Download Module 36: Uncertainty relation Lecture 36: Uncertainty relation

Module 36: Uncertainty relation
Lecture 36: Uncertainty relation
The Heisenberg’s uncertainty talks about the errors involved in the simultaneous measurement of two physical observables in Quantum mechanics. For
example it say that the x components of the position and momentum cannot
be measured simultaneously with infinite precision or accuracy. If the errors
associated are denoted by ∆x and ∆px then ∆x∆px ≥ h̄/2. As shown in
the previous chapter it would be true for any two physical variables if the
corresponding quantum mechanical operators do not commute. Hence a very
accurate measurement of one quauntity will destroy the accuracy in the other.
This should not be thought as the incompetance of the observer or the weaknesses of the measuring instruments. One has to beleive that the Nautre is
so. This principle can be used for calculating the ground state energies of various quantum mechanical systems. In this chapter we shall study some simple
systems and estimate the energies associated with them.
36.1
Particle in a box
We think of a box of siz L and a confine a quantum particle of mass m in it and
would like to estimate the least enegy of the particle. According to classical
mechanics the least energy is zero when the particle is at rest. But that is
not possible in quantum mechanics since if the particle were at rest then its
momentum would prcisely be zero and since it would be at rest its position
could also be measured with infinite accuracy and that violates Heisenberg’s
principle. Hence the particle is never at rest inside the box and since it moves
it would have certain kinetic energy (in the absence of any potential inside
the box). The least enrgy, permitted by the uncertainty, is the ground state
energy of the system and is known as the ‘zero point energy’. We shall now
try to estimate this energy using the uncertianty, ∆x∆px ∼ h̄.
Since the particle is confined inside the box of length L then the maximum
possible error in its position measurement cannot exceed the size of the box.
Hence we have, ∆x = L. The maximum error in postion would introduce the
minimum error in the momemtum measurement, so we have ∆px |min ∼ h̄/L
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CHAPTER 36. UNCERTAINTY RELATION
using the uncertainty relation. Now the actual momentum of the particle
cannot be less than the error in the momentum, so the least momemtum
possible for the particle is px ∼ h̄/L. This least momentum would produce
the least energy of the particle,
p2x
h̄2
E=
∼
.
2m
2mL2
(36.1)
When this problem is exactly solved using the Schrödinger equation one obh̄2 π 2
tains the ground state energy as 2mL
2 which is an order of magnitude larger.
But the important result is that the simple calculation using uncertainty shows
the right dependence of energy on the box size, L, which scales as L−2 . Now
we can have some rough quantitative estimates of the energy in couple of
situations.
i) We can assume an electron trapped in an atom of size L. In this case
we use m = 9.1 × 10−31 kg (mass of the electron) and L ∼ 1 A◦ (rough size of
an atom). Substituting the value of h̄ = h/2π = 1.05 × 10−34 , we have
h̄2
E =∼
∼ 6.1 × 10−19 J ∼ 4eV.
2
2mL
(36.2)
We know that the ionisation potential (ground state) of the simplest atom
(Hydrogen) is approximately 13.6 eV and it is of the same order of magnitude
as that of our rough estimate.
ii) The second example we take of a nucleus which has a size of the order
of few fermis (1 fm= 10−15 m). We assume an alpha particle trapped in the
nucleus and try to estimate its energy using the uncertainty principle. We
roughly take the size of the nucleus (box), L = 5 × 10−15 m. The mass of the
alpha particle is 2(mn + mp ) ∼ 4 × 1.67 × 10−27 kg. So the rough energy of
the alpha inside the nucleus would be
h̄2
E =∼
∼ 0.33 × 10−13 J ∼ 0.2MeV.
2
2mL
(36.3)
Again we find that the very crude estimates made exploting the uncertainty
principle gives us an order of magintude lower than that of typical alpha particle energies which are usually a few MeV. Of course one can check that a
slight change in the size of the box would change this energy drasticlly due to
the inverse sqare dependence. For example a nucleus of size 3 fm would make
the alpha particle energy approximately 1 MeV.
36.2
The Hydrogen atom
In the above cases we did not have any potential. We can make our calculations
finer by incorporating the potential in the system. In the following we show
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36.2. THE HYDROGEN ATOM
one such example of one dimensional Hydrogen atom. The total energy of the
electron in the Hydrogen atom is written as,
E = K.E. + P.E. =
p2
e2
−
.
2m 4πǫ0 r
(36.4)
Now as argued earlier the maximum possible error in the position r ∼ ∆r and
that corresponds to minimum error ∆p ∼ h̄/∆r. The minimum p ∼ ∆p, since
it cannot be less than the error associated with it. Substituing these in (36.4)
we get,
h̄2
e2
E=
.
(36.5)
−
2m(∆r)2 4πǫ0 ∆r
Note that we have two competing terms in the energy expression, (36.5). Now
for obtaining the least energy we have to optimise the expression (36.5) with
∂E
= 0. Minimising the energy we have,
respect to the error ∆r, i.e. ∂∆r
∂E
h̄2
e2
=−
+
= 0.
∂∆r
m(∆r)3 4πǫ0 (∆r)2
(36.6)
The expression (36.6) finds the error ∆r which minimises the energy and which
is equal to
4πǫ0
⇒ ∆r =
.
(36.7)
me2
When we evalute the ∆r by substituing the values of m, e etc. we find that it is
equal to 0.528 Å and which is nothing but the Bohr radius defining the size of
the Hydrogen atom. Substituting the value of ∆r from (36.7) in the expression
of total energy (36.5) we get the ground state energy of the Hydrogen atom
as,
me4
E=−
.
(36.8)
2 × (4πǫ0 )2 h̄2
The expression obtained in (36.8) is the exact expression for the ground state
energy of the Hydrogen atom. The negative sign shows that the system
(electron-proton) is bound and one has to provide at least this much of energy
to tear of the system. When evaluated its magnitude is approximately 13.6 eV
and which is nothing but the ionisation potenstial of the Hydrogen atom. It
is indeed surprising that by the above crude calculation we produce the exact
answers to the Hydrogen atom problem which in reality needs a considerable
amount of calculations for soving the Schrödinger equation to arrive at correct
answers matching with the experimental values.
Question: In fact we made two mistakes in the above calculations and by
a sheer stroke of luck these two mistakes cancelled each other and produced
the right result. The students are encouraged to point out these mistakes.
The other uncertainty which has important consequences is the EnergyTime uncertainty, i.e.
∆E∆t ≥ h̄/2.
(36.9)
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CHAPTER 36. UNCERTAINTY RELATION
One of its consequences is that the energy conservation could be violated for
a very short spell of time. If one has infinite time to measure the energy of a
system then there is no error in the energy measurement and the measurements
are accurate. But suppose there is very little time available for measuring the
energy ( as in the case of atomic transitions where the transitions last for only
order of 10−8 secs) then there are large errors associated with these energy
measurements as in the case of the atomic levels. So the atomic levels are not
blade sharp but they are fuzzy with a mean value of the energy.
Problems
1. Estimate the ground state energy of a quantum harmonic oscillator
using uncertainty relation for which the total energy is given by,
E=
p2x
1
+ kx2 ,
2m 2
(36.10)
where m is the mass
q of the oscillator and k is the spring constant.
[Ans. h̄ω = h̄ k/m. If you use the minimum uncertainty you would
produce half the above value and which is again an exact result for the zero
point energy of the quantum one dimensional harmonic ocillator.]