S2 Variation 2011

1

SECTION 2: VARIATIONAL METHOD

1. Introduction 2

2. The ground state 2

2.1 The Variational method 2

2.2 The ground state of the one-dimensional harmonic oscillator 3

3. Energy bound for excited states 6

3.1 The Variational method 6

3.2 First-excited state of the one-dimensional harmonic oscillator 7

4. Calculation of expectation value of kinetic energy 8

5. The ground-state of hydrogen 9

5.1 Application of the variational method 9

5.2 Alternative trial wave functions 10

6. Further Examples 13

6.1 One-dimensional example 13

6.2 Three-dimensional example 14

References

The original version of these notes was based on chapter 8 of the textbook by Mandl, with additional material

from section 7.1 of the book by Griffiths; specific references are given as footnotes to the text. The important

application of the variational method to the atom helium, discussed in both references, is considered later in

this course.

Original Version: 1995; Last Major Revision: July 2002

Revision Date: 16 October 2009

Printed 23 October 2009

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1. INTRODUCTION

Most problems encountered in quantum mechanics (as in classical physics) cannot be solved exactly, so

appropriate approximation techniques must be developed.

• The variational method can be used to determine an upper bound on the energy of the ground state of a

quantum system.

• Suitably modified, the method can also be used to produce a bound on the energy of certain excited states.

• It does not provide any reliable information about wave functions or quantities other than the energy,

even for the ground state.

• The variational method is a standard technique of approximation in molecular physics and chemistry.

In this course, we consider only the simplest version of the variational approach, due to Ritz (1908). The basic

theory for ground states is first developed and then applied to the ground state of the one-dimensional

harmonic oscillator. The theory is extended to cover excited states, with an application to the first-excited

state of the harmonic oscillator. The more realistic case of the ground state of the hydrogen atom is finally

considered in some detail.

2. THE GROUND STATE

2.1 The Variational method

We begin1 by considering a system described by a Hamiltonian H, with eigenstates u1, u2, u3 and so on with

corresponding eigenvalues E1, E2, E3 etc. We assume that the energies are ordered: E1 ≤ E2 ≤ E3 etc.

Let Ψ represent some arbitrary state of the system, not necessarily normalized. Since the set of eigenstates {un}

is complete, we can write

.1n n

n

c u∞

=

Ψ = ∑

where the cn are complex expansion coefficients. The expectation value of the energy in the state Ψ is

( ),

,

n n n nn n

n n n nn n

c u H u cHE

c u u c

∗′ ′ ′

∗′ ′ ′

∑Ψ ΨΨ = =

Ψ Ψ ∑

This expression can be simplified using the orthogonality of the eigenstates:

nn n nu u δ′ ′=

and the fact that the un are eigenstates of the Hamiltonian H:

n n n nn n n nu H u E u u E δ′ ′ ′= =

giving

( )2

2

n n n

n n

c EE

c

∑Ψ =

∑.

1 See section 8.1 of the textbook of Mandl.

3

But, for all n, En ≥ E1. Hence

( )2

1

12 ,n n

n n

c EE E

c

∑Ψ ≥ =

∑

i.e. the expectation value of H for any state Ψ is an upper bound to the ground-state energy E1.

The procedure to be followed in applying the method to a particular problem is as follows.

1) Using physical criteria choose a family of trial wave functions ( )1 2, , Nα α α αΨ = Ψ … that depend on a set

of N variational parameters {α} and incorporate the correct qualitative features of the actual ground

state.

2) Calculate the energy functional E(Ψα)

( ) ,H

Eα α

α

α α

Ψ ΨΨ =

Ψ Ψ

which will be a function of the variational parameters {α}.

3) Minimize the energy functional with respect to the variational parameters {α} by solving the set of N

coupled equations

1 2( , , )0N

i

E α α α

α

∂=

∂…

to give the set of N parameters {α0} that minimizes the energy functional E(α).

4) Calculate the minimum energy E(α0), which provides an upper bound on the true ground-state energy.

This process can be repeated with other trial wave functions to produce an even lower upper bound on the

energy.

2.2 The ground state of the one-dimensional harmonic oscillator2

To illustrate the method, we look at an example that can actually be solved exactly, so that the accuracy of

the result can be judged.

