Quantum Mechanics

(1)

P.J. Mulders

Department of Physics and Astronomy, Faculty of Sciences, Vrije Universiteit Amsterdam

De Boelelaan 1081, 1081 HV Amsterdam, the Netherlands email: mulders@few.vu.nl

November 2011 (vs 3.01)

Lectures given in the academic year 2011-2012

1

(2)

Voorwoord

Het college Quantummechanica wordt dit najaar verzorgd door Prof. Piet Mulders geassisteerd door Wilco den Dunnen bij het werkcollege.

Het college volgt in grote lijnen het boek Introduction to Quantum Mechanics, second edition van D.J.

Griffiths (Pearson). Parallel aan het college zullen er aantekeningen verschijnen, die samenvattend van aard zijn.

Het gehele vak beslaat 8 studiepunten en wordt gegeven in periodes 2, 3 en 4. Wekelijks worden 2 uur hoorcollege gegeven, 1 uur wordt besteed aan een presentatie van een van de studenten en vragen, terwijl er 2 uur werkcollege zijn. Daarnaast moeten er opgaven worden ingeleverd, die worden beoordeeld.

We plannen om na blok 3 een deeltentamen af te nemen (schriftelijk op 28-1-2011). De opgaven en het deeltentamen vormen onderdeel van de toetsing. Het geheel wordt afgesloten met een (mondeling) tentamen (in principe in de week 21 t/m 26 maart 2011). Dit tentamen, waarbij boek en uitgewerkte opgaven geraadpleegd mogen worden, gaat behalve over theoretische aspecten (afleidingen niet van buiten leren!) ook over opgaven die tijdens het werkcollege zijn behandeld. De hierboven genoemde regeling geldt voor studenten die actief deelnemen aan colleges en werkcolleges.

Piet Mulders November 2011

(3)

Schedule (indicative)

Material in book Material in notes Week 44 1.1 - 1.3 (p. 1-15) 1

Week 45 1.4 - 1.6 (p. 12-23) 1 Week 46 2.1, 2.2 (p. 24-40) 2, 3.1 Week 47 2.3 (p. 40-59), 2, 3.2

Week 48 2.4, 2.5, 2.6 (p. 59-92) 2, 3.3, 3.4, 3.5 Week 49 3 (p. 92-130) 4, 5

Week 50 3 (p. 92-130) 5, 6, 7

Week 2 4.1 8

Week 3 4 9

Week 4 4 10

Week 5 deeltentamen

Week 6 4 11

Week 7 4, 5.1 12

Week 8 5.1, 5.2, 6.1 12, 13

Week 9 6 13, 14

Week 10 6, (12.1, 12.2) 15, 17

Week 11 7 16

Week 12 9.1 18, 19

Literature

1. D.J. Griffiths, Introduction to Quantum Mechanics, Pearson 2005 2. F. Mandl, Quantum Mechanics, Wiley 1992

3. C. Cohen-Tannoudji, B. Diu and F. Lalo¨e, Quantum Mechanics I and II, Wiley 1977 4. J.J. Sakurai, Modern Quantum Mechanics, Addison-Wesley 1991

5. E. Merzbacher, Quantum Mechanics, Wiley 1998

6. B. Bransden and C. Joachain, Quantum Mechanics, Prentice hall 2000

(4)

1 Basic concepts of quantum mechanics

1.1 Introduction

A classical system is determined by x(t) from which we get the velocity v(t) = ˙x(t), etc. The motion is obtained from Newton’s equation,

md²x dt² = −dV

dx.

It determines how we get from x(0) −→ x(t) (given sufficient boundary conditions x(0) and ˙x(0)).

Quantum mechanically, one works with a wave function ψ(x, t) satisfying the Schr¨odinger equation, i~∂ψ

∂t = −~² 2m

∂²ψ

∂x² + V (x, t) ψ(x). (1)

This determines how the system evolves from ψ(x, 0) −→ ψ(x, t). The constant appearing in this equation is Planck’s constant,

h = 6.626 × 10⁻³⁴ J s.

or more precise the reduced Planck’s constant ~ = h/2π = 1.055 × 10⁻³⁴ J s = 6.582 × 10⁻¹⁶ eV s. This also determines the domain where quantum mechanics is needed to get a good description of nature, namely the domain where appropriate quantities are of the order of this constant. In this respect, note that energy × time and/or to momentum × distance or angular momentum have the same dimension as

~.

To go any further, one needs to know what to do with ψ(x, t). It turns out that its physical significance comes in by doing something with it. In particular |ψ(x, t)|²dx is the probability to find the system at time t in an interval of size dx around the point x, or for a finite domain,

Pab= Z b

a dx |ψ(x, t)|². (2)

The function itself and the various ways to denote it, like

ψ(x, t) = hx|ψ(t)i = hx, t|ψi, (3)

is referred to as probability amplitude (NL: waarschijnlijkheidsamplitudo) for state (NL: toestand) ψ (or

|ψi) to be at place x at time t. The use of the notation |ψi will be specified later. Where the system was before time t cannot be answered, one would have to do a measurement for that at the earlier time.

But as soon as a measurement is done and the particle is found at position a, the state is no longer described by the wave function ψ but by ψa, which would be a function peaking around point a. This sharp function will then spread according to the Schr¨odinger equation. Pretty weird!

1.2 Operators and their expectation values

Suppose we know (we’ll see later how) that we have a lot of identical particles, all being described by the same wave function ψ, of which we measure the position x. If these positions would be discrete, i.e. just a collection of possibilities xiwe would measure how many (ni) particles end up in each of the positions.

