1 2018-03-01

An Introduction to the Virial1 Theorem in Quantum Mechanics

Spiros Konstantogiannis

Physicist, M.Sc.

spiroskonstantogiannis@gmail.com

1 March 2018

1 The word virial comes from the Latin vis, meaning “force”, “energy”, “strength”, it is not the name of

the scientist who formulated the theorem.

http://wordinfo.info/unit/4285/ip:4/il:V

https://en.wikipedia.org/wiki/Virial_theorem

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Copyright © 2018 by Spiros Konstantogiannis.

All rights reserved.

Only non-commercial use of content from the present document is permitted, provided

that a link to the respective blog post is included.

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Contents

An Introduction to the Virial Theorem in Quantum Mechanics .......................... 1

Contents ............................................................................................................ 3

Introduction ....................................................................................................... 4

1. The virial theorem in one dimension .............................................................. 5

Examples ..........................................................................................................13

2. The virial theorem in three dimensions..........................................................17

3. References ....................................................................................................23

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Introduction

We first calculate three useful commutators and we use them to derive, by means of

the Ehrenfest theorem, the expression of the Newton’s second law in quantum

mechanics.

Next, we use again the previous commutators, and the Ehrenfest theorem, to prove the

virial theorem in one dimension.

Then, we apply the virial theorem to the harmonic oscillator, to a general attractive

potential, to the attractive delta potential, and to the one dimensional hydrogen atom,

for which we show that its bound energies are negative and the respective wave

functions vanish at zero.

In the last section, in a way similar to that used in one dimension, we prove the virial

theorem in three dimensions.

Keywords: virial theorem, Ehrenfest theorem, one dimensional hydrogen atom,

bound eigenstates

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1. The virial theorem in one dimension

Firs, we’ll calculate the commutators ( )ˆ ˆ,p V x , ˆˆ ,p H , and ˆˆ,x H .

A commutator does not depend on the representation we use to calculate it, and

thus we can choose the representation that makes the calculation easier.

We choose the position space (position representation), where x x= and ˆ dp idx

= − h .

1. ( )ˆ ˆ,p V x

Let ( )xψ be an arbitrary wave function in position space.

The action of the previous commutator on ( )xψ yields

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ), ,d d d di V x x i V x x i V x x V x xdx dx dx dx

ψ ψ ψ ψ − = − = − − = h h h

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )i V x x V x x V x x i V x xψ ψ ψ ψ′ ′ ′ ′= − + − = −h h

That is

( ) ( ) ( ) ( ),di V x x i V x xdx

ψ ψ ′− = − h h

Since the wave function ( )xψ is arbitrary,

( ) ( ),di V x i V xdx

′− = − h h

Since the commutator does not depend on the representation, we end up to

( ) ( )ˆ ˆ ˆ,p V x i V x′= − h (1)

Notes

i) For ( )ˆ ˆV x ax= , (1) becomes

[ ]ˆ ˆ,p ax i a= − h

For 1a = , we obtain

[ ]ˆ ˆ,p x i= − h

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or

[ ]ˆ ˆ,x p i= h ,

i.e. we end up to the well-known position-momentum commutator.

ii) For ( )ˆV x a= , (1) becomes

( )ˆ ˆ, 0p V x = ,

and since

{

2

Kineticenergyoperator

ˆˆˆ ˆ, , 02pp T pm

= =

,

we have

ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ, , , , 0p H p T V p T p V = + = + =

Thus

ˆ ˆ, 0H p =

Using the Ehrenfest theorem, the time evolution of the momentum expectation value

in the constant potential ( )ˆV x a= is, since the momentum operator does not depend

on time,

ˆ0t

d pdt

=

That is

ˆ constanttp =

The momentum expectation value is then constant in time, as it also happens in

classical mechanics for a particle moving in a spatially constant potential.

Note

The Ehrenfest theorem states that the time evolution of the expectation value of a

Hermitian operator A is given by

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ˆ ˆˆˆ ,t

tt

d A i AH Adt t

∂ = + ∂h

where H is the Hamiltonian of the system.