The Hamiltonian of the one-dimensional harmonic oscillator of mass m with oscillator parameter ω is

2 22 2

2

1,

2 2

dH m x x

m dxω= − + − ∞ < < +∞

ℏ

To construct a suitable set of trial wave functions, we use the following criteria:

• We can build into the trial wave function any symmetries we might expect the ground state wave function

to have. In this case, since H is an even function of x we will try an even function of x as the ground-state

wave function.

• In addition, for the wave function to be normalizable, it must remain finite everywhere. In particular we

require

0 as xψ → → ∞

In fact, for many realistic potentials the bound-state wave function decreases exponentially at large

distances.

2The notes in this section are based primarily on pages 1151-3 of Quantum Mechanics Vol II by C Cohen-Tennoudji, B Diu and F Laloë, published by Wiley, 1977. See also Example 1 in section 7.1 of the book by Griffith.

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So we will use as the set of trial wave functions

2( ) exp( ), 0x xαψ α α= − > (1)

where α is the single variational parameter. This is the simplest wave function that satisfies the criteria.

• Note that since the Gaussian function eq.(1) is so easy to handle mathematically, it is often used as a trial

function in situations where it bears little resemblance to the true ground-state wave function.

We must calculate the energy of this trial function from

( )H K V

Eα α α α

α

α α α α

Ψ Ψ Ψ + ΨΨ = =

Ψ Ψ Ψ Ψ

where K and V are the kinetic energy and potential energy operators, respectively.

Now

( )2exp 22

dx xα α

πψ ψ α

α

+∞

−∞= − =∫

and

2 22 2 2

22 2 2

x xdK dx e e

m dx m

α α

α α

πψ ψ α

α

+∞ − −

−∞= − =∫

ℏ ℏ

and

( )2 2 2 21 1 1exp 2

2 8 2V m dx x x mα α

πψ ψ ω α ω

α α

+∞

−∞= − =∫

Hence

221 1

( ) .2 8

HE m

m

α α

α α

ψ ψα α ω

ψ ψ α= = +

ℏ

To minimize this with respect to α we put

0E

α

∂=

∂

to find the value of α that minimizes the energy:

0

1.

2

mωα α= =

ℏ (2)

Substituting this back into the expression for E(α) we find that the minimum energy for this set of trial

functions is

0

1( )

2E α α ω= = ℏ (3)

This is in fact the exact energy, because the exact wave function is of the form eq.(1) with α being given by

eq.(2).

Alternative wave function

To provide a better test of the method, we will try a different trial wave function3. Instead of the function

eq.(1), we will use

3This example forms the basis of problem 7.2 of the textbook of Griffith. See also page 1153 of the book by Cohen-Tennoudji et al.

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2

1for 0a a

x aψ = >

+

where the variational parameter is now denoted a. This trial form also satisfies the first two criteria used above,

but it does not decrease exponentially for large x.

With this trial wave function

( )22 2a a

dx

a ax a

πψ ψ

+∞

−∞= =

+∫

and4

( )

2 2 2

2 2 2 2

22 2

22

1 1

2 2 4

1 1

2 2 2

a a

a a

dK dx

m x a dx x a m a a

xV m dx m

ax a

πψ ψ

πψ ψ ω ω

+∞

−∞

+∞

−∞

= − = + +

= =+

∫

∫

ℏ ℏ

giving

221

( ) .4 2

E a m ama

ω= +ℏ

This has a minimum for

0

1

2

ma a

ω= =

ℏ

with minimum energy

0

1( )

2E a ω= ℏ .

This is a factor of 2 larger than the energy obtained before.

Note:

• In those few cases in which the exact ground-state energy E1(exact) is known, a good indicator of the

accuracy of the approximation is provided by calculating the ratio

0 1

2 1

( ) (exact)

(exact) (exact)

E a E

E E

−−

where E2(exact) is the exact energy of the first excited state. In this case the ratio is 20%; these energies

are illustrated in Fig.1.

• As stated earlier, although the energy may be calculated with reasonable accuracy, this does not

guarantee that the approximate wave function is accurate.

Fig.2 compares the exact and approximate wave functions used above, calculated for the particular case

mω/ℏ = 2.

