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Canonical Commutation: Is there a flaw in this logic, and if so, where?
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Jay R. Yablon
May 20, 2012, 10:01:38 AM5/20/12
to
A few years ago I spent a lot of time dealing with the perplexity that the
Canonical Commutation Relation between the space operators X^i and the
momentum operators P^j (in h-bar=1 units):
[X^i, P^j] = i delta^ij with i,j=1,2,3 (1)
does not form a covariant relationship with time x^0=t and energy p^0=E ,
i.e., that i and j <> 0 in (1). And, of course, energy is an eigenvalue not
an operator, and when I last looked, there was no known time operator.
But, we do know as a general rule, in the Heisenberg picture, that a dynamical
operator variable O commutes with the Hamiltonian H according to :
[O, H] = i dO/dt (2)
And, of course, the Hamiltonian, which has energy Eigenvalues, H|psi> =
E|psi>, transforms as a time component in generally-covariant equations.
So, setting aside how to construct this, suppose we have a time operator T
which, like the space operators, admits of a continuous spectrum of
eigenvalues T|psi>=t|psi> because time is continuous (I am using
capitalization for operators and lower case for eigenvalues / numbers).
Setting O=T, then (2) becomes:
[T, H] = i dT/dt (3)
If we can find a way to set dT/dt = "1," then (3) can be turned into:
[T, H] = i (4)
and can be put into a covariant relationship with (1). So, how to we create a
dT/dt = "1"?
Starting with T|psi>=t|psi>, we can differentiate to obtain
dT|psi>=dt|psi> at least as an operator equation. Then we can write:
(dT/dt)|psi>="1"|psi> (5)
Which would say that all the eigenvalues of dT/dt would have to be 1, and
perhaps this gives us a shoehorn into how to construct the time operator T.
Were we to achieve this, then we could define the Hamiltonian four-vector
formed as H^u == (H, P^1, P^2, P^3) and perhaps extend (1) to read:
[X^u, H^v] = i delta^uv with u,v=0,1,2,3 (6)
The 0k and k0 relationships introduced by this would then be:
[T, P^k] = 0 (7)
and
[X^k, H] = 0 (8)
Am a barking up a reasonable tree? If there is a flaw here, where is it?
Thanks,
Jay
_________________________________________________
Jay R. Yablon
910 Northumberland Drive
Schenectady, New York 12309-2814
Phone / Fax: 518-377-6737
Email: jya...@nycap.rr.com
Co-moderator: sci.physics.foundations
Blog: http://jayryablon.wordpress.com/
Canonical Commutation Relation between the space operators X^i and the
momentum operators P^j (in h-bar=1 units):
[X^i, P^j] = i delta^ij with i,j=1,2,3 (1)
does not form a covariant relationship with time x^0=t and energy p^0=E ,
i.e., that i and j <> 0 in (1). And, of course, energy is an eigenvalue not
an operator, and when I last looked, there was no known time operator.
But, we do know as a general rule, in the Heisenberg picture, that a dynamical
operator variable O commutes with the Hamiltonian H according to :
[O, H] = i dO/dt (2)
And, of course, the Hamiltonian, which has energy Eigenvalues, H|psi> =
E|psi>, transforms as a time component in generally-covariant equations.
So, setting aside how to construct this, suppose we have a time operator T
which, like the space operators, admits of a continuous spectrum of
eigenvalues T|psi>=t|psi> because time is continuous (I am using
capitalization for operators and lower case for eigenvalues / numbers).
Setting O=T, then (2) becomes:
[T, H] = i dT/dt (3)
If we can find a way to set dT/dt = "1," then (3) can be turned into:
[T, H] = i (4)
and can be put into a covariant relationship with (1). So, how to we create a
dT/dt = "1"?
Starting with T|psi>=t|psi>, we can differentiate to obtain
dT|psi>=dt|psi> at least as an operator equation. Then we can write:
(dT/dt)|psi>="1"|psi> (5)
Which would say that all the eigenvalues of dT/dt would have to be 1, and
perhaps this gives us a shoehorn into how to construct the time operator T.
Were we to achieve this, then we could define the Hamiltonian four-vector
formed as H^u == (H, P^1, P^2, P^3) and perhaps extend (1) to read:
[X^u, H^v] = i delta^uv with u,v=0,1,2,3 (6)
The 0k and k0 relationships introduced by this would then be:
[T, P^k] = 0 (7)
and
[X^k, H] = 0 (8)
Am a barking up a reasonable tree? If there is a flaw here, where is it?
Thanks,
Jay
_________________________________________________
Jay R. Yablon
910 Northumberland Drive
Schenectady, New York 12309-2814
Phone / Fax: 518-377-6737
Email: jya...@nycap.rr.com
Co-moderator: sci.physics.foundations
Blog: http://jayryablon.wordpress.com/
Anon E. Mouse
May 22, 2012, 4:27:54 AM5/22/12
to
> If we can find a way to set dT/dt = "1," then (3) can be turned into:
>
> [T, H] = i (4)
>
> and can be put into a covariant relationship with (1). So, how to we create a
> dT/dt = "1"?