That probability is Pi= ni/N , where N =P

ini is the total number of measurements. One has

1 = X

i

Pi, (4)

hxi = X

i

xiPi, (5)

hx²i = X

i

x²_i Pi, (6)

(10)

which states that the probabilities add up to one, and weighing with xi or x²_i gives the average of x, x², etc., which can be extended to the average of any weight function w(x). The standard deviation (squared) is found by weighing with (xi− hxi)². Since hxi is a number, one finds

σ²= ∆x²=X

i

(x²_i − hxi)²Pi = hx²i − hxi². (7)

For the quantum mechanical outcome of an x-measurements one has the continuous version of the above with all (real) possibilities for x, with probabilities ρ(x, t) = |ψ(x, t)|². This implies the normalization and leads to specific expectation values (NL: verwachtingswaarden),

1 =

Z

dx ρ(x, t) = Z

dx ψ^∗(x, t)ψ(x, t), (8)

hxi = Z

dx x ρ(x, t) = Z

dx ψ^∗(x, t) x ψ(x, t), (9)

hx²i = Z

dx x²ρ(x, t) = Z

dx ψ^∗(x, t) x²ψ(x, t), (10) which can be extended to any weight function and for which one has a standard deviation (NL: standaard- afwijking) ∆x of which the square is

∆x²= Z

dx (x − hxi)²ρ(x, t) = hx²i − hxi². (11) The notation on the right hand side with x between the functions ψ^∗and ψ can be conveniently generalized to any operator

h ˆOi = Z

dx ψ^∗(x, t) ˆOψ(x, t), (12)

where Ô is an operator acting on the function ψ. The left-handside is referred to as the expectation value of O. In the simple case above we would have the position operator ˆx that acts like ˆxψ(x, t) = x ψ(x, t), but also the derivative d/dx is an example of an operation working on ψ and producing a new function (the derivative function). The normalization is in fact just the expectation value of the unit operator, hˆ1i = 1. Usually the hat on the operators will be omitted. Finally we note that an expectation value does depend on the state ψ, so the better notation would have been h Ôi^ψ or hψ| Ô|ψ instead of just h Ôi.

Just as one talks about the expectation value of ˆx, one now can talk about the expectation value of ˆO (in state ψ).

A first consistency check of quantum mechanics is the time independence of the normalization. This is straightforward, although it involves a bit of manipulation and doing partial integrations. One proves

that d

dt Z

dx ψ^∗(x, t)ψ(x, t) = 0, (13)

for any normalizable function (Eq. 8) that satifies the Schr¨odinger equation (Eq. 1). A normalized continuous function must go to zero at x → ±∞. For the probability in any finite interval one has a modified equation,

dPab

dt = J(a, t) − J(b, t), (14)

where

J(x, t) = −i~

2m

ψ^∗ ∂ψ

∂x −∂ψ^∗

∂x ψ

(15) is the probability current (NL: waarschijnlijkheidsstroom) or flux.

(11)

If we look at the time dependence of hxi, it is (in general) nonzero. We obtain d

dthˆxi = −i~

m Z

dx ψ^∗(x, t)∂ψ(x, t)

∂x =hˆpxi

m , (16)

with the momentum operator (NL: impulsoperator) ˆp given by ˆ

px= −i~ ∂

∂x. (17)

The position and momentum are actually the basic quantities in quantum mechanical problems, as they are in classical mechanics. E.g. classically the energy can be written as a function of p and x. That same function is actually the righthand-side of the Schr¨odinger equation in Eq. 1,

i~∂ψ(x, t)

∂t = ˆHψ(x, t), (18)

with

H =ˆ pˆ²

2m+ V (ˆx, t) = −~² 2m

∂²

∂x² + V (x, t). (19)

This is the energy operator or Hamiltonian, which is the most important operator because its role as time evolution operators. Other important operators are e.g. the angular momentum operators ˆℓ = ˆr× ˆp, which in 3 dimensions play a key role. Explicitly, they are

ℓx= −i~

y ∂

∂z− z ∂

∂y

, (20)

ℓy= −i~

z ∂

∂x− x ∂

∂z

, (21)

ℓz= −i~

x ∂

∂y − y ∂

∂x

. (22)

1.3 Commutators and uncertainty relations

At this point, we just want to add a word of caution on the order of operators. Calculating ˆA ˆB ψ can be different from ˆB ˆA ψ. Trying this for a simple wave function, for instance

ψ(x) = N e^{−α x}²^/2e^−iωt. We have

ˆ

x ψ(x) = x N e^{−α x}²^/2e^−iωt ˆ

pxψ(x) = −i~dψ

dx = +i~ αx N e^{−α x}²^/2e^−iωt Applying the operatos once more, we get

ˆ

x²ψ(x) = x²N e^{−α x}²^/2e^−iωt ˆ

pxx ψ(x) = −i~ˆ d dx

x e^{−α x}²^/2e^−iωt

= −i~ 1 − αx²

N e^{−α x}²^/2e^−iωt, ˆ

x ˆpxψ(x) = −i~ x d dx

e^{−α x}²^/2e^−iωt

= i~ αx²N e^{−α x}²^/2e^−iωt, ˆ

p²_xψ(x) = −~²d²ψ

dx² = ~² α − α²x²

N e^{−α x}²^/2e^−iωt,

(12)

The various expectation values (familiarize yourselves with these type of integrals and see how you get them using Mathematica) are

h1i = |N|² Z _∞

−∞

dx e^{−α x}²= |N|²r π

α= 1 ⇒ N = α π

1/4

, hˆxi = |N|²

Z _∞

−∞

dx x e^{−α x}²= 0, hˆpxi = i~|N|²α

Z _∞

−∞

dx x e^{−α x}²= 0, hˆx²i = |N|²

Z _∞

−∞

dx x²e^{−α x}² = 1 2α, hˆp²_xi = |N|²α~²

Z _∞

−∞dx 1 − α x²

e^{−α x}² = α~² 2 .

