Download Slide show "Notes_15" - Department of Physics | Oregon State

The Heisenberg Uncertainty Relationship (HUR)
(under construction. When expected to be ready? – it’s uncertain)
The Heisenberg Uncertainty Principle
But one thing is not uncertain to me – you have
certainly heard of this important relationship.
It states:
h
x  p x   (where  
2
)
Here, x is the “uncertainty of the particle position”
– in other words, the precision with which the particle
position can be determined (the uncertainty we are
talking about now is NOT that resulting from the imperfect measuring equipment; assume that it is “infinitely precise”).
And px is the “uncertainity of the particle momentum” – or, the precision with which the momentum
component in the x direction can be measured.
The Heisenberg Uncertainty Relationship
states that the position x and the momentum
px cannot be both determined with an urestrictibly good precision. Even the best possible apparatus would not help here: there is
always a tradeoff! High precision in position
determination means that the momentum cannot be precisely determined – and vice versa.
The two uncertainties are such that their
product will always be x  px   . And this is
not because our apparatus is not perfect. This
is a LAW OF NATURE.
A story from Dr. Tom’s own life experience:
When I was an undergraduate student,
we were told many times by our instructors
on various occasions:
“As shown by Heisenberg, px  x   ….”
or:
“As is well known, px  x   ….”
or:
“As the Uncertainty Relation states, p x  x   ...”
I was feeling frustrated, because they always were
giving us that information “like a rabbit from a
magician’s tall hat”– it is so, you have to believe!
Finally, at last term of my junior year, I was
taking a “Quantum Mechanics One” course,
and only then the professor showed us how
to derive the Heisenberg Uncertainty Relation
from the “first principles”. The procedure is
based on a fundamental mathematical theorem called “the Schwartz Inequality”.
But I still remember how frustrated I was when
I had to believe in the HUR only “because wise
men had shown that it is so”.
Therefore, my sincere wish is that my students
never have such odd feelings – and therefore I
always want to show them ASAP where this
famous formula comes from.
The method based on the “Schwartz Inequality”
is too advanced for this course because one has
to first get enough knowledge of the foundations
of Quantum Mechanics.
But there is a very instructive method of deriving the HUR for wavepackets composed of de
Broglie Waves, and I want to show you that.
Note: some authors of textbooks on introductory Quantum Mechanics
show this method, and they certainly think it is “general enough”
because they don’t discuss the method based on the Schwartz
inequality.
With wavepackets, there is a “trade-off”: to get a narrow
one, you have to take waves from a broad range of k ;
and narrower range produces a wider packet:
Now there will be several pages of calculations (mostly, integrals).
We will not discuss this math stepby-step in class, but we will scroll through
slides #10 to #18 with brief explanations
only. The material is given here for you
to know that the final result CAN BE derived in a fully rigorous manner – and if you
wish, you may check! But it is not necessary, if you prefer to accept the results
without proof, this is also OK. But you always may check, if you change your mind.
In order to show that rigorously , we will construct a packet
by summing elementary waves whose spectrum of k - values
is a Gaussian function centered at k0 :
 ( k  k 0 ) 2 / 2 2
G (k )  Ae
where  is the standard deviation of the distributi on
and is a measure of the spread of the packet.
If the packet is a superposit ion of a finite number of
waves with a discrete set of k - values, then the
function describing the entire packet can be written as :
 ( x) 
 G(k ) sin( k x)
i
all waves
i
However, as we said earlier, in order to obtain
a SINGLE packet representi ng a SINGLE
particle, we have to superpose an infinite number
of elementary waves - it means, we have to integrate
over a continuum of states. Then :

 ( x)   G (k ) sin( xk)dk

However, for " mathematic al convenienc e" , it is
better to use a complex function for describing a
ikx
single wave : not A sin( kx), but Ae .
The simple equation of a plane wave we use is
A sin( kx  t ). As I say, we can use instead
Aei ( kx t ) . Here we take advantage of the Euler
i
formula : e  cos   i sin  . So, our wave
becomes : A cos( kx  t )  iA sin( kx  t ). There
is an imaginary term - however, we can " forget"
about it and use only the real one.
But you will say : Dr .Tom! The real term is cos(kx  t ),
it is not the same as sin( kx  t )!!!
Yes, this is true - but it makes essentiall y no difference
wheter one uses a sine or a cosine function t o describe
an elementary wave.
OK, so let' s put everything together,
and let' s integrate!