The subscript t indicates that the expectation values are obtained at time t .

If the operator A does not explicitly depend on time, i.e. if ˆ

0At

∂=

∂, then the

Ehrenfest theorem is written as

ˆˆˆ ,t

t

d A i H Adt

= h

2. ˆˆ ,p H

We have

( ) ( ) ( )2

0

ˆˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , , , , ,2pp H p T V p T p V p p V x p V x i V xm

′= + = + = + = = − h

14243

where, in the last equality, we used (1).

Thus

( )ˆˆ ˆ,p H i V x ′= − h (2)

Using again the Ehrenfest theorem, and the commutator (2), the time evolution of the

momentum expectation value in an arbitrary potential ( )ˆV x is, since the momentum

operator does not depend on time,

}

( ) ( )ˆ ˆˆ ˆ, ,

ˆˆ ˆ

H p p H

tt t

d p i i V x V xdt

=−

′ ′= = −hh

That is

( )ˆ

ˆtt

d pV x

dt′= −

If ( ) ( ) ( )ˆ

ˆ ˆˆ

dV xF x V x

dx′= − = − is the force operator, the previous equation becomes

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( )ˆ

ˆtt

d pF x

dt= (3)

We thus end up to the Newton’s second law, but with the difference that, in

quantum mechanics, it holds for the expectation values of the force and

momentum.

3. ˆˆ,x H

We have

( ) [ ]2

2

0

ˆ 1 1ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , , , , ,2 2 2px H x T V x T x V x x x p x ppm m m

= + = + = = = = 14243

[ ] [ ]( ) ( )1 1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , 22 2 2

ix p p p x p i p pi i p pm m m m

= + = + = =h

h h h

That is

ˆˆ ˆ, ix H pm

= h (4)

Using again the Ehrenfest theorem, and the commutator (4), the time evolution of the

position expectation value in an arbitrary potential ( )ˆV x is, since the position

operator does not depend on time,

}ˆ ˆˆ ˆ, ,

ˆ ˆˆ

H x x H

t t

t

d x pi i pdt m m

=−

= − =h

h

That is

ˆˆ t

t

d xp m

dt= (5)

or, since ˆ

ˆtt

d xv

dt= ,

ˆ ˆt tp m v=

Differentiating (5) with respect to time and substituting (3) yields

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( )2

2

ˆˆ t

t

d xF x m

dt= (6)

The relation (6) is an equivalent expression of the Newton’s second law, for the

expectation values of the force and position.

Now, we proceed to the proof of the virial theorem in one dimension.

We’ll use again the Ehrenfest theorem to write the time evolution of the expectation

value of the operator ˆ ˆpx .

Since the operator ˆ ˆpx does not depend on time,

ˆ ˆ ˆ ˆ ˆ,t

t

d px i H pxdt

= h (7)

Note

The operator ˆ ˆpx is not Hermitian, since ( )† † †ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆpx x p xp px= = ≠ .

However, the operator ˆ ˆ ˆ ˆxp px+ is Hermitian, as it is easily seen.

Then, applying the Ehrenfest theorem for the operator ˆ ˆ ˆ ˆxp px+ yields

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ,t

t

d xp px i H xp pxdt+

= + h,

since the operator ˆ ˆ ˆ ˆxp px+ does not depend on time.