We note that for this wave function the expectation value of the position and momentum are zero, but the standard deviations are nonzero, ∆x = 1/√

2α and ∆px= ~p

α/2. This shows that in this particular case a wave function which is strongly peaked around zero (α is large) has a large standard deviation if one looks at the momentum. The product of expectation values is constant, however, ∆x ∆px= ~/2. In general, we will find for any wave function that

∆x ∆px≥ ~

2. (23)

This is an example (and the most famous one) of an uncertainty relation and as we will discover later, it is a property that is intimately connected with the fact that the operators ˆx and ˆpx do not commute,

[ˆx, ˆpx] = ˆx ˆpx− ˆpxx = i~.ˆ (24) This commutator is actually the most basic one in quantum mechanics, referred to as the canonical commutation relation. That it holds for any function is easily seen by calculating for an arbitrary function

φ h

x, d dx

i

φ(x) = x d dxφ − d

dx(xφ(x)) = xdφ dx− xdφ

dx − φ(x) = −φ(x).

1.4 Eigenstates

Very special wave functions are those for which

Aφˆ n= anφn. (25)

These functions φn are called eigenfunctions of ˆA with eigenvalues an. The nice thing is that for these functions (assuming they are normalized) the expectation value of ˆA is

h ˆAi = Z

dxφ^∗_n(x) ˆAφn(x) = an

Z

dxφ^∗_n(x)φn(x) = an, (26)

and similarly h ˆA²i = a²n, thus one has ∆A = 0, i.e. there is no uncertainty in A.

As an example we look at the functions

φk(x) = e^{i kx}, (27)

(13)

known as plane waves (NL: vlakke golven). It is easy to show that they are eigenfunctions of the momentum operator,

ˆ

pxφk(x) = −i~ ∂

∂xφk = ~k φk(x), (28)

with eigenvalue ~k (k is usually referred to as wave number, NL: golfgetal). Thus there is no momentum uncertainty ∆px= 0. It is actually a functions for which ρ = φ^∗_kφk = 1, thus one finds that its position is undetermined (consistent with the uncertainty relation). It is straightforward to calculate the flux for a plane wave. It is given by

Jx=~k

m = ρ vx= vx. (29)

If you want something more complex, show that the function of three coordinates (and r²= x²+ y²+ z²), Y+1(x, y, z) = x + iy

r f (r),

is an eigenfunction of ℓz in Eq. 22. The index of the function actually refers to this eigenvalue. Also calculate the flux for this wave function. What do you note about the direction of the flux in a particular point (x, y, z).

(14)

2 Time evolution

2.1 Stationary states

We return to our important Schr¨odinger equation, i~∂ψ(x, t)

∂t = ˆH(ˆx, ˆpx)ψ(x, t) = −~² 2m

d²ψ(x, t)

dx² + V (x)ψ(x, t).

In the case that the potential in the Hamiltonian is independent of the time, one can easily see that the Schr¨odinger equation allows separable solutions of the form

ψ(x, t) = φ(x) f (t).

Inserting this ansatz in Eq. 1, one finds with ∂ψ(x, t)/∂t = φ(x) df /dt and ∂ψ(x, t)/∂x = f (t) dφ/dx, i~^df_dt

f (t) = E =

−~² 2m

d²φ

dx² + V (x)φ(x)

φ(x) , (30)

where E must be t-independent but also x-independent, thus constant. We obtain two equations

−~² 2m

d²φ

dx² + V (x)φ(x) = Eφ(x), (31)

the time-independent Schr¨odinger equation and a time dependence i~df

dt = E f (t) =⇒ f (t) ∝ e^{−i Et/~}. (32)

Thus we get (in general many) stationary states (NL: stationaire toestanden),

ψEn(x, t) = φn(x) e^{−i E}ⁿ^t/~, (33) where φn(x) are eigenfunctions (NL: eigenfuncties) of the (time-independent) Hamiltonian with eigenvalues En,

H(x, p)φˆ n(x) = Enφn(x). (34)

It is easy to see that if φn is normalized, then

h ˆHi^φn = En, h ˆH²i^φn = E_n², and thus ∆Hφn= 0. (35) Thus, for an eigenstate φn of ˆH the standard deviation is zero and the state has a precise energy. To find real energies the Hamiltonian must satisfy certain conditions, to be precise it must be hermitean (to be discussed later). Actually, when measuring an energy, it must be one of the eigenvalues of ˆH and the system is immediately after that measurement in a state φn(x). The set of eigenvalues of ˆH is known as its energy spectrum.

2.2 General time evolution

The eigenvalues of the Hamiltonian form the energy spectrum of a quantum theory. One can show that normalizable solutions φn(x) of the time-independent Hamiltonian can be choosen real, while the eigenvalues of the Hamiltonian En must be real. Furthermore the eigenfunctions satisfy

Z

dx φ^∗_m(x) φn(x) = δmn, (36)

(15)

where δmn is known as the Kronecker delta, which is 1 if m = n and δmn= 0 if m 6= n. Such functions are called orthogonal (NL: orthogonaal).

The stationary states as in Eq. 33 have a well-determined energy and a known time dependence which is a simple phase. The probability ρ(x, t) is for a stationary state independent of the time. This is not true in general. But starting with a given function at t = 0, one can expand it in the set of eigenfunctions, which turn out to form a complete set of functions. This means that we can write

ψ(x, 0) =X

n

cnφn(x) with cn= Z

dx φ^∗_n(x) ψ(x, 0). (37)

Since the Schr¨odinger equation is a linear equation and we know the time-dependence of ψEn(x, t) we immediately deduce that

ψ(x, t) =X

cnφn(x) e^{−i E}ⁿ^t/~. (38)

Such solutions in general are complex and the probability is not time-independent. They show oscillations.

If the initial state ψ(x, 0) is a linear combination of φ1(x) and φ2(x) the oscillations will be of the form with sine or cosine with argument cos((E1− E2)t/~), hence have a frequency f = (E1− E2)/h or an angular frequency ω = 2π f = (E1− E2)/~.