4 Useful integrals:( ) ( )

2

4 35/2 5/22 2

3,

16 8

x dx dx

a ax a x a

π π∞ ∞

−∞ −∞= =

+ +∫ ∫

6

Figure 1: Comparison of the variational energy with the exact energies of the ground and first-excited states

of the Harmonic Oscillator.

Figure 2: Exact and approximate wave functions for the ground state of the Harmonic Oscillator

3. ENERGY BOUND FOR EXCITED STATES

We now discuss how the method can be applied to give a bound on the energy of some excited states of

quantum systems5.

3.1 The Variational method

The exact wave function of any excited state of a quantum system is orthogonal to the exact ground-state

wave function. So to adapt the variational method to the calculation of excited-state energies, we choose a trial

wave function that is orthogonal to the exact ground-state wave function, assuming this to be non-degenerate

(the generalization is straight-forward), i.e.

5See section 8.3 of the textbook by Mandl; also problem 7.4 of Griffith.

7

1 0.uαΨ =

As before, we expand the trial wave function in the complete set of eigenstates:

.1n n

n

c uα

∞

=

Ψ = ∑

Substitution of this expansion into the scalar product gives

1 1 1 11 1

,n n n n

n n

u c u u c cα δ∞ ∞

= =

Ψ = = =∑ ∑

The requirement that Ψα be orthogonal to u1 therefore leads to c1 = 0 and the expansion of the wave function

becomes

.2n n

n

c uα

∞

=

Ψ = ∑

Hence

( )2

2

222

.n n n

n n

c EHE E

c

≥

≥

∑Ψ ΨΨ = = ≥

Ψ Ψ ∑

where we have used the fact that for all n ≥ 2, En ≥ E2.

However:

• The exact ground-state wave function is not known, and use of a trial function orthogonal to the ap-

proximate wave function of the ground state does not necessarily lead to an upper bound on E2.

• To force orthogonality to the true ground-state wave function, we can choose a trial wave function Ψα that

has different symmetries to the ground state (for example, different parity, different angular momentum

etc.).

• This will provide a bound on the energy of the lowest state with the chosen symmetries.

3.2 First-excited state of the one-dimensional harmonic oscillator

We require6 a trial wave function which is orthogonal to the ground state, which has even parity. We therefore

choose a trial function which has odd parity (we shall thus get an estimate of the energy of the lowest state

of odd-parity, which may or may not be the first-excited state).

The simplest wave function that satisfies the specified criteria is

( )2( ) exp , 0x x xαψ α α= − >

where α is the single variational parameter.

The normalization integral7 is

2 2exp( 2 )dx x xα αψ ψ α+∞

−∞= −∫

The expectation value of the kinetic energy is

6See pages 1152-3 of the book by Cohen-Tannoudji et al. 7We do not need to compute this integral, since it cancels out in calculating the energy.

8

( )

2 22 2

2

22 2

exp( ) exp( )2

3 exp 22

dK dx x x x x

m dx

dx x xm

α αψ ψ α α

α α

+∞

−∞

+∞

−∞

= − − −

= −

∫

∫

ℏ

ℏ

and similarly for the potential energy

( )

( )

2 4 2

2 2 2

1exp 2

21 3

exp 2 .2 4

V m dx x x

m dx x x

α αψ ψ ω α

ω αα

+∞

−∞

+∞

−∞

= −

= −

∫

∫

Hence the energy of the state with this trial wave function is

223 3 1

( )2 8

HE m

m

α α

α α

ψ ψα α ω

ψ ψ α= = +

ℏ

To minimize this with respect to α we put 0E

α

∂=

∂ to find

0

1

2

mωα α= =

ℏ

We substitute this value of the parameter back into the expression for the energy to find that the minimum

energy for this set of trial functions is

0

3( )

2E α ω= ℏ

Again, this is the exact result for the energy of the first excited state, which has the symmetry built into the

trial function.

4. CALCULATION OF EXPECTATION VALUE OF KINETIC ENERGY

To solve problems using the variational method, we need to calculate the integral

( )2 22 2 32 2 ( ) ( )m m r r d rα αψ ψ ψ ψ∗− ∇ = − ∇∫ℏ ℏ

� � (4)

In a one-dimensional problem this is straight-forward, but in realistic three dimensional cases, both ψα and the

operator ∇2 are functions of three variables, and the calculation of the integral can be quite lengthy (mainly

because of the form of the operator ∇2). The following simplifies the calculation considerably in such cases8.