 ( x)   Ae
 ( k  k 0 ) 2 / 2 2 ikx
e dk
-

 Ae
ik0 x
e
 ( k  k 0 ) 2 / 2 2 i ( k  k 0 ) x
e
d (k  k0 )
-

 Ae
ik0 x
e

 k 2 / 2 2 ikx
e dk
-
Let' s focus now on the integral.
Let' s switch from the " ei notation" to the
" exp( i ) notation" - it' s easier to follow :

k
e

-
2
/ 2

2
2
k
eikx dk   exp[  2  ikx]dk
2
-
Now, we have to make a small " trick" :
multiply t he integrand by a " well - chosen one" :
 x 2 2 x 2 2 
exp 

  1, and then contimue :
2 
 2

 k2
x 2 2 x 2 2 
  exp  2  ikx 

dk
2
2 
 2

 x 2 2  
 1
2
2
2 4
 exp 
exp

k

2
ikx


x


2

2    2




dk





 x 
 1
2
2
2 4
 exp 
exp

k

2
ikx


x


2
 2
2

  
2
2
 x 
 1
2 2
 exp 
  exp  2 k  ix dk
2    2


1
Use a " dummy" : u 
(k  ix 2 );
 2
2
2
then dk  ( 2 )dk ;

 x 2 2 
2
 exp 
  ( 2 )  exp  u du
2 


 



dk



 x 
2
 exp 
  ( 2 )  exp  u du
2 


2
2

But :
 
 
2
exp

u
du




so finally we obtain :
 x 2 2 
 2  exp 
, which we can write as :
2 

2


x
 2  exp 
2
 21 /   
Now, look : the initial spectrum of waves was :
2


(
k

k
)
 ( k  k 0 ) / 2
0
G (k )  Ae
 A exp 

2
2


where  - standard deviation, or the " spread"
2
2
of the k  values in the packet.
And we obtained a wavefunct ion described
by another Gaussian :
2


x
ik0 x
 ( x)  
2
  e  exp 
2

A


21 /   
wave

constant

of mean 
k - value
Gaussian"envelope"
Compare the two Gaussians :


 (k  k0 ) 
x
exp 
 and exp 
2
2
2


 21 /   
The standard deviation of k (" spread in k " ) is  ;
The standard deviation of x (" spread in x" ) is 1/ ;
2
2
So :
(" spread in k " )  (" spread in x" )  1
This is illustrate d in the next slide :
If we think of the standard deviations as of
“uncertainties”, and we use for them the
symbols k and x , we can write:
k  x  1
But for the de Broglie waves the wavenumber
and the momentum are related as:
px    k so that
px    k
Which leads to the final result:
px  x  
The “Gaussian wave packet” is also known
as “the minimum uncertainty wave packet”.
For a wave packet whose k spectrum is described by any other function than a Gaussian, the uncertainties are always such that:
px  x  
(please accept without proof).
What we did here should not be treated as a
“derivation of the Heisenberg relationship”–
HUR can be derived from more fundamental
assumptionts – it was only an illustration of
“how the HUR works” in wave packets.
Last question – what is the physical
Interpretation of de Broglie waves?
In sound waves, it is the air molecules that oscillate;
A sound wave consists of areas of higher and lower
density (or pressure).
In EM radiation, it is the electric and magnetic field
that oscillate.
In waves on water surface, it’s the water that moves
periodically up and down.
In seismic waves reaching the Earth’s surface – everybody knows, let’s better not talk about sad things.
But what is oscillating in de Broglie waves?
What is the “undulating agent” in such waves?
Certainly, nothing material! (the often used term
“waves of matter” is highly misleading!)
A field? – no, surely, there is no field of any kind
associated with the de Broglie waves.
THEN, WHAT?!
Well, to answer this question, we have to clear up
certain things.
Note that we always observe particles “as particles”.
In an act of observation, or in an “act of particle detection”, we never see a wave. Things we see are always
“manifestations” of the particle-like nature of the particles we observe. We see tiny specs on a photographic
film, tiny flashes on a fluorescent screen, “tracks” in a
cloud or bubble chamber.
Wave-like properties of particles are always manifested
indirectly.
Now, consider the double slit experiment with electrons.
Extremely valuable for our understanding of de Broglie
waves are experiments in which electrons are shone at
the double-slit apparatus one at a time, which enables
us to see individual flashes on a fluorescent screen.
In a famous experiment done in Bologna,
Italy, in 1974, the flashes from electrons
reaching the screen
were detected by an
ultrasensitive photodetector, and after
each flash the film
In a camera focused
on the screen was
advanced by one
frame. Then, the data
were combined, showing the time evolution
of the interference pattern.
In an analogous experiment done in Japan in
1989 a more modern technique of electronic
recording was used – but the results were
essentially the same as in Bologna.
To make the long story short: after many years
debating and many disputes (often heated), physicists finally found an answer that was accepted by a broad majority: namely, that the de Broglie waves are “waves of probability” – meaning
that the value of the de Broigle wavefunction of
a particle expresses the probability of finding
the particle at a given point:
 Probabilit y of finding

 the particle at x



 (x)
2
It should be noted that Albert Einstein did not like the “probabilistic
interpretation”, and he died unconvinced. Although the majority of
physicists now accept this interpretation, the discussion is not yet
finished – the problem is still the subject of ontological debate.