2.3 Spectrum and basic behavior of solutions

This is best illustrated by considering the one-dimensional hamiltonian H = −ˆp²_x/2m + V (ˆx), stationary solutions of the form ψ(x, t) = φ(x) exp(−iEt/~), are obtained by solving

−~² 2m

d²

dx² + V (x)

φ(x) = E φ(x), rewritten as

d²

dx²φ = −2m

~² (E − V (x))

| {z }

k²(x)

φ(x). (39)

κ

²

(x)

~ k

²

(x)

~ V(x)

E

(x) oscillatory exponential

x x

φ

We distinguish two situations:

(i) k²(x) ≥ 0 in the region where E ≥ V (x). In that case the change of slope is opposite to the sign of

(16)

the wave function, which implies that the solution always bends towards the axis. Looking at the case that k²(x) = k² = 2m(E − V⁰)/~²(constant potential V0) one has:

φ(x) = A sin kx + B cos kx, (40)

or

φ(x) = A^′e^{i kx}+ B^′e^{−i kx}. (41)

(ii) k²(x) = −q²(x) ≤ 0 in the region where E ≤ V (x). In that case the change of slope has the same sign as the wave function, which implies that the solution always bends away from the axis. Looking at the case that k²(x) = −q² = −2m(V0− E)/~² (constant potential V0) one has:

φ(x) = A sinh qx + B cosh qx, (42)

or

φ(x) = A^′e^qx+ B^′e^−qx. (43)

2.4 Boundary conditions and matching conditions

In order to find proper solutions of the Schr¨odinger equation, which is a second order (linear) differential equation, one needs d²φ/dx², hence φ(x) and dφ/dx must be continuous. The second condition implies

dφ dx

_a+ǫ− dφ dx

a−ǫ

= Z a+ǫ

a−ǫ

dx d²φ dx² =2m

~² Z a+ǫ

a−ǫ dx (V (x) − E) φ(x) = 0

Thus continuity not necessarily requires a continuous potential, but it is necessary that the potential remains finite. In points where the potential becomes infinite one must be careful, e.g. by studying it as a limiting case.

Cases in which the potential makes a jump are studied by imposing matching conditions from both sides, namely equating the wave function and its derivative. Because one in quantum mechanics often takes some freedom in the normalization (to be imposed later), one often chooses to equate the ratio of derivative and wave function, the socalled logarithmic derivative,

ǫ→0lim

(dφ/dx) φ(x)

a−ǫ

= lim

ǫ→0

(dφ/dx) φ(x)

a+ǫ

. (44)

The behavior at infinity (let’s simply assume the potential to be zero) depends on the energy, e.g. if E = −~²q²/2m < 0 one must have

x→∞lim φ(x) = C e^−qx−→ 0, and lim

x→−∞φ(x) = C^′e^qx−→ 0. (45) If the energy E = ~²k²/2m > 0 one could have in case of an wave from the left with incoming flux ~k/m (allowing to be general for a reflected wave) the boundary condition

x→−∞lim φ(x) = e^{i kx}+ ARe^{−i kx}. (46) In a particular physics problem, one might look for a solution with only a transmitted wave

x→∞lim φ(x) = ATe^{i kx}. (47)

with a transmitted flux |AT|²~k/m and thus a transmission probability (transmitted/incoming flux) T = |AT|². With the above ingredients several one-dimensional problems can be solved. Some examples

(17)

of potential wells and barriers will be discussed below. Given a particular potential we note three domains:

(i) No solutions exist when E ≤ V^min. In that case one has everywhere k²(x) ≤ 0 and there is no way that the exponentially behaving solutions at ±∞ can be matched without a region where the solution bends back towards the axis, i.e. a region where E ≥ V (x).

(ii) In the regin Vmin < E < Vasym (where Vasym is the lowest asymptotic value of the potential) one has a discrete energy spectrum corresponding to a discrete set of solutions: starting with a vanishing exponential at, say, x = −∞, one will find in general a solution that at x = +∞ behaves as a linear combination of two exponential functions (as in Eq. 43). Only for very specific energies the coefficient of the growing exponential e^qxwill be zero and one finds a normalizable solution. These localized solutions, found for discrete energies, are referred to as bound states.

(iii) In the region E ≥ Vasymone has a continuous spectrum of which the solutions at infinity oscillate or equivalently are complex-valued waves (with definite momentum). One always can find such a solution.

This is even true if the asymptotic values of the potential at x = ±∞ are not equal.

Finally, considering an infinitely high potential (e.g. for x > 0) as the limit of a large potential V0 for x > 0 one must match on to the solution e^−qx with q²= 2mV0/~², leading to

limx↑0

(dφ/dx) φ(x) = lim

V0→∞

r2mV0

~² = ∞, (48)

which requires φ(0) = 0 and a finite derivative as only nontrivial possibility.

2.5 Three elementary properties of one-dimensional solutions

Property 1: In one dimension any attractive potential has always at least one bound state.

Property 2: For consecutive (in energy) solutions one has the node theorem, which states that the states can be ordered according to the number of nodes (zeros) in the wave function. The lowest energy solution has no node, the next has one node, etc.

Property 3: Bound state solutions of the one-dimensional Schr¨odinger equation are nondegenerate.

Proof: suppose that φ1 and φ2 are two solutions with the same energy. Construct W (φ1, φ2) = φ1(x)dφ2

dx − φ2(x)dφ1

dx, (49)

known as the Wronskian. It is easy to see that d

dxW (φ1, φ2) = 0.

Hence one has W (φ1, φ2) = constant, where the constant because of the asymptotic vanishing of the wave functions must be zero. Thus

(dφ1/dx)

φ1 =(dφ2/dx)

φ2 ⇒ d

dxln φ1= d dxln φ2

⇒ d

dxln φ1

φ2

= 0 ⇒ ln φ1

φ2

= constant ⇒ φ1∝ φ², and hence (when normalized) the functions are identical.

(18)

3 One-dimensional problems

3.1 The infinite square well

We start by considering the potential

V (x) = 0 for |x| < a/2 and V (x) = ∞ for |x| ≥ a/2.

As discussed in the previous section, the wave function is zero for |x| > a/2. For E ≤ 0 one would have solutions of the form

φ(x) = A e^qx+ B e^−qx,

with q²= 2m|E|/~², which does not have a solution (one finds A = B = 0 from the conditions φ(a/2) = φ(−a/2) = 0).

