Re: 10 questions on QM postulates

> I am an undergrad physics student who just completed a course in
> modern physics (mostly an introduction to QM). I have several
> questions related to the postulates of Quantum mechanics which I find
> intriguing and which my textbook imo did not address adequately. I
> have listed my questions below. I know there are a lot of questions;
> answering just one at a time would be great.

I'd be interested to know which book you used in your course.
>
>
> 1) How can the mathematical concept of hermitian operators be
> translated into physical terms? What are all the conditions which
> ultimately restrict operators such that they can be only hermitian as
> opposed to any other type?

That's a very deep question, and I also asked it to my professor when he
introduced the concept. His answer was: "The operators, describing
observables, must have only real eigenvalues, and Hermitean operators
have only real eigenvalues."

In my opinion, that's a short but not really satisfactory answer. After
thinking about the foundations of quantum theory from time to time (I
apply quantum theory every day, but one does not think about its
foundations too often in this kind of work, so it's more a hobby of
mine to think about it from time to time, especially when I hear
something about modern quantum-optics experiments, which help a lot to
understand the foundations of quantum theory), I think the real
mathematical reason that the observables can be described without any
inconsistencies by Hermitean (or to be more precise by essentially
self-adjoint) operators is the spectral theorem, i.e., each quantum
state can be described by a superposition of a complete set of
generalized eigenvectors of an self-adjoint operator.

I do not claim that an operator must be necessarily self-adjoint to meet
this property. I'm not sure wheter it is or not, it might be sufficient
that an operator commutes with its adjoint, which is usually called a
"normal operator". Perhaps somebody more familiar with the mathematical
details may answer this question.

Anyway, the point of the spectral theorem is that you can interpret the
(generalized) components of any Hilbert-space vector |psi> with respect
to the (generalized) eigenvectors |a> of an self-adjoint operator A as
probability (distribution) amplitudes. In more physical terms, this
means that, provided the system in question is perpared in the state,
described by the Hilbert-space vector |psi>, the probability
(distribution) to find the value a when measuring the observable A is
given by

P(a|psi)=3D|<a|psi>|^2 (|psi> normalized, i.e. <psi|psi>=3D1).

Here I simplify the argument a little bit by assuming that each
measurement of A is complete, i.e., for each (generalized) eigenvalue a
there is only one linearly independent (generalized) eigenvector.

Now the |a> fulfil two important properties:

(1) they are mutually orthogonal to each other (in the generalized
sense, if there are continous parts in the spectrum of A), i.e.,

<a|a'>=3D\delta(a-a'),

where a and a' run over the set of (real) generalized eigenvalues of the
operator A. \delta is a Dirac-Delta distribution if a is in the
continous part of the operator's spectrum or a Kronecker delta if it is
in the discrete part.

Note that all thinkable cases appear to be common in quantum theory: An
self-adjoint operator can have only a continous spectrum (e.g.,
positions and momenta of a particle), only a discrete spectrum (e.g.
the modulus of angular momentum can have the eigen values j(j+1)
\hbar^2, where j \in {0,1/2,1,...}, and the component in one direction
of the angular momentum then is also discrete with the eigenvalues
\sigma \hbar with \sigma \in {-j,-j+1,...,j-1,j}) or both (e.g. the
spectrum of the Hamilton operator of the hydrogen atom as a discrete
part, describing energy levels of the bound proton and electron, and a
continuous part, describing scattering between a proton and an
electron).

That means, if a system is prepared in the eigenstate |a>, one can never
find another value than a, when measuring the observable A. This makes
the whole thing consistent: If by prepartion of a system, the
observable A is assigned a definite value a, you must always measure
theis value a, otherwise it would not make sense to assign this value.
Thus, the self-adjointedness of the operator A ensures this necessary
consistency condition in the interpretation of quantum states and
preparation of a system in a state through measurement of the
observable.

(2) The set of eigenvectors |a> is complete, i.e., you have

\sum_{a} |a><a|=3D1, (completeness relation)

where for simplicity, I assume from now on the spectrum to be only
discrete. If there are also continuous parts in the spectrum, you must
just add the corresponding integral.

Also this is of profound importance for the consistency of the physical
interpretation, because above we have stated that the probability to
find the value a, when measuring the observable A, is given by

P(a|psi)=3D|<a|psi>|^2=3D<psi|a><a|psi>,

when the system is prepared in the state |psi>.