We consider the following identity (this is actually an operator identity, integrated over all space):

( ) ( )3 2 3 3

all space

( ) ( ) ( ) ( ) ( ) ( )r r d r r r d r r r d rψ ψ ψ ψ ψ ψ∗ ∗ ∗∇ ⋅ ∇ = ∇ + ∇ ⋅ ∇∫ ∫ ∫� � � ��

(5)

Computation of the kinetic energy integral involves evaluation of the first integral on the right-hand side of

this equation. Consider first the integral on the left-hand side of the equation; from Gauss' theorem,

( ) ( )3

all space infinite surface

( ) ( ) ( ) ( ) .r r d r r r dAψ ψ ψ ψ∗ ∗∇ ⋅ ∇ = ∇ ⋅∫ ∫��

We assume that the wave function satisfies the following asymptotic condition:

8See pages 188-189 of the textbook of Mandl.

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( ) ( ) 0 asr r rψ ψ∗∇ → → ∞��

so that the integrand vanishes on the surface at infinity, and

( )infinite surface

( ) ( ) 0r r dAψ ψ∗∇ ⋅ =∫��

In practice, all wave functions describing real situations satisfy this equation. Hence the left-hand side of the

eq.(5) is identically zero and we get

2 3 3( ) ( ) ( ) ( )r r d r r r d rψ ψ ψ ψ∗ ∗∇ = − ∇ ⋅ ∇∫ ∫� ��

In other words

2

2 22 3

2

all space

( )2

m r d rm

α αψ ψ ψ− ∇ = + ∇∫ℏ� �ℏ

(6)

This integral is usually much easier to calculate than that in eq.(4), since it involves only first derivatives.

5. THE GROUND-STATE OF HYDROGEN

For a more realistic application of the method, we consider the ground state of the hydrogen atom9.

5.1 Application of the variational method

The Hamiltonian of the hydrogen atom is, with the usual notation,

2 22

02 4

eH

m rπε

−= − ∇ +

ℏ

The trial wave function must satisfy these criteria:

• The ground state of a particle in a central potential is expected to have orbital angular momentum zero,

so that its wave function has no angular dependence, ( ) ( )r rψΨ =�

, for 0 ≤ r < ∞.

• For a state of zero angular momentum, the radial wave function is non-zero at the origin10,

( 0) 0rψ = ≠ .

• 2( )rψ must be finite for all r, so that we require

0 for rψ → → ∞

We therefore choose the set of trial functions, with a single variational parameter α,

( ) exp( ), 0r rαψ α α= − >

The normalization integral is

22 2

0

3

sin 4 exp( 2 )d d r dr r r drα αψ ψ φ θ θ π α

π

α

∞= Ψ = −

=

∫ ∫ ∫ ∫

9 This discussion is based on section 8.2.1 of the textbook of Mandl. 10 For potentials V(r) less divergent than 1/r2, the radial wave function obeys R(r) → Arl as r → 0 where A is some constant.

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where the volume element in spherical coordinates is 2 sindV r d d drθ θ φ= and we have used the fact that the

wave function does not depend on the variables θ and φ.

For the potential energy:

2 2 22

0 00 0 0

2

20

4 exp( 2 ) 4 exp( 2 )4 4 4

4

e e er r dr r r dr

r r

e

α αψ ψ π α π απε πε πε

π

πε α

∞ ∞ − − − = − = −

= −

∫ ∫

and for the kinetic energy, using the technique described above:

22 2 222 3 2

0

2 2 22

0

( ) 4 exp( )2 2 2

4 exp( 2 )2 2

r d r r r drm m m r

r r drm m

α αψ ψ ψ π α

α ππ α

α

∞

∞

∂− ∇ = ∇ = −

∂

= − =

∫ ∫

∫

ℏ ℏ ℏ

ℏ ℏ

Hence the total energy is

2 2 3 2 22

20 0

( )2 4 2 4

e eE

m m

π π αα α α

α πε α π πε

= − = − ℏ ℏ

To minimize this with respect to α we put 0E

α

∂=

∂ to find

2

0 20 0

1

4

e m

aα

πε= =

ℏ

where a0 is the combination of fundamental constants called the Bohr radius. Hence the minimum energy for

this set of trial functions is

2

00 0

1( )

2 4

eE

aα

πε= −

This is the exact result since the exact wave function has the form chosen for the trial function.