For E > 0 one has solutions, which in general are of the form φ(x) = A sin(kx) + B cos(kx),

with k² = 2mE/~², which leads to either A = 0 or B = 0. Check that the solutions separate into (including normalisation) the even solutions

φn(x) = r2

a cos nπ a x

with En= n² π²~²

2ma² (odd n values), (50) and the odd solutions

φn(x) = r2

a sin nπ a x

with En= n² π²~²

2ma² (even n values). (51) The number of nodes is given by n − 1. It is easy to check that the solutions are orthogonal,

Z ∞

−∞

dx φ^∗_n(x) φm(x) = δmn. (52)

3.2 Harmonic oscillator

The harmonic oscillator potential is encountered in very many applications. It is a natural approximation for a system in equilibrium. Around the minimum of any potential function one can write

V (x) = V (xmin) +1

2k (x − xmin)²+ . . . ,

and after a redefinition of the energy (shift) and the coordinates (choose xmin = 0), one has V (x) =

1

2k x²≡¹₂mω²x². The latter involves the angular frequency ω =p

k/m, which appears in the classical solutions. From m¨x = F = −dV/dx = −kx one finds as classical solution

x(t) = A cos(ωt + ϕ),

p(t) = m ˙x(t) = −mω A sin(ωt + ϕ) The energy

E = p²

2m+ V (x) = mω²A², is time-independent. We note that it can be written as

E =(mωx + i p)(mωx − i p)

2m = A₋(t) A+(t)

2m ,

(19)

where A_±(t) = (mω x ± i p) = mω A e^±i(ωt+ϕ).

For the time-independent Schr¨odinger equation for the harmonic oscillator in one dimension the energy expression is taken as the Hamiltonian with x and p operators. It is written as

H φ(x) = p²

2m+ V (x)

φ(x) =

−~² 2m

d² dx² +1

2mω²x²

φ(x) = Eφ(x). (53)

To handle it mathematically, it is conventient to make this equation dimensionless by introducing ξ = α x, leading to

−~²α² 2m

d²

dξ² +mω² 2 α² ξ²

φ(ξ) = E φ(ξ).

To fix α one can equate the multiplicative factors in both terms, ~²α²/m = mω²/α² which gives α² = mω/~, leading to

~ω

−1 2

d² dξ² +1

2ξ²

φ(ξ) = E φ(ξ). (54)

Rather than solving this as a differential equation (done in the appendix below), we use an algebraic approach to solve this problem. For this we write the Hamiltonian as a product of two (conjugate) operators, but one must be careful in this case, since the operators x and p (or equivalently ξ and d/dξ do not commute. To be precise we have the basic canonical commutation relation

[x, p] ≡ xp − px = i~

(see Eq. 24). This also means that the classical quantities A₋and A+mentioned above, do not commute after replacing the position and momentum by operators. Introducing

a_±= 1

√2

ξ ∓ d

dξ

= 1

√2mω~(mω x ∓ i p) , (55)

one then finds that these operators satisfy

[a₋, a+] = 1 (56)

and one must be careful to rewrite H. The correct result is H = ~ω

a+a₋+1 2

. (57)

Finding the eigenvalues of this is done in a few steps. It is sufficient to study the eigenvalues of N ≡ a+a₋, which means that we are looking for (normalized) eigenfunctions φn satisfying

N φn= a+a₋φn≡ nφn. (58)

It is easy to see that the eigenvalue n satisfies n =

Z

dx φ^∗_nN φn = Z

dx φ^∗_n(a+a₋φn) = Z

dx (a₋φn)^∗(a₋φn) ≥ 0.

Here one must realize that whileR

dx φ^∗(xφ) =R

dx (xφ)^∗φ andR

dx φ^∗(pφ) =R

dx (pφ)^∗φ (x and p are hermitean operators) this is not true for the a_±operators, for which one hasR

dx φ^∗(a_±φ) =R

dx (a_∓φ)^∗φ.

Returning to the eigenvalue n, we thus know it is real and non-negative.

The beauty of a_± is that they can be used to construct new eigenfunctions,

N (a+φn) = a+a₋a+φn= a+(a+a₋+ 1)φn= (n + 1)a+φn, (59) N (a₋φn) = a+a₋a₋φn= (a₋a+− 1)a−φn= (n − 1)a+φn. (60)

(20)

Thus a_±φn = c_±φ_n±1. From Eq. 3.2 we see that c₋ =√

n. Similarly one easily sees that c+=√ n + 1.

Thus if φn is a solution of the Hamiltonian, a+φn=√

n + 1 φn+1 and a₋φn=√

n φ_n−1 (61)

are also eigenfunctions of the Hamiltonian with energies increased or lowered by ~ω. But this poses a problem with Eq. 3.2. We may end up with negative eigenvalues, unless there is a solution such that a₋φ = 0. This solution exist, namely φ0. Thus n = 0 is necessarily one of the eigenvalues of N . All other solutions then can be constructed from this. Summarizing, we have found that

Hφn =

n +1

2

~ω φn, (62)

with n = 0, 1, 2, . . . and eigenfunctions satisfying φn= (a+)ⁿ

√n! φ0, (63)

where φ0 satisfies a₋φ0= 0 of which the solution is easily found, a₋φ0=

ξ + d

dξ

φ0(ξ) = 0 =⇒ φ0(ξ) ∝ e^−ξ²^/2. (64) Note actually that the functional form of the solutions is not needed to find the spectrum (eigenvalues of Hamiltonian) or to find expectation values of x, p, or combinations of them. The operators x and p can simply be expressed in a_± of which the action on the φn’s is known.