The completeness relation ensures that the probality to find some value
in the spectrum when measuring A is 1, i.e., that one finds always a
value for A, and it must be in the spectrum of the corresponding
self-adjoint operator:

\sum_a P(a|psi)=3D\sum_a <psi|a><a|psi>=3D<psi|psi>=3D1.

Also the expectation values of A are easily calculated:

<A>:=3D\sum_a a*P(a|psi)=3D\sum <psi|A a><a|psi>=3D<psi|A|psi>

>
> 2) how is it possible (or: how come it does not contradict the
> principles of QM) that a continuum of eigenstates (as opposed to
> quantized eigenstates) can exist for unbound states?

Also this is a delicate question. Admittedly, what I wrote above is a
mess, and most textbooks also write a mess about this point. That's
just to simplify the maths a little bit for us poor physicists. The
point is that continuum "states" are not true Hilbert-space states, but
"generalized" states. Mathematically they belong not to Hilbert space
but to a larger space, which is the dual space of a smaller subspace of
the Hilbert space, namely those vectors, where the self-adjoint
operator, describing the observable, is defined.

Let's take the example of the x-coordinate of a spinless particle. We
can describe the Hilbert space by the complex functions whose square
can be integrated over whole R, i.e.,

<psi|psi>=3D\int dx psi^*(x) psi(x)=3D\int dx |psi(x)|^2<\infty.

Now there are a lot of functions in this Hilbertspace, known as L^2,
where the position operator x is not defined. Take, e.g., the
(unnormalized) wave function

psi(x)=3Dsin(x)/x,

which for sure is square-integrable. Here, in position representation,
the x-operator is just given by the multiplication of the wave function
with x, but

x psi(x)=3Dsin(x)

is not square-integrable.

The x operator is defined on a smaller sub-space \Omega of L^2, which is
dense in L^2, i.e., each wave function of L^2 can be represented as the
limit of a sequence of functions in \Omega (that's a mathematical
theorem about self-adjoint operators in Hilbert space!).

Now the "eigenvectors" of the position operator are not functions, but
distributions of generalized functions, namely

phi_{x0}(x)=3D\delta(x-x0),

where phi_{x0} denotes the generalized eigenvector of the position
operator with the generalized eigenvalue x0.

\phi_{x0} is defined as a functional on the subset \Omega:

\psi \in \Omega |-> \int dx phi_{x0}^*(x) psi(x)=3Dpsi(x0).

The advantage of this rather abstract formalism is that it legitimates
Dirac's brillant bra-ket calculus: we can write without much headaches:

psi(x)=3D<x|psi>,

where |psi> is a Hilbert-space vector, providing a mapping from the
abstract Hilbert space H of kets to the Hilbert space L^2 of wave
functions.
>
> 3) How can it be shown that the gradient (of psi when operating) times
> =96ihbar (i.e. the momentum operator) is equivalent to h/lambda? Can an=
y
> equivalence between the momentum operator and the classical mechanical
> definition of momentum (mv)be shown at all ( I know, it's called the
> wave equn, but still)?

That's also a very good question. For me, the most satisfactorial answer
is provided by Noether's theorem, which you hopefully know from your
classical-mechanics course: There you learn that momentum is the
generator of space translations.

>
> 4)How can it be shown directly (that is, without reverting to KE=3D p^2
> /2m) that the kinetic energy operator actually is a kinetic energy
> term?
>
> 5)How can it be shown directly (again, without reverting to H=3Dp^2 /2m=
+
> PE) that the the time derivative (of psi when used in operation) times
> ihbar is actually
> equivalent to the hamiltonian? I am asking these
> questions because I find it intriguing that the time derivative of psi
> should be proportional to its second spatial derivative (assuming
> constant potentials). This does not strike me as something obvious,
> even though playing around with the equations suggests something like
> this (which I still do not conceptually understand):
>
> Assume we can set H=3D-W (Can we really do this? Why or why not?)
>
> then: -W=3D int(F.dx)=3Dint (dp/dt dx)
> (here, F and dx would obviously have to point in opposite directions
> for the work to be negative)
> finally, substitute the momentum operator, and we magically recover
> the left side of Schroedinger's equation.
> (BTW I have never seen a "force operator". I imagine that's because it
> is closely linked with particles rather than waves. But is that a
> strong enough reason to probhibit its use? Why?)

The answer to these questions is a little bit too long to find space
here in the news group. The answer is again Noether's theorem, and may
be found in the very good textbook

L. Ballentine, Quantum Mechanics,

which somebody, who asks such deep questions as you, should read,
although it's not easy!