>
> Starting with T|psi>=t|psi>, we can differentiate to obtain
> dT|psi>=dt|psi> at least as an operator equation. Then we can write:
>
> (dT/dt)|psi>="1"|psi> (5)
>
> Which would say that all the eigenvalues of dT/dt would have to be 1, and
> perhaps this gives us a shoehorn into how to construct the time operator T.
I am quite certain I do not completely follow your posting.
>
> [T, H] = i (4)
>
> and can be put into a covariant relationship with (1). So, how to we create a
> dT/dt = "1"?
>
> Starting with T|psi>=t|psi>, we can differentiate to obtain
> dT|psi>=dt|psi> at least as an operator equation. Then we can write:
>
> (dT/dt)|psi>="1"|psi> (5)
>
> Which would say that all the eigenvalues of dT/dt would have to be 1, and
> perhaps this gives us a shoehorn into how to construct the time operator T.
Having admitted that as a preface, I think that setting dT/dt = i = 1
describes a four vector for normal rectilinear Euclidean space with no
dilation. A Newtonian space-time. A solid foundation, but not an
integrative solution.
Jos Bergervoet
May 22, 2012, 9:05:44 PM5/22/12
to
On 5/20/2012 4:01 PM, Jay R. Yablon wrote:
...
"multplication by t"? That would give similar
results as multiplication by x, y, or z, which
give the well-known nice commutation relations
with p_x, p_y and p_z.
> and can be put into a covariant relationship with (1). So, how to we create a
> dT/dt = "1"?
The choice I suggested gives you exactly that
relation! Somehow I feel that you left out an
extra requirement you implicitly are making..
Multiplication by t does exactly what you ask
for!
--
Jos
...
> Setting O=T, then (2) becomes:
>
> [T, H] = i dT/dt (3)
>
> If we can find a way to set dT/dt = "1," then (3) can be turned into:
>
> [T, H] = i (4)
So why don't you choose your operator T as
>
> [T, H] = i dT/dt (3)
>
> If we can find a way to set dT/dt = "1," then (3) can be turned into:
>
> [T, H] = i (4)
"multplication by t"? That would give similar
results as multiplication by x, y, or z, which
give the well-known nice commutation relations
with p_x, p_y and p_z.
> and can be put into a covariant relationship with (1). So, how to we create a
> dT/dt = "1"?
relation! Somehow I feel that you left out an
extra requirement you implicitly are making..
Multiplication by t does exactly what you ask
for!
--
Jos
Hendrik van Hees
May 23, 2012, 12:01:41 PM5/23/12
to
On 23/05/12 03:05, Jos Bergervoet wrote:
>> [T, H] = i (4)
>
> So why don't you choose your operator T as
> "multplication by t"? That would give similar
> results as multiplication by x, y, or z, which
> give the well-known nice commutation relations
> with p_x, p_y and p_z.
>
>> and can be put into a covariant relationship with (1). So, how to we create a
>> dT/dt = "1"?
>
> The choice I suggested gives you exactly that
> relation! Somehow I feel that you left out an
> extra requirement you implicitly are making..
>
> Multiplication by t does exactly what you ask
> for!
If you could describe time as an observable in quantum theory, then it
>> [T, H] = i (4)
>
> So why don't you choose your operator T as
> "multplication by t"? That would give similar
> results as multiplication by x, y, or z, which
> give the well-known nice commutation relations
> with p_x, p_y and p_z.
>
>> and can be put into a covariant relationship with (1). So, how to we create a
>> dT/dt = "1"?
>
> The choice I suggested gives you exactly that
> relation! Somehow I feel that you left out an
> extra requirement you implicitly are making..
>
> Multiplication by t does exactly what you ask
> for!
must fulfill Eq. (4) since by definition the Hamiltonian generates time
translations. Of course, then you could define H as -i \partial_t and T
as the multiplication with t, if acting on wave functions (e.g., in
position or momentum representation). Up to the breaking of the usual
sign convention this looks good at first glance, but it is bad for (at
least) two reasons.
If you assume time to be an observable and obeying the above commutation
relation then by the same arguments as for position and momentum
operators in the Heisenberg algebra [x_i,p_j]=i delta_{ij} one has whole
R as the spectrum of H, and thus energy wouldn't be bounded from below,
and there was no stable matter, contrary to everyday experience.
Moreover, I do not see, how you could get any real dynamical description
of quantum systems, since it is not clear, which equation you would
solve. The equation of motion i\partial_t psi(t,x)=H psi(t,x) would be a
tautology (despite the fact that with our sign choice of (4) I'd have to
write -i\partial_t on the left-hand side).