But

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆt t t

d xp px d xp d pxdt dt dt+

= +

and

ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆˆ ˆ ˆ, , , , ,t t t t

H xp px H xp H px H xp H px + = + = +

Thus

( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆˆ ˆ ˆ ˆˆ ˆ ˆ, , , ,t t

t t t t

d xp d px i i iH xp H px H xp H pxdt dt

+ = + = + h h h

That is

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ, ,t t

t t

d xp d px i iH xp H pxdt dt

+ = + h h,

from which we see that

ˆ ˆ ˆ ˆˆ,t

t

d xp i H xpdt

= h

and

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ˆ ˆ ˆ ˆ ˆ,t

t

d px i H pxdt

= h

The commutator ˆ ˆ ˆ,H px is written as

ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , , ,H px p H x H p x p x H p H x = + = − −

Substituting the commutators ˆˆ ,p H and ˆˆ,x H from (2) and (4), respectively,

yields

( )( ) ( )2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, i iH px p p i V x x p i V x xm m

′ ′= − − − = − + h h

h h

That is

( )2ˆ ˆ ˆ ˆ ˆ ˆ, iH px p i V x xm

′= − + h

h

The commutator ( )ˆ ˆ,V x x′ vanishes, since the position operator commutes with

itself, and thus we obtain

( ) ( )ˆ ˆ ˆ ˆV x x xV x′ ′=

Then the commutator ˆ ˆ ˆ,H px is written as

( )2ˆ ˆ ˆ ˆ ˆ ˆ, iH px p i xV xm

′= − + h

h

Substituting into (7) yields

( ) ( )2 2ˆ ˆ 1ˆ ˆ ˆ ˆ ˆ ˆttt

t

d px i i p i xV x p xV xdt m m

′ ′= − + = −h

hh

That is

( )2ˆ ˆ 1 ˆ ˆ ˆttt

d pxp xV x

dt m′= − (8)

But

22ˆ 1ˆ ˆ

2 2 ttt

pT pm m

= =

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Thus

2 ˆˆ 2t t

p m T=

Substituting into (8) yields

( )ˆ ˆ ˆ ˆ ˆ2t

tt

d pxT xV x

dt′= − (9)

Now, we assume that the previous expectation values are taken in an arbitrary

bound energy eigenstate.

Then

ˆ ˆ0t

d pxdt

= (10)

Proof

Since the state – let us denote it by ψ – is bound, the probability of finding the

particle at infinity is zero and thus the expectation value of the operator ˆ ˆpx is, at any

time, finite.

Then, the expectation value of the previous operator, at time t , in position space

(position representation), is written as

( ) ( ) ( ) ( )( )* *ˆ ˆ , , , ,t

dpx dx x t i x x t i dx x t x x tdx

ψ ψ ψ ψ∞ ∞

−∞ −∞

′= − = − ∫ ∫h h

That is

( ) ( )( )*ˆ ˆ , ,tpx i dx x t x x tψ ψ∞

−∞

′= − ∫h

where the prime denotes differentiation with respect to the position x .

Thus

( ) ( )( )*ˆ ˆ, ,td px di dx x t x x t

dt dtψ ψ

∞

−∞

′= − ∫h (11)

Note

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The expectation value of an operator does not depend on the representation we use

to calculate it, and thus we can choose the representation that makes the calculation

easier.

Since the state ψ is an energy eigenstate, its time evolution ( )tψ is

( ) exp iEttψ ψ = − h

where E is the energy of the state.

Then, the wave function ( ),x tψ describing the state ( )tψ in position space is

( ) ( ), exp iEtx t xψ ψ = − h

where ( )xψ is the wave function at time 0t = .

Using the previous relation, the integrand of (11) is written as

( ) ( )( ) ( ) ( )*

* , , exp expiEt iEtx t x x t x x xψ ψ ψ ψ′ ′ = − − = h h

( ) ( )( ) ( ) ( )( )* *exp expiEt iEtx x x x x xψ ψ ψ ψ ′ ′= − = h h

That is

( ) ( )( ) ( ) ( )( )* *, ,x t x x t x x xψ ψ ψ ψ′ ′=

We see that the integrand does not depend on time, thus the integral in the right-hand

side of (11) is also time independent, and thus

( ) ( )( )* , , 0d dx x t x x tdt

ψ ψ∞

−∞

′ =∫

and then from (11) we derive (10), i.e.

ˆ ˆ0t

d pxdt

=

By means of (10), (9) becomes

( )ˆ ˆ ˆ2tt

T xV x′= (12)

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The relation (12) is the virial theorem in one dimension.