> This brings me also to my next question:
>
> 6) What is the reason for distinguishing in the Schroedinger equation
> between the Hamiltonian and total Energy? Is this related to the view
> that Planck's constant can be considered to be the fundamental unit of
> action? Under what circumstances is it valid to substitute total
> energy for the Hamiltonian?

In which sense was this distinguished in your lecture? A reason might be
the following:

The total energy (i.e., kinetic + potential energy) has only a certain
fixed value for systems, which are prepared in an energy-eigenstate.
For all other states, the Hamiltonian has not a fixed value.

Nevertheless, the Hamiltonian H represents the observable "total
energy", and it is the generator of time evolution, because according
to Noether's theorem it's the conserved quantity from the symmetry of
the physical laws under time translations.

>
> 7)What examples, other than the simple harmonic oscillator, can be
> used to illustrate the correspondence principle, that is, the idea
> that in the limit of large scales quantum mechanics reduces to
> classical mechanics? In particular, how can we get from the
> Schroedinger equation to the classical Hamiltonian (or vice versa)?

Look in a good textbook under "WKB approximation" or semi-classical
approximation. It's again a little bit difficult to squeeze the
interesting answer into a newsgroup posting, which becomes already
quite lengthy ;-)).
>
> 8) Is there any experimental evidence which would answer the following
> question: does the probability of finding a particle in a volume
> element (psi*psi) change if the shape of the volume element is changed
> (but the volume is kept the same as before)? Have any experiments been
> conducted which examine the predicted and actual probabilities of
> finding particles in volume-elements that are not cubical?

I don't know. For sure, one has not found a contradiction of quantum
theory with experiment. To the contrary: The more hard the quantum
theory is tested by experiment, the more accurate we find it's
consistency with the outcomes of these experiments. A lot of
experiments, which one could think of only as "gedanken experiments"
some years ago, can now be done with real experiments in the lab,
thanks to the advancement in quantum optics. That's the reason, why a
quite new branch of quantum studies has arised in the last view years
like "quantum computing" and "nano technology".
>
> 9)Does psi have a dimension (like length?) Can psi itself ever end up
> being orthogonal to the imaginary axis (i.e. be real, rather than
> complex)? If so, under what circumstances? Would that have any effect
> on what we observe? I ask this question because if its phase shift
> ever caused psi to lie exactly on the real axis, this would seem to
> introduce a dimensional problem when we square it: A complex number
> with some dimension multiplied by its complex conjugate gives a real
> number with same dimension. A real number with some dimension squared
> obviously gives a number with the square of the dimension.

The wave function, i.e., the states in the position representation

psi(x)=3D<x|psi>

must have dimension length^{-d/2}, where d is the number of position
observables needed to describe the system (i.e., 3 for one particle, 6
for two particles and so on), both, real and imaginary part. Of course
|psi(x)|^2 is always a real number >=3D0, as it must be for a probability
distribution.

This must be so, since |psi(x)|^2 d^d x is the probability distribution
to find the particles in a little volume element d^d x around x. The
shape of the volume element is unimportant. As I stated above, I am not
aware whether this prediction of quantum theory ever has been tested in
the lab.
>
> 10) Some versions of the postulates also mention this one:
>
> The total wavefunction must be antisymmetric with respect to the
> interchange of all coordinates of one fermion with those of another.
> Electronic spin must be included in this set of coordinates
>
> I understand that this postulate leads to the Pauli exclusion
> principle, but I don=92t know how it does so (i.e. how do you derive th=
e
> pauli exclusion principle from this postulate?). Also, the postulate
> itself is not clear to me, can someone explain this to me?.
>
>
> Again, I know this is a lot, so answering just a part at a time would
> be fine.

Well, I take the opportunity to leave the last question unanswered,
since this would again be a long one. The answer, using path-integral
methods, can be found in the original paper

M. G. G. Laidlaw, C. M. DeWitt, Feynman Functional Integrals for Systems
of Indistinguishable Particles, Phys. Rev. D 3 (1970) 1375, URL
http://link.aps.org/abstract/PRD/v3/i6/p1375

but that's REALLY a tough one!

--
Hendrik van Hees Texas A&M University
Phone: +1 979/845-1411 Cyclotron Institute, MS-3366
Fax: +1 979/845-1899 College Station, TX 77843-3366
http://theory.gsi.de/~vanhees/ mailto:hees@xxxxxxxxxxxxx

.

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