# Energy conservation low in quantum mechanic.

2 views

### kurd...@yahoo.com

Aug 22, 2005, 4:38:43 AM8/22/05
to
Hello,

let us suppose that we have a quantum harmonic oscillator being in the
first exciting state. It means that system has some definite energy
E_1. Than we change force constant of of the oscillator. As a
consequence eigen states and there energy are changed. I think that
after this change system cannot be in some new eigen state. Because in
this case the system would have some new definite value of energy and
we would have contradiction with conservation low. So I conclude that
after force constant is changed system will be in some superposition of
new eigen states. It means that making measurement we can with some
probability find the system in some state having some energy. I have
two question. Does the energy conservation low implies that sum of
eigen-energies multiplied on the probability to find system in the
corresponding state should be equal to the initial energy of the system
(e_1*v_1 + e_2*v_2 + ... e_n*v_n = E_1)? Let us suppose that we made
measurement and the find system in some new eigen-state with energy
e_k. How to be with the conservation low? I mean e_k is not equal to
E_1 and it should be a problem?

----------------------------------------------------------------------------

I think my questions can be formulated in shorter way:

1. If state of a system is a superposition of the eigen state, how is
defined total energy of the system?

2. Should energy conservation low work for this quantum mechanical
energy?

3. If energy conservation low is satisfied, is it satisfied
statistically (by averaging over many measurements)?

### Bjoern Feuerbacher

Aug 22, 2005, 9:30:04 AM8/22/05
to

Essentially, in QM there are only conservation laws for the
*expectation values* of observables. And expectation values are
defined essentially as you wrote above, e_1*v_1 + ... = E_1.

Bye,
Bjoern

### Puppet_Sock

Aug 22, 2005, 10:24:03 AM8/22/05
to

Maybe in QM. In quantum field theory, conserved quantities are
conserved at every vertex, and by each propagator. So, they are
conserved microscopically.

To the OP: You talk about "changing the force." Think about how
that might happen. For example, you might have a charged particle
in an electric field that had the right shape. (I don't know just
how you'd create that field, but whatever. Maybe you can do it in
one dimension and ignore the other two dimensions?) How would you
change it? You'd have to move charges around. You'd have to change
the electromagnetic field around the charge in the eigenstate.

In other words, you'd have to fling photons around. And these
photons would have to carry energy of comparable amount to the
energy of the eigenstate you are considering.

Take an example that is more achievable. Consider an atom with
6 happy electrons and six happy protons, and enough neutrons
to make everything stable. (Homework: What chemical element is it?)
Now suppose you look at the energy eigenstate of the electron
that is highest in energy inside the atom. And suppose you follow
what happens to it when you smack out one of the other electrons
by hitting it with a photon. You can have as high energy a photon
as you like, making the other electron go away very quickly.

What you wind up with is the electrons that stay have now got
less negative charge to compete with. They are goiong to see
the six charges on the nucleus with less screening. So they
will all have to adjust their energy levels. So, what happens
is, the atom will emit one or more photons, to readust its
energy levels to be in the ground state for the new system.
During that process, there's a lot of complicated jostling,
and a lot of virtual photons exchanged, and just generally a
lot of complexity. After, you have some few photons flying away
and the atom in a the new ground state with 5 electrons around
Socks

### kurd...@yahoo.com

Aug 22, 2005, 11:25:19 AM8/22/05
to
If my system is in a superposition of the eigen states (and I know the
corresponding wave function), I can say with which probability given
eigen energy will be found in the experiment. But if measurement is
already made and I found system in the first excited state, does it
means that I destroyed the previous state and system is know in the
first excited state? Or I need to say that after measurement system is
in the same superposition of states as it was before measurement and if
I do new measurement I can find the system in the second excited state?
Can one speak about measurements which do not change wave function of
measured system?

### kurd...@yahoo.com

Aug 22, 2005, 11:38:50 AM8/22/05
to
Does your example with charged particle in an electric field imply that
one cannot change the force constant of the quantum harmonic oscillator
without changing its energy?

In the second example (electron in atom) I do not understand what you
mean under "adjust their energy levels". I can imagine an arbitrary
system with some energy levels and particle which whose state is
superposition of eigen states. What I do not understand is which
superposition of eigen states (or simply which state) more adjust
energy levels. I would say, that in the case of absence of external
perturbations system will always stay in the initial superposition of
states. And in the case of atom, electron will always be in excited
state and never emit photon (if there are no external perturbations).