Appendix: Hermite polynomials

The problem of the one-dimensional harmonic oscillator in essence reduces to the differential equation y^′′+ g0(x) y = 0 with g0(x) = 2n + 1 − x² (65) for which the solutions are given by

y(x) = e^−x²^/2Hn(x). (66)

where Hn are polynomials of degree n. They are normalized as Z _∞

−∞

dx e^−x²[Hn(x)]²= 2ⁿn!√

π, (67)

and satisfy the differential equation

d²

dx² − 2x d dx + 2n

Hn(x) = 0. (68)

Some useful properties are

Hn(x) = (−)ⁿe^x² dⁿ

dxⁿ e^−x² (Rodrigues formula), (69) x Hn(x) = 1

2Hn+1(x) + n H_n−1(x), (70)

d

dxHn(x) = 2n H_n−1(x). (71)

(21)

Some explicit polynomials are H0(x) = 1, H1(x) = 2x, H2(x) = 4x²− 2, H3(x) = 8x³− 12x.

-1 -0.5 0.5 1

-4 -2 2 4

Plots of the Hn(x) or HermiteH[n,x] functions for n = 0, 1, 2, and 3.

3.3 Free particle states

The next example is the case that V (x) = 0. In this case one only has positive energy solutions that do not vanish asymptotically, but oscillate. We will see that there is a continuous spectrum. Introducing Ek ≡ ~ωk≡ ~²k²/2m or ~k =√

2m E, the Schr¨odinger equation takes the form d²

dx²φ(x) = −k²φ(x), (72)

with as in Eq. 40 or 41 simple solutions, for the case V = 0 generally referred to as plane waves,

φk(x) = A e^{i kx}. (73)

The stationary time-dependent solutions are ψk(x, t) = A exp

i kx − ~k² 2mt

= A e^{i kx−i ω}^k^t)= A e^{i k(x−v}^phase^t), (74) where vphase= ωk/k = ~k/2m is the phase velocity. We note that the density of this solution is constant, ρ = |A|². The functions correspond to definite momenta p = ~k (they are eigenfunctions of the momentum operator). Using the expression for the current (Eq. 15) one gets J(x, t) = |A|²~k/m = ρ p/m = ρ v, where v is the ’normal’ classical velocity, v = p/m = ~k/m, not to be confused with the phase velocity.

As discussed before, we now also know the evolution of any system, described by an arbitrary wave function ψ(x, 0) = φ(x) at time t = 0. We have to expand this solution into plane waves,

φ(x) = Z _∞

−∞

dk

2π φ(k) e˜ ^{i kx} with φ(k) =˜ Z _∞

−∞

dx φ(x) e^{−i kx}, (75)

where the second equation tells us how to find the coefficient. This is just the continuous version of Eq. 37 with P

n ↔R

dk/2π and cn ↔ ˜φ(k). The procedure is know as Fourier transformation and its inverse.

The use of 2π versus √

2π in the integrations is just a matter of convention. The full time-dependent solution becomes the wave packet

ψ(x, t) = Z dk

2π φ(k) e˜ ^{i kx−i ω}^k^t. (76)

One might ask what the velocity is of the corresponding density |ψ(x, t)|². This can be studied, general- izing the case of ωk = ~k²/2m to an arbitrary situation where ω(k) is the dispersion relation. Suppose

(22)

the wave packet is strongly peaked around k ≈ k⁰, in that case hpi ≈ ~k⁰. By expanding the frequencies around k0one finds

ω(k) ≃ ω(k0) + dω dk

k0

(k − k0) ≃ ω0+ vg(k − k0). (77) Here the group velocity vg = dω/dk = dE/dp has been introduced, which is also the ’classical’ velocity from basic classical mechanics. We get

ψ(x, t) ≃ Z _∞

−∞

dk

2π φ(k) e˜ ^{i kx−i(ω}⁰^+v^g^(k−k⁰^))t

≃ e^−i(ω⁰^−v^g^k⁰^)t Z _∞

−∞

dk

2π φ(k) e˜ ^{i k(x−v}^g^t)

≃ e^−i(ω⁰^−v^g^k⁰^)tφ(x − vgt) (78)

which up to a phase that doesn’t affect |ψ|²is the same wave function as at t = 0, but shifted as if moving with the group velocity vg.

3.4 The delta potential

First, let’s discuss the properties of the delta-function. The Dirac delta-function is in fact a distribution (mapping functions into numbers), defined via

Z

dx f (x) δ(x) = f (0), (79)

or R

dx f (x) δ(x − a) = f(a). It can be considered as the limit of a peaked function, e.g.

δ(x) = lim

ǫ→0φǫ(x), (80)

where φǫ(x) = 0 if |x| > ǫ/2 and φ^ǫ(x) = ǫ if |x| ≤ ǫ/2. Other examples are δ(x) = lim

Λ→∞

√Λ

πe^−Λ²^x², (81)

δ(x) = lim

Λ→∞

sin Λx π x = lim

Λ→∞

Z Λ

−Λ

dk 2πe^{i kx}=

Z _∞

−∞

dk

2πe^{i kx}. (82)

Some properties are

δ(−x) = δ(x), δ(ax) = 1

|a|δ(x), x δ(x) = 0, Z

dx f (x) δ(x − a) = f(a), d

dxθ(x) = δ(x),

where θ(x) is the Heaviside function, θ(x) = 0 for x < 0 and θ(x) = 1 for x ≥ 0. Note that Mathematica can work with this function, e.g. to define a square well potential with depth −V0 using the expression V (x) = −V⁰θ(¹₂a + x) θ(¹₂a − x).

(23)

Bound state(s) of the delta potential

Given a one-dimensional potential of the form

V (x) = − ~²

m aδ(x). (83)

The Schr¨odinger equation becomes

−φ^′′(x) − 2

aδ(x)φ(x) = 2mE

~² φ(x). (84)

By integrating from (0 − ǫ) tot (0 + ǫ) one shows that for a solution of the Schr¨odinger equation limx↓0φ^′(x) − lim

x↑0φ^′(x) = −2 φ(0)

a . (85)

For this potential there (always) exist one bound state solution. For E = −~²q²/2m we get φ(x) = e^qx voor x < 0,

φ(x) = e^−qx voor x > 0.

The condition in Eq. 85 gives q = 1/a. Thus we find a bound state with energy E = −~²/2ma².