It is obvious that the relation (12), i.e. the virial theorem, also holds for bound

states which are not energy eigenstates, provided that for those non-energy-

eigenstates the relation (10) still holds.

Examples

Ι. ( ) 2 212

V x m xω= (harmonic oscillator)

Then

( ) ( ) ( )2 2 2 2V x m x xV x m x V xω ω′ ′= ⇒ = =

That is

( ) ( )2xV x V x′ =

Thus, the virial theorem gives

T V=

where, for simplicity, we omit the subscript t and the operator symbol (“hat”).

ΙΙ. ( ) 2nV x kx= , 0k > , 1, 2,...n =

Then

( ) ( ) ( )2 1 22 2 2n nV x nkx xV x nkx nV x−′ ′= ⇒ = =

That is

( ) ( )2xV x nV x′ =

Then, the virial theorem gives

T n V=

Also

( )1E T V n V V n V= + = + = +

That is

( )1E n V= +

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If the system is in an energy eigenstate, then E E= , and thus

( )1E n V= +

ΙΙΙ. ( ) ( ) , 0V x xλδ λ= − > (attractive delta potential)

In order to calculate the derivative of the potential, i.e. the delta function derivative,

we use the relation ( ) 0x xδ = .

Differentiating both sides of the previous equation with respect to x , we obtain

( )( ) ( ) ( ) ( ) ( )0 0x x x x x x x xδ δ δ δ δ′ ′ ′= ⇒ + = ⇒ = −

Thus

( ) ( ) ( ) ( )xV x x x x V xλ δ λδ′ ′= − = = −

That is

( ) ( )xV x V x′ = −

Thus, the virial theorem gives

2 0T V+ =

We remind that the attractive delta potential has only one bound state, which is also

its ground state.

IV. One dimensional hydrogen atom

In this case, the potential is

( ) , 0V xxλ λ= − >

For 0x > , ( )V xxλ

= − , and thus

( ) ( ) ( )2V x xV x V xx xλ λ′ ′= ⇒ = = −

That is

( ) ( )xV x V x′ = −

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For 0x < , ( )V xxλ

= , and thus

( ) ( ) ( )2V x xV x V xx xλ λ′ ′= − ⇒ = − = −

That is

( ) ( )xV x V x′ = −

Thus

( ) ( )xV x V x′ = − , 0x ≠

Then, the virial theorem gives

2 0T V+ =

or

2V

T = −

If the system is in an energy eigenstate, then E E= , and thus

2 2V V

E T V V= + = − + =

Then

2V E= (13)

i. By means of (13), i.e. using the virial theorem, we’ll show that the bound energies

of the one dimensional hydrogen atom are negative.

Proof

In position space, the expectation value of the potential in an arbitrary energy

eigenstate ψ is written as,

( ) ( ){

( ) ( ) ( ) ( ) 22*

Scalarfunction

xV dx x V x x dx x V x dx

xψ

ψ ψ ψ λ∞ ∞ ∞

−∞ −∞ −∞

= = = −∫ ∫ ∫

That is

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( ) 2x

V dxx

ψλ

∞

−∞

= − ∫ (14)

The wave function ( )xψ , as an eigenfunction, is linearly independent, thus it is not

identically zero, i.e. it takes and non-zero values, and then there exists 0x ∈¡ such

that ( )0 0xψ ≠ . Then, since ( )xψ is continuous, it takes non-zero values in a

neighborhood of 0x and there ( ) 2

0xx

ψ> , thus

( ) 2

0x

dxx

ψ∞

−∞

>∫ , and then, from

(14), 0V < , and then from (13) (virial theorem), 0E < , i.e. the bound energies of

the one dimensional hydrogen atom are negative.

ii. We’ll use again the expression of the virial theorem, i.e. the relation (13), to show

that the wave function of an arbitrary bound state of the one dimensional hydrogen

atom vanishes at zero.