### Puppet_Sock

Aug 22, 2005, 11:42:19 AM8/22/05
to
kurda...@yahoo.com wrote:
> Does your example with charged particle in an electric field imply that
> one cannot change the force constant of the quantum harmonic oscillator
> without changing its energy?

What it implies is, you can't learn quantum mechanics by
asking vague questions and getting people to stroke you
for an extended time. If you want to understand this stuff,
then go crack open a book and read about it. Do the homework
questions. If you still have questions after you've read
at least three textbooks on QM, and done all the homework
questions, then come back. Or if you have specific questions
on individual questions you want help with, come back.
Socks

### Zigoteau

Aug 22, 2005, 3:22:18 PM8/22/05
to

Hi, Kurda,

> If my system is in a superposition of the eigen states (and I know the
> corresponding wave function), I can say with which probability given
> eigen energy will be found in the experiment. But if measurement is
> already made and I found system in the first excited state, does it
> means that I destroyed the previous state and system is know in the
> first excited state?

That depends on whether you subscribe to the Copenhagen interpretation
of QM. There are other interpretations in which the wave function never
collapses.

> Or I need to say that after measurement system is
> in the same superposition of states as it was before measurement and if
> I do new measurement I can find the system in the second excited state?
> Can one speak about measurements which do not change wave function of
> measured system?

No. Measurement of a quantum system always perturbs it.

Cheers,

Zigoteau.

### Gregory L. Hansen

Aug 22, 2005, 3:45:39 PM8/22/05
to

A classic problem is a particle in the ground state of an infinite
potential well with some given width. Then "suddenly" the well expands to
twice that width.

The new wavefunction is the same as the old because of the "sudden"
assumption-- it's half of a sine wave sitting in the middle of a potential
well. You can Fourier transform that to express it as a superposition of
energy eigenstates of the new system, and give the probability that the
result will be E1 or E2 or E3...

It might seem at first that energy isn't conserved, because the system
could be in any state. But it wasn't explained HOW the well was expanded.
It wasn't explored whether the particle did work on the moving walls of
the well, or the walls did work on the particle. It was simply declared
"Let it be so."

A more complete problem might have involved something like an electron
trapped between boundaries on a carbon chain, and then a photon knocks
into the side of it, distorting the molecule.

At any rate, energy is conserved in quantum mechanics, although a given
problem may not be specified completely enough to explore that.

--
"The polhode rolls without slipping on the herpolhode lying in the
invariable plane." -- Goldstein, Classical Mechanics 2nd. ed., p207.

### kurd...@yahoo.com

Aug 22, 2005, 3:48:49 PM8/22/05
to
> What it implies is, you can't learn quantum mechanics by
> asking vague questions and getting people to stroke you
> for an extended time.
My question is not vague and I do not understand why I getting people
to stroke.

> If you want to understand this stuff,

I think that communication with people is much more productive than
reading of books. Moreover, I think it is much easier to ask my
question to a person who really knows the answer than to find answer in
textbooks.

> Do the homework
> questions.
I did.

> If you still have questions after you've read
> at least three textbooks on QM,

I have read more than three textbooks on QM.

> and done all the homework
> questions,

I did.

> then come back.
I am here.

> Or if you have specific questions
> on individual questions you want help with, come back.

I think my question is quit specific.

With the usage of my knowledge of QM I have concluded that a harmonic
oscillator cannot be in eigenstate, after an external changes of the
force constant was performed (under condition than at t=0 harmonic
oscillator was in eigenstate). And I asked whether my derivation is
correct. You did not give me a direct answer but pointed you example
where a change of the force constant cannot be performed without
photons and corresponding energy flow to/from the system. I asked
whether this is a fundamental property. And you send me to read
textbooks.

By the way, you did not explain why do you think that system (electrons
in atom) should go to ground state? I think that I was right when I say
that without external perturbation a system will always stay in an
excited state (within QM, I do not speak about quantum electrodynamic).

### Zigoteau

Aug 22, 2005, 4:17:54 PM8/22/05
to

Hi, Kurda,

> My question is not vague and I do not understand why I getting people
> to stroke.

They think that the question you are asking is homework. I myself
wasn't quite sure.