5.2 Alternative trial wave functions11

To provide a more stringent test of the variational method, we repeat the calculation for the ground state of

hydrogen using trial functions with different functional forms.

We follow the usual procedure for solving the Schrödinger equation in spherical coordinates.

• We change the radial variable r to the dimensionless quantity ρ = r/a0 where a0 is the Bohr radius defined

above.

• We separate off the angular part of the wave function ( ),lmY θ φ from the radial part Rnl(r).

• The radial part of the wave function is written ( )/nlR u ρ ρ= .

The wave function is therefore written:

( ) ( )3/20

( ), , ,lm

ua Y

ρρ θ φ θ φ

ρ

−Ψ =

11This section is based primarily on pages 767-9 of Quantum Mechanics Vol II, by A Messiah, published by North Holland, 1965.

11

where the factor 3/20a− has been included to simplify the normalization.

As discussed above, for the ground state the orbital angular momentum l = 0 with angular wave function

00( , ) constantY θ φ = , and suitable trial functions for the radial part of the wave function are:

(i) [ ]( ) expu bρ ρ ρ= −

(ii) 2 2

( )ub

ρρ

ρ=

+

(iii) [ ]2( ) expu bρ ρ ρ= −

Note that the variational parameter is now denoted b. Each of these trial functions satisfies some of the criteria

listed above.

• The choice (i) is just the trial function already used, and it satisfies all the criteria.

• The choice (ii) does not decrease exponentially at infinity. It has the correct behaviour at the origin but

a very different behaviour asymptotically.

• The choice (iii) gives a radial wave function Rnl that is not non-zero at the origin, but correctly falls off

exponentially at infinity.

All three functions have zero radial nodes; they should therefore resemble the ground-state wave function

rather than that of an excited state, and the resulting variational energy should be closer to the ground-state

energy than to any of the excited states.

The table below shows the results of variational calculations for these three trial wave functions.

(i) (ii) (iii)

N2 1/4b3 π/4b 3/4b5

E(b) in Ry b(b − 2) 2

8

2

b

b

π

π

−

13( 1)b b−

bmin 1 π/4 3/2

E(bmin) in Ry −1 −0.81 −0.75

• 2N is the normalization integral for the wave function; the wave function must be divided by N.

• The quantity Ry is a unit of energy called the Rydberg. It is equal to the magnitude of the exact

ground-state energy of hydrogen; i.e.

2

0 0

1 Ry 13.6 eV.8

e

aπε= =

To assess the accuracy of the calculation, we consider the quantities in the following table.

(i) (ii) (iii)

min 1

2 1

( )E b E

E E

−−

0 0.25 0.33

2

1 min1 ψ ψ− 0 0.21 0.05

min minrψ ψ 1.5 a0 ∞ 1.66 a0

12

• E1 and E2 are the exact energies of the ground and first-excited states of hydrogen. The ratio min 1

2 1

( )E b E

E E

−−

is a measure of the accuracy of the variational approximation for the energy; this is illustrated in Fig.3.

Figure 3: Exact and variational energies for hydrogen atom (in Rydbergs)

• The quantities 2

1 minψ ψ and min minrψ ψ are numerical measures of the accuracy of the wave

function. Fig.4 compares the approximate radial wave functions ( )u ρ with the exact wave function.

Figure 4: Exact and approximate wave functions for the ground state of the hydrogen atom

It is seen that choice (ii) provides a somewhat better approximation to the ground-state energy than choice

(iii), although the corresponding wave function is certainly not very accurate.

• As noted above, the function u2 has the proper behaviour at the origin but asymptotically is very different

from the exact solution; nonetheless, the energy predicted with this function is still very satisfactory.

• The function u3, on the other hand, is very inaccurate near the origin but has the correct exponential

behaviour asymptotically; however, it gives a worse result for the energy.

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6. FURTHER EXAMPLES

6.1 One-dimensional example

We consider an example of a particle confined to a one-dimensional space12, the anharmonic oscillator. The

particle, of mass m, is bound in the ground state of the potential

4( ) , , 0V x Ax x A= − ∞ < < +∞ >

We take as a trial function for the ground state:

2exp( ), 0,xαψ α α= − >

where α is the single variational parameter. This trial wave function has the following features:

• It has no nodes apart from that at the origin (the number of nodes in the wave function tends to increase

with excitation energy);

• It has even parity;

• It is normalizable, since ψ(x) → 0 as x → ±∞.