Proof

The wave function of an arbitrary bound state is linear combination of bound energy

eigenfunctions, thus it suffices to show that an arbitrary bound energy eigenfunction

( )xψ vanishes at zero.

Let us assume that ( )xψ does not vanish at zero. Then, since ( )xψ is continuous, it

takes non-zero values in a neighborhood of 0 0x = , thus there exists 0r > such that

( ) 2x rψ ≥ and then

( ) 2x rx x

ψ≥ , in the neighborhood of 0 0x = .

Thus, if the interval [ ],ε ε− belongs – i.e. it is a subset – of the previous

neighborhood,

( )

{

( )2

0

Evenfunction

1 12 2 ln ln 0x

dx r dx r dx rx x x

ε ε ε

ε ε

ψε

− −

≥ = = − = ∞∫ ∫ ∫

since ln ε is finite and ln 0 = −∞ .

Then

( ) 2x

dxx

ε

ε

ψ

−

= ∞∫

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Since the integrand ( ) 2xx

ψ is non-negative,

( ) ( )2 2x x

dx dxx x

ε

ε

ψ ψ∞

−∞ −

≥∫ ∫

Thus

( ) 2x

dxx

ψ∞

−∞

= ∞∫

Then, from (14), V = −∞ , and from (13) (virial theorem), E = −∞ , i.e. the energy of

the eigenstate is −∞ . However, this is impossible, because the minimum energy of the

one dimensional hydrogen atom is finite (for the proof, see reference 4).

Therefore, the wave function ( )xψ vanishes at zero, and since it is arbitrary, every

bound energy eigenfunction vanishes at zero, and then, every bound-state wave

function – not necessarily energy eigenfunction – of the one dimensional hydrogen

atom vanishes at zero.

As a consequence, the ground-state wave function of the one dimensional hydrogen

atom has a zero, at zero.

2. The virial theorem in three dimensions

Similarly to the one dimensional case, we’ll calculate the commutators ( )ˆ ˆ,p V r r r ,

ˆ ˆ,p H r , and ˆ ˆ,r H

r , and then we’ll use the Ehrenfest theorem.

In three dimensions, the following commutation relations hold

ˆ ˆ, 0j kr r =

ˆ ˆ, 0j kp p =

ˆ ˆ,j k jkr p i δ = h

where , 1,2,3j k = , and the index 1 refers to the coordinate x , the index 2 refers to the

coordinate y , and the index 3 refers to the coordinate z .

The symbol i in the right-hand side of the last relation is the imaginary unit.

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The symbol jkδ is the Kronecker delta, i.e.

1, 0, jk

j kj k

δ=

= ≠

As in the one dimensional case, we choose the position space (position representation)

to calculate the previous commutators.

Let 1 2 3, ,e e er r r be the unit vectors of the Cartesian axes , ,x y z , respectively.

The action of the commutator ( )ˆ ˆ,p V r r r on an arbitrary wave function ( )rψ r yields

( ) ( ){

( ) ( ) ( ) ( )( ) ( ) is a summation

index, taking thevalues 1,2,3

ˆ ˆ ˆ ˆ ˆ, ,j j j j j j

j

p V r r p e V r r p e V r V r p e rψ ψ ψ

= = − =

r r r r r r r r r r r

( ) ( ) ( ) ( )( ) { ( ) ( )( ) ( ) ( )

1 23

ˆ

, ,

ˆ ˆj

j

j j j jj jp i

rr x r yr z

p V r r V r p r e i V r r V r i r er r

ψ ψ ψ ψ∂

=−∂

≡ ≡≡

∂ ∂= − = − − − = ∂ ∂ h

r r r r r r r r r rh h

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( )

( )j j jj j j j

V r

r V r V ri V r r V r e i r e i e r

r r r rψ

ψ ψ ψ

∇

∂ ∂ ∂∂= − − = − = − = ∂ ∂ ∂ ∂

r r

r r rr r r r r r r r

h h h

14243

( ) ( )i V r rψ= − ∇r r r

h

That is

( ) ( ) ( ) ( )ˆ ˆ,p V r r i V r rψ ψ = − ∇ rr r r r r

h

Since the wave function ( )rψ r is arbitrary,

( ) ( )ˆ ˆ,p V r i V r = − ∇ rr r r

h (15)

Next, we’ll calculate the commutator ˆ ˆ,p H r .