> With the usage of my knowledge of QM I have concluded that a harmonic
> oscillator cannot be in eigenstate, after an external changes of the
> force constant was performed (under condition than at t=0 harmonic
> oscillator was in eigenstate).

A lot depends on how the force constant was changed. Your question
sounds very much like a homework question, in that it is rather
artificial. In real experimental situations, any perturbation that
changes the force constant will most likely also interact directly with
the electron. As Greg says, the "obvious" assumption would be that the
wavefunction is left unaltered by the instantaneous change of force
constant. If it were an eigenstate of the system with the original
force constant, then it would be a superposition of the eigenstates of
the new one.

> <...>

> By the way, you did not explain why do you think that system (electrons
> in atom) should go to ground state? I think that I was right when I say
> that without external perturbation a system will always stay in an
> excited state (within QM, I do not speak about quantum electrodynamic).

Yes, you're right, without external perturbation a system can stay
indefinitely in any of its eigenstates. However real systems radiate,
and otherwise interact with their environment, ending up with an energy
of the order of kT = 4e-21 J = 0.025 eV at room temperature. If you are
talking about electronic transitions of small systems like atoms, that
means the ground state.

Cheers,

Zigoteau.

### kurd...@yahoo.com

Aug 22, 2005, 5:29:23 PM8/22/05
to
Hi, Zigoteau,

As far as I know different interpretations of QM are vary often just a
matter of wording (searching of appropriate combination of words) since
on the level of experiments they give the some results. However, I
agree that sometimes different formulations predict different results
of experiments and such experiments are rather tricky (complicated) and
were performed only very recently (something related with EPR paradox).
What I want to say is that measurement of energy distribution should
not be the case where different interpretations are mater. May be I am
wrong.

My question was about interpretation of superposition. If a system is
in a superposition of two states we can with some probability find a
system in one of these states. However, I think that it would be
incorrect to say that superposition of states means that we just do not
know in which of the two states is the considered system. It means that
if we performed experiment and during experiment we found the system in
the first eigenstate we still cannot say that system noe is in the
first state (system is still in the superposition of states). The
probability to find the system during the second measurement in the
second eigenstate is still not equal to zero, I think. And it is true
(what I said) under two suppositions:
1. superposition cannot be interpreted as lack of knowledge about the
system,
2. measurement does not influence on the system.

But experiment always influences on the system and my question is
whether what I said is true even if we do not neglect by this influence
(interaction with meassuring device).

For example:
1. Psi = C_1*phi_1 + C_2*phi_2 - before the first measurement.
2. During measurement we found that energy of the system is E_1 (what
corresponds to state phi_1).

Question:
What is the state of the system after measurement?

1. C_k_new = C_k_old?
2. C_1_new =1, C_2_new=0?
3. C_k_new not equal to C_k_old?

Kurda.

### kurd...@yahoo.com

Aug 22, 2005, 5:46:09 PM8/22/05
to
Hi Gregory,

System that I am speaking about is a large molecule (about 50 atoms),
and harmonic oscillators correspond to normal modes. System is not in
equilibrium (in the sense of atomic position) partially because of the
influence of the solvent. It means that geometry of the molecule is
changed and as a consequence frequency of normal modes is also changed.
I cannot said that change of frequency (force constant, potential
energy) is vary fast, moreover it is very slow in comparison to the
period of normal mode vibration. Therefore, I do not think that after
force constant is changed the wave function will be the same, it will
somehow follow changes. Moreover, I cannot say that oscillator performs
some work which changes potential energy (in terms of your example,
movement of particle is not a reason why potential energy (width of
box) was changed). One of the problems is that something that changes
potential energy of oscillator can also change the total energy of
oscillator. But I do not understand, what is a mechanism of these
changes? Te only way how particle can "see" the external world is the
time dependent potential energy which was already taken into account. I
suppose that I need to solve time dependent SchrÃ¶dinger equation and I
think that in this case total energy will be automatically constant,
isn't?

Kurda.

### Gregory L. Hansen

Aug 23, 2005, 10:05:37 AM8/23/05
to

I think your question can be answered simply on the conceptual level-- the
molecule exchanges energy with the solvent. If your wavefunction doesn't
include the solvent, then it won't include the energy of the solvent.