The relevant integrals are:

2exp( 2 )2

dx xα α

πψ ψ α

α

∞

−∞= − =∫

and

1/24 2

5

3 2exp( 2 )

32

AV A dx x xα α

πψ ψ α

α

∞

−∞

= − = ∫

and also

2 1/22 2 2 2 22 2 2 2

2exp( ) 4 exp( 2 )

2 2 2 2 2

d ddx x dx x x

m dx m dx m mα α

παψ ψ α α α

∞ ∞

−∞ −∞

− = − = − = ∫ ∫ℏ ℏ ℏ ℏ

The energy of the state ψα is therefore

2 2

2 2

2

( )2 3 1

( )2 16

dV x

m dx AE

m

α α

α α

ψ ψ

α αψ ψ α

− +

= = +

ℏ

ℏ

The value of α which minimizes this is given by

( )30 2

3

4

Amα =

ℏ

with minimum energy

2/ 32 2 23

0 0 2 20

1/ 3 2/ 34 4 21/3

4 2

3 1 4 3 3( )

2 16 3 2 4 16

30.6814 .

4

A Am AE

m Am m

AA

m m

α αα

= + = + = =

ℏ ℏ ℏ

ℏ

ℏ ℏ

12 This example, which is based on problem 8.1 of the textbook by Mandl, will not be discussed in the lectures. It should be regarded as a worked example to further illustrate the technique.

14

This problem can actually be solved exactly13; the exact ground-state energy is

2/321/3

gs

1.060

2E A

m

= ℏ

showing that the approximation is accurate to about 2% in this case.

6.2 Three-dimensional example

We consider now a further example where the particle moves in three-dimensional space14. The particle, of

mass m, is bound in the ground state of the potential

0( ) exp( / ), 0V r V r a a= − − >

where the strength V0 of the potential is related to its range a:

2

0 2

4

3V

ma=

ℏ.

We take as a trial function for the ground state:

( ) exp( /2 ), 0,r r aβψ β α= − >

where β > 0 is the variational parameter. The factor 2a is included to make β dimensionless and to simplify

the algebra later. This trial wave function has the following features:

• It has angular momentum zero.

• It is normalizable, since ψ(r) → 0 as r → ∞.

• It is non-zero but finite as r → 0.

The normalization integral is:

322 2

0sin 4 exp( / ) 8

ad d r dr r r a drβ β βψ ψ φ θ θ ψ π β π

β

∞ = = − = ∫ ∫ ∫ ∫

where the volume element in spherical coordinates is 2 sindV r d d drθ θ φ= and we have used the fact that the

wave function does not depend on the variables θ and φ. The expectation value of the potential energy

operator is

3

20 0

04 exp( / )exp( / ) 8

1

aV V dr r r a r a Vβ βψ ψ π β π

β

∞ = − − − = − + ∫

and that of the kinetic energy operator is

22 2 22 2

04 exp( /2 ) 2

2 2 2

d ar dr r a

m m dr mβ βψ ψ π β π

β

∞− ∇ = − =∫ℏ ℏ ℏ

The energy of the state ψβ is therefore

22

322

0

( )2

( )2 2 1

V rm

E Vm a

β β

β β

ψ ψβ β

ββψ ψ

− ∇ + = = − +

ℏ

ℏ

13 The result is quoted on page 298 of Quantum Physics by S Gasiorowicz, published by Wiley, 1974 14 This example is based on problem 8.3 of the textbook of Mandl.

15

From 0E

β

∂=

∂ we find either 0β = or

( )4 4

1

21

β

β=

+.

The first solution 0β = is clearly not physical since we require β > 0. The second equation has four solutions:

two are complex and not physical, one is β = 1 (found by inspection) and the fourth is β ≈ 0.087 (found by

using the binomial approximation). The last solution actually gives a maximum, so the solution that

minimizes the energy is β0 = 1 and the minimum energy is

2 2

0 0 02 2

1 1( )

8 8 24 32E V V

ma maβ = − = − = −

ℏ ℏ

S2 Variation 2011

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