We have

( )2 2ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ, , , , , , ,

2 2p pp H p T V p T p V p p V p i V rm m

= + = + = + = − ∇

r rrr r r r r r r r

h

where, in the last equality, we used the relation (15).

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Thus

( )2ˆˆ ˆˆ, ,

2pp H p i V rm

= − ∇

rrr r r

h (16)

But

22 2

ˆ 1 1ˆ ˆ ˆ ˆ ˆ, , ,2 2 2j j k j k jpp p e p p p em m m

= =

rr r r

where the indices ,j k are summation indices, taking the values 1,2,3.

Since the commutator ˆ ˆ,j kp p vanishes, the commutator 2ˆ ˆ,j kp p also vanishes,

thus the commutator 2ˆˆ ,

2ppm

rr vanishes too, and (16) becomes

( )ˆ ˆ,p H i V r = − ∇ rr r

h (17)

Next, we’ll calculate the commutator ˆ ˆ,r H r .

We have

{( )

2

is asummationindex

ˆˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , , , ,2j j j j

j

pr H r T V r T r V r e r e V rm

= + = + = + =

rr r r r r

{( )2 2

is a 0summationindex

1 1ˆ ˆ ˆ ˆ ˆ, , ,2 2j j k j j j k j

k

r e p r V r e r p em m

= + =

r

14243

That is

21ˆ ˆ ˆ ˆ, ,2 j k jr H r p e

m = r (18)

where the indices ,j k are summation indices, taking the values 1,2,3.

But, we have

2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , , , 2 2j k j k k k j k j k k k jk jk k jk k jr p r p p p r p r p p p i i p i p i pδ δ δ = = + = + = = h h h h

That is

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2ˆ ˆ, 2j k jr p i p = h

Note

Since 2ˆ ˆ ˆk k kp p p= , the index k in the commutator 2ˆ,j kr p is repeated, thus it is

a summation index. On the contrary, the index j is not repeated, thus it is not a

summation index.

In the right-hand side of (18), both indices ,j k are repeated, thus they are both

summation indices.

Generally, in an expression, when an index is repeated, it is a summation

index.

The summation indices do not appear in the final result.

Thus, for instance, the commutator 2ˆ,j kr p depends only on the index j .

Then, (18) is written as

1ˆ ˆˆ ˆ ˆ ˆ ˆ, 22 j j j j

i ir H i p e p e pm m m

= = = h hr r

h

That is

ˆ ˆˆ, ir H pm

= hr r (19)

Besides, (17) is written as

( ) ( )ˆ ˆˆ ˆ, ,j j j j j jj j

V r V rp e H i e p H e i e

r r∂ ∂ = − ⇒ = − ∂ ∂

r rr r r r

h h

Since the unit vectors jer , 1, 2,3j = , are linearly independent, the last equation gives

( )ˆˆ ,jj

V rp H i

r∂ = − ∂

r

h (20)

In the same way, (19) gives

ˆ ˆ,j jir H pm

= h (21)

Now, we proceed to the calculation of the time evolution of the expectation value of

the operator ˆ ˆp r⋅r r , which is the inner product of the momentum operator, which, in

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three dimensions, is a vector operator, with the position operator, which is also a

vector operator in three dimensions.

Attention!

The inner product of two vector operators is NOT commutative.

Since the operator ˆ ˆp r⋅r r does not depend on time, the time evolution of the

expectation value of ˆ ˆp r⋅r r , given by the Ehrenfest theorem, is

ˆ ˆˆ ˆˆ ,t

t

d p r i H p rdt

⋅ = ⋅

r r

r r

h (22)

As in the one dimensional case, the subscript t indicates that the expectation values

are obtained at time t .