Scattering states of the delta potential

For positive energies we have plane wave solutions except for x = 0 and we can calculate the transmission and reflection coefficient (AT en AR) for a wave coming from the left (from x → −∞). The wave function corresponding to this problem is ( taking E = ~²k²/2m)

φ(x) = e^{i kx}+ ARe^{−i kx} for x < 0, φ(x) = ATe^{i kx} for x > 0.

The boundary conditions at x = 0 become (using continuity and for derivatives Eq. 85), 1 + AR = AT,

ika AT − (ika − ika A^R) = 2AT. From this one obtains

AR= 1 1 − i ka, AT = i ka

1 − i ka.

One can easily show that the flux is conserved. Calculating the flux for plane waves e^±ikx one has J(x) = ±~k/m, we find that flux for x < 0 is ~k(1 − |AR|²)/m and for x > 0 is ~k |AT|²/m. Indeed the results satisfy 1 − |AR|²= |AT|².

3.5 Bound states and scattering solutions for a square well potential

We consider the potential

V (x) = −V0 for |x| < a and V (x) = 0 for |x| ≥ a.

(24)

One does not find a finite solution for E < −V⁰. One would have exponential behavior in all x-ranges and since k²(x) < 0 everywhere, one must have a solution which everywhere either has a positive or a negative derivative. For −V⁰< E < V0, finite solutions are of the form

φ(x) = C^′e^qx for x ≤ −a, (86)

φ(x) = A sin(kx) + B cos(kx) for |x| ≤ a, (87)

φ(x) = C e^−qx for x ≥ a (88)

(see Eq. 45), where E = −~²q²/2m and E + V0= ~²k²/2m. The matching conditions for wave functions and derivatives give

C^′e^−qa = −A sin(ka) + B cos(ka), A sin(ka) + B cos(ka) = C e^−qa,

(q/k) C^′e^−qa = A cos(ka) + B sin(ka), A cos(ka) − B sin(ka) = (q/k) C e^−qa.

It is easy to convince one self that there are two classes of solutions,

• even solutions with A = 0 and C^′ = C,

• odd solutions with B = 0 and C^′ = −C.

For the solutions one obtains from the matching of logarithmic derivatives in the point x = a

k tan(ka) = q (even), (89)

k cot(ka) = −q (odd). (90)

Introducing the dimensionless variable ξ = ka and writing ξ0 =p

2mV0a²/~², one has qa =p ξ₀²− ξ² and E = −~²(ξ₀²− ξ²)/2ma². The variable ξ runs from 0 ≤ ξ ≤ ξ⁰. The conditions become

tan(ξ) =

pξ²₀− ξ²

ξ (even), (91)

tan(ξ) = − ξ

pξ₀²− ξ² (odd). (92)

One sees that there always is an even bound state in the region 0 ≤ ξ ≤ minimum{ξ⁰, π/2}, and then depending on the depth of the potential (i.e. as long as ξ ≤ ξ⁰) a first odd bound state (with one node) between π/2 ≤ ξ ≤ π, a second even bound state (with two nodes) between π ≤ ξ ≤ 3π/2, etc.

Next one might look for solutions with positive energy, in particular of the kind with boundary conditions as in Eqs. 46 and 47,

φ(x) = e^{i kx}+ ARe^{−i kx} for x ≤ −a, (93)

φ(x) = A sin(Kx) + B cos(Kx) for |x| ≤ a, (94)

φ(x) = ATe^{i kx} for x ≥ a, (95)

where E = ~²k²/2m and E + V0= ~²K²/2m. The matching conditions become e^{−i ka}+ ARe^{i ka} = −A sin(Ka) + B cos(Ka), A sin(Ka) + B cos(Ka) = ATe^{i ka},

i k

Ke^{−i ka}− i A^R k

Ke^{i ka} = A cos(Ka) + B sin(Ka), A cos(Ka) − B sin(Ka) = i A^T k

Ke^{i ka}.

(25)

Eliminating A and B one obtains for the (complex) amplitudes of the reflected and transmitted waves, AR = e^{−2i ka} i(K²− k²) sin(2Ka)

2kK cos(2Ka) − i (k²+ K²) sin(2Ka), (96)

AT = e^{−2i ka} 2kK

2kK cos(2Ka) − i (k²+ K²) sin(2Ka). (97) In contrast to negative energies, one has always solutions for positive energy. The interpretation of the specific ansatz of the wave function is an incoming wave from the left, φi(x) = e^{i kx}(with flux ji = k/m), a reflected wave, φr(x) = ARe^{−i kx}(with flux jr= −|A^R|²k/m), and a transmitted wave, φr(x) = ATe^{i kx} (with flux jt= |A^T|²k/m). The probabilities for reflection and transmission are

R = |A^R|² = (K²− k²)²sin²(2Ka)

4k²K² cos²(2Ka) + (k²+ K²)² sin²(2Ka)

= (K²− k²)² sin²(2Ka)

4k²K²+ (K²− k²)² sin²(2Ka) (98)

T = |AT|² = 4k²K²

4k²K² cos²(2Ka) + (k²+ K²)² sin²(2Ka)

= 4k²K²

4k²K²+ (K²− k²)² sin²(2Ka), (99) satisfying 1 − R = T (flux from the left = flux to the right). Note that there exists particular energies, for which the wave vector K satisfies 2Ka = nπ, in which case R = 0 and T = 1, i.e. the potential is

’invisible’ for those particular waves.

3.6 Reflection and transmission through a barrier

We consider the situation of a positive square potential, also called a barrier potential, V (x) = +V0 for |x| < a and V (x) = 0 for |x| ≥ a.

In this case we only have scattering solutions for E > 0. The case E ≥ V⁰ is even completely similar to the scattering solutions in the previous section. We have the same expression for k²= 2mE/~², but now K²= 2m(E −V⁰)/~². The quantities k and K are the wave numbers in the different regions, respectively.

In terms of k and K the expressions for reflection and transmission remain the same as in Eqs. 98 and 99. One again has the situation that there are energies for which 2Ka = nπ and the barrier is ’invisible’.