The potential well problem I gave is easy to work, but not very realistic.
The other one was barely more realistic, but still solvable. That's the
way it usually goes. We can hope that a simple problem gives some insight
into a more complicated one, even if it doesn't solve it. The sudden
approximation apparantly isn't valid in your problem. Maybe the
adiabatic aproximation is good enough. That is, if we go back to the
potential well problem, the walls move away slowly enough that we can say
the particle is always in the ground state. If you're between those two
extremes, it's hard.

You don't need to solve the time-dependent Schroedinger equation unless
you want to work out the time-dependent behavior. If you just want to see
energy conserved you can solve the time-independent equation for a
wavefunction that includes the molecule and every molecule of solvent.
But large molecules in solvents are complicated systems, and I'm not very
good at working with them.

--
"The average person, during a single day, deposits in his or her underwear
an amount of fecal bacteria equal to the weight of a quarter of a peanut."
-- Dr. Robert Buckman, Human Wildlife, p119.

### Zigoteau

Aug 24, 2005, 9:11:45 AM8/24/05
to

Hi, Kurda,

> As far as I know different interpretations of QM are vary often just a
> matter of wording (searching of appropriate combination of words) since
> on the level of experiments they give the some results. However, I
> agree that sometimes different formulations predict different results
> of experiments and such experiments are rather tricky (complicated) and
> were performed only very recently (something related with EPR paradox).
> What I want to say is that measurement of energy distribution should
> not be the case where different interpretations are mater. May be I am
> wrong.

There are two cases where the Copenhagen interpretation (CI) of QM
gives results in contradiction to experiment. My favorite is Hund's
paradox. The Hamiltonian of a chemical molecule is for all practical
purposes invariant with respect to reflection in a plane. The are no
observable differences between the properties of chiral entantiomers.
Even if the weak interaction is chiral, there is no way an enantiomer
represents a pure eigenstate of the Hamiltonian. Yet it is possible to
prepare a sample of essentially pure enantiomer, and the sample does
not racemize on observation of its chirality e.g. by measuring its
rotatory power, or by measuring the interaction of a single molecule
with e.g. a suitable antibody.

The second case was recently published by Shahriar Afshar, both in the
serious and popular scientific press. His experiment was mainly
optical. You will find mention of his work by doing a Google groups
search on "Shariar Afshar" and group:sci.physics.

The Everett multiple-worlds interpretation of QM says that the
wavefunction never collapses. This way of resolving the paradox leaves
you with Schroedinger's cat. CI wants to limit the amount of structure
in the universe to the - still staggering - amount of information
conceived of in classical mechanics. However it ends up by being
solipsistic: you and everybody else in the universe are figments of my
imagination. I am quite happy that I can only ever be aware of a very
small part of the universe. I know, for example, that there are many
mobile phone conversations and TV programs going through my brain as EM
waves, of which I am totally oblivious. The fact that I am only aware
of the cat being alive, and not aware of it being dead at the same
time, just shows how limited my perception is.

> My question was about interpretation of superposition. If a system is
> in a superposition of two states we can with some probability find a
> system in one of these states. However, I think that it would be
> incorrect to say that superposition of states means that we just do not
> know in which of the two states is the considered system. It means that
> if we performed experiment and during experiment we found the system in
> the first eigenstate we still cannot say that system noe is in the
> first state (system is still in the superposition of states). The
> probability to find the system during the second measurement in the
> second eigenstate is still not equal to zero, I think. And it is true
> (what I said) under two suppositions:
> 1. superposition cannot be interpreted as lack of knowledge about the
> system,

Superposition is one thing, our knowledge about the system is another.
I take it we are talking about the real world here. Are you talking,
for example, about an atom constrained in a UHV system by optical
molasses? I think you will find that your little mathematical
expression is a vast simplification for teaching purposes only.

> 2. measurement does not influence the system.

No, that's not right. Quantum systems are very readily perturbed, and
it is not possible either to predict or to compensate for this
perturbation.

> But experiment always influences on the system and my question is
> whether what I said is true even if we do not neglect by this influence
> (interaction with meassuring device).
>
> For example:
> 1. Psi = C_1*phi_1 + C_2*phi_2 - before the first measurement.
> 2. During measurement we found that energy of the system is E_1 (what
> corresponds to state phi_1).
>
> Question:
> What is the state of the system after measurement?
>
> 1. C_k_new = C_k_old?
> 2. C_1_new =1, C_2_new=0?
> 3. C_k_new not equal to C_k_old?

Definitely 3.

Cheers,

Zigoteau.