _{E < V}

In this case _{V(x) − E > 0} everywhere and the solution would diverge to infinity In this case no solution is acceptable, consistently with the intuitive expectation that no states can have energy smaller than _V

2) _V

_{< E < 0}

To summarise, one can show that:

There exists at least one bound state with no zeros The spectrum of bound states is discrete

The spectrum of bound states is non degenerate

When there are multiple bound states, the state has nodes _ψ

_n 3) _{E > 0}

In this case the solution are never exponentially damped, and they

become those of the free particle at infinity: there are no bound states