The case 0 ≤ E ≤ V⁰appears at first sight different because the wave functions in the region −a ≤ x ≤ a become exponentional, e.g. linear combinations of cosh(qx) and sinh(qx) with q²= 2m(V0−E)/~². But this in fact is nothing else than using complex wave numbers, K → i q (note that sign doesn’t matter), leading instead of Eqs. 96 and 97 to

AR = e^{−2i ka} −i (k²+ q²) sinh(2qa)

2kq cosh(2qa) − i (k²− q²) sin(2qa), (100)

AT = e^{−2i ka} 2kq

2kq cosh(2qa) − i (k²− q²) sinh(2qa). (101) leading to the reflection and transmission probabilities through a barrier,

R = |A^R|² = (k²+ q²)²sinh²(2qa)

4k²q²+ (k²+ q²)² sinh²(2qa) (102) T = |AT|² = 4k²q²

4k²q²+ (k²+ q²)² sinh²(2qa). (103)

(26)

In this situation one always has T < 1. In the case that qa ≫ 1, which is the case if E ≪ V⁰ and 2m V0a²/~²≫ 1, the transmission coefficient reduces to the tiny probability

T ≈ 16 k²q²

(k²+ q²)²e^−4qa≈ 16 E V0

e^−4qa.

In fact, only the exponential is the important part in this probability. For a barrier with a variable potential, the result becomes the famous WKB formula for tunneling, which is the product of the consecutive exponential factors for small barriers,

T ≈ exp −2 Z x2

x1

dx 2m

~²

1/2

[V (x) − E]^1/2

!

, (104)

where x1 and x2 are the points where V (x1) = V (x2) = E with V (x) ≥ E in the region x¹≤ x ≤ x².

Hitting a barrier

Considering the situation of a potential

V (x) = +V0 for x < 0 and V (x) = 0 for x ≥ 0.

For the solutions with 0 ≤ E = ~²k²/2m ≤ V⁰, one can find solutions of the form

φ(x) = C e^qx for x ≤ 0, (105)

φ(x) = e^{−i kx}− ARe^{i kx} for x ≥ 0, (106)

where ~²q²= 2m(V0−E) and we have already eliminated one overall constant by choosing the coefficient before e^{−i kx}, the incoming wave (from the right), to be unity. By equating the logarithmic derivatives from right and left, one finds the solution for AR, which can be written as

AR=q + i k

q − i k = e^{2i δ} with tan δ(E) = k q =

r E

V0− E. (107)

We see as expected from flux conservation that |A^R|²= 1 (there is no transmission) and can be expressed in terms of a phase shift δ(E). When V0→ ∞, the phase shift disappears and the wave function on the right-hand side becomes a sin(kx) with φ(0) = 0. For energies E > V0 one will have both reflection and transmission.

(27)

4 The Hilbert space of QM using Dirac notation

4.1 Space of states = ket-space (Hilbert space)

• Quantum mechanical states are denoted |ui and form a linear vector space over the complex numbers (C), the Hilbert space H = {|ui},

|u1i ∈ H

|u2i ∈ H c1, c2∈ C





−→ c¹|u¹i + c²|u²i ∈ H (108)

• Given a basis {|u1i, . . . , |u1i} for an N-dimensional Hilbert-space (N can be infinite!) consisting of a collection linearly independent kets, we can express every ket in this basis (completeness),

|ui ∈ H −→ |ui = XN n=1

cn|uni. (109)

4.2 Scalar product and the (dual) bra-space

For elements |ui, |vi ∈ H we can construct the complex number hu|vi ∈ C, for which

• hu|vi^∗= hv|ui,

• If |ui = c1|u1i + c2|u2i then hv|ui = c1hv|u1i + c2hv|u2i.

Note that this implies

hu|vi = hv|ui^∗ = c^∗₁hv|u¹i^∗+ c^∗₂hv|u²i^∗

= c^∗₁hu1|vi + c^∗2hu2|vi.

• hu|ui ≥ 0.

Beside the ket-space we can also introduce the dual bra-space, H^∗= {hu|}, which is anti-linear meaning that

|ui = c1|u1i + c2|u2i ←→ hu| = c1^∗hu1| + c^∗2hu2|. (110) The scalar product is constructed from a bra-vector and a ket-vector (”bra(c)ket”).

4.3 Orthonormal basis

• Basis {|u1i, |u2i, . . .} with hum|uni = δmn.

• If |ui =P

ncn|uⁿi, then cⁿ = huⁿ|ui (proof) and we can write

|ui =X

n

|uni hun|ui

| {z }

cn

=



 c1

c2

...



 . (111)

• Note that

hu|ui = 1 ←→X

n

|cⁿ|²= 1, (112)

hence the name probability amplitude for cn.

Quantum Mechanics

P.J. Mulders

Department of Physics and Astronomy, Faculty of Sciences, Vrije Universiteit Amsterdam

De Boelelaan 1081, 1081 HV Amsterdam, the Netherlands email: mulders@few.vu.nl

November 2011 (vs 3.01)

Contents

1 Basic concepts of quantum mechanics

1.1 Introduction

1.2 Operators and their expectation values

1.3 Commutators and uncertainty relations

1.4 Eigenstates

2 Time evolution

2.1 Stationary states

2.2 General time evolution

2.3 Spectrum and basic behavior of solutions

κ

(x)

~ k

(x)

~ V(x)

E

(x) oscillatory exponential

x x

φ

2.4 Boundary conditions and matching conditions

2.5 Three elementary properties of one-dimensional solutions

3 One-dimensional problems

3.1 The infinite square well

3.2 Harmonic oscillator

Appendix: Hermite polynomials

3.3 Free particle states

3.4 The delta potential

Bound state(s) of the delta potential

Scattering states of the delta potential

3.5 Bound states and scattering solutions for a square well potential

3.6 Reflection and transmission through a barrier

Hitting a barrier

4 The Hilbert space of QM using Dirac notation

4.1 Space of states = ket-space (Hilbert space)

4.2 Scalar product and the (dual) bra-space

4.3 Orthonormal basis