The commutator ˆ ˆˆ ,H p r ⋅ r r is written as

{{Position space is a

summationindex

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ, , , , ,j j j j j j j j jjj

i VH p r p r H p r H p r H p H r p p i rm r

∂ ⋅ = − ⋅ = = + = − ∂

hr r r rh

where, in the last equality, we used (20) and (21).

Thus

( ) ( )( )2 2ˆˆˆ ˆˆ ˆˆ ˆ, 2 2

2j

j j j jj j

pi V V pH p r p p i r i i r i i r V r i T r V rm r m r m

∂ ∂ ⋅ = − = − = − ⋅∇ = − ⋅∇ ∂ ∂

rr rhr r r r r r

h h h h h h

That is

( )( )ˆ ˆˆ ˆ, 2H p r i T r V r ⋅ = − ⋅∇ rr r r r

h (23)

By means of (23), (22) becomes

( )( ) ( ) ( )ˆ ˆ

ˆ ˆ ˆ2 2 2ttt tt

d p r i i T r V r T r V r T r V rdt

⋅= − ⋅∇ = − − ⋅∇ = − + ⋅∇

r rr r rr r r r r r

hh

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That is

( )ˆ ˆ

ˆ2ttt

d p rT r V r

dt

⋅= − + ⋅∇

r rrr r (24)

In position space, the time evolution of the expectation value of ˆ ˆp r⋅r r is written as

( ) ( )3 *ˆ ˆ

ˆ ˆ, ,td p r d d r r t p r r t

dt dtψ ψ

∞

−∞

⋅= ⋅∫

r r

r r r r r (25)

If, as in the one dimensional case, we assume that the state ψ is a bound eigenstate

of energy E , then

( ) ( ), exp iEtr t rψ ψ = −

r r

h

Thus

( ) ( ) ( ){

( )* *

It does notdepend ontime

ˆ ˆ ˆ ˆ, , exp expiEt iEtr t p r r t r p r rψ ψ ψ ψ ⋅ = ⋅ − =

r r r r r r r r

h h

( ) ( ) ( ) ( )* *ˆ ˆ ˆ ˆexp expiEt iEt r p r r r p r rψ ψ ψ ψ = − ⋅ = ⋅

r r r r r r r r

h h

That is

( ) ( ) ( ) ( )* *ˆ ˆ ˆ ˆ, ,r t p r r t r p r rψ ψ ψ ψ⋅ = ⋅r r r r r r r r

We see that the integrand ( ) ( )* ˆ ˆ, ,r t p r r tψ ψ⋅r r r r does not depend on time, thus the

integral in the right-hand side of (25) does not depend on time either, thus

( ) ( )3 * ˆ ˆ, , 0d d r r t p r r tdt

ψ ψ∞

−∞

⋅ =∫r r r r r

Then, (25) gives

ˆ ˆ0t

d p r

dt

⋅=

r r

(26)

Thus, (24) gives

( )ˆ2tt

T r V r= ⋅∇rr r (27)

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The relation (27) is the virial theorem in three dimensions.

As in the one dimensional case, the relation (27), i.e. the virial theorem, also

holds for bound states which are not energy eigenstates, provided that for those

non-energy-eigenstates the relation (26) still holds.

3. References

[1] Eugen Merzbacher, Quantum Mechanics (Wiley, Third Edition, 1998).

[2] David J. Griffiths, Introduction to Quantum Mechanics (Prentice Hall, Inc., 1995).

[3] S. LeBohec, Quantum mechanical approaches to the virial, 30 June 2015,

http://www.physics.utah.edu/~lebohec/ScaleRelativity/Virial/virial.pdf.

[4] Guillermo Palma and Ulrich Raff, The One Dimensional Hydrogen Atom

Revisited, https://arxiv.org/abs/quant-ph/0608038.