In usual nonrelativistic quantum mechanics, you rather start from the
Galileo group and construct unitary ray representations, leading to the
usual non-relativistic quantum theory with time as a parameter, not an
observable.
--
Hendrik van Hees
Frankfurt Institute of Advanced Studies
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/
carlip...@physics.ucdavis.edu
May 23, 2012, 4:42:04 PM5/23/12
to
Jay R. Yablon <jya...@nycap.rr.com> wrote:
[...]
exist in a theory in which the energy is unbounded below. To see this,
note that the unitary operator U=e^{-iaT} would be an energy translation
operator (just as e^{iap} is a position translation operator); acting on
any energy eigenstate with energy E, it would create a new state with
energy E-a.
There is also a theorem by Unruh and Wald, Phys. Rev. D 40 (1989) 2598,
that in a theory in which energy is bounded below, there is no variable
that even increases monotonically with Schrodinger time t: that is, any
"clock" has a finite probability of sometimes running backwards relative
to t.
It's conceivable that there is a way to escape this in quantum gravity,
where in some sense the total energy is zero for a physical state (because
gravitational potential energy balances all other forms of energy). But
that would require a quantum theory of gravity. Short of that, I'm afraid
it can't be done.
(Note that relativistic quantum theory takes the opposite approach --
rather than promoting t to and operator, it demotes x to a parameter.)
Steve Carlip
[...]
> So, setting aside how to construct this, suppose we have a time operator T
> which, like the space operators, admits of a continuous spectrum of
> eigenvalues T|psi>=t|psi> because time is continuous (I am using
> capitalization for operators and lower case for eigenvalues / numbers).
> Setting O=T, then (2) becomes:
> [T, H] = i dT/dt (3)
> If we can find a way to set dT/dt = "1," then (3) can be turned into:
> [T, H] = i (4)
> and can be put into a covariant relationship with (1). So, how to we create a
> dT/dt = "1"?
As Hendrik van Hees wrote, a self-adjoint operator obeying (4) can only
> which, like the space operators, admits of a continuous spectrum of
> eigenvalues T|psi>=t|psi> because time is continuous (I am using
> capitalization for operators and lower case for eigenvalues / numbers).
> Setting O=T, then (2) becomes:
> [T, H] = i dT/dt (3)
> If we can find a way to set dT/dt = "1," then (3) can be turned into:
> [T, H] = i (4)
> and can be put into a covariant relationship with (1). So, how to we create a
> dT/dt = "1"?
exist in a theory in which the energy is unbounded below. To see this,
note that the unitary operator U=e^{-iaT} would be an energy translation
operator (just as e^{iap} is a position translation operator); acting on
any energy eigenstate with energy E, it would create a new state with
energy E-a.
There is also a theorem by Unruh and Wald, Phys. Rev. D 40 (1989) 2598,
that in a theory in which energy is bounded below, there is no variable
that even increases monotonically with Schrodinger time t: that is, any
"clock" has a finite probability of sometimes running backwards relative
to t.
It's conceivable that there is a way to escape this in quantum gravity,
where in some sense the total energy is zero for a physical state (because
gravitational potential energy balances all other forms of energy). But
that would require a quantum theory of gravity. Short of that, I'm afraid
it can't be done.
(Note that relativistic quantum theory takes the opposite approach --
rather than promoting t to and operator, it demotes x to a parameter.)
Steve Carlip
Jos Bergervoet
May 24, 2012, 5:09:01 PM5/24/12
to
On 5/23/2012 6:01 PM, Hendrik van Hees wrote:
> On 23/05/12 03:05, Jos Bergervoet wrote:
...
> On 23/05/12 03:05, Jos Bergervoet wrote:
> If you could describe time as an observable in quantum theory, then it
> must fulfill Eq. (4) since by definition the Hamiltonian generates time
> translations. Of course, then you could define H as -i \partial_t and T
> as the multiplication with t,
Well, we could call H just p_0. No need to give it a
> must fulfill Eq. (4) since by definition the Hamiltonian generates time
> translations. Of course, then you could define H as -i \partial_t and T
> as the multiplication with t,
special name in that case!
> .. but it is bad for (at
> least) two reasons.
>
> If you assume time to be an observable and obeying the above commutation
> relation then by the same arguments as for position and momentum
> operators in the Heisenberg algebra [x_i,p_j]=i delta_{ij} one has whole
> R as the spectrum of H, and thus energy wouldn't be bounded from below,
> and there was no stable matter, contrary to everyday experience.
That is not completely obvious.. The assumption seems to be
>
> If you assume time to be an observable and obeying the above commutation
> relation then by the same arguments as for position and momentum
> operators in the Heisenberg algebra [x_i,p_j]=i delta_{ij} one has whole
> R as the spectrum of H, and thus energy wouldn't be bounded from below,
> and there was no stable matter, contrary to everyday experience.
that the energy would tend to go to -infinity. But (since
we're comparing the coordinates) we do not see that all
particles in our universe tend to get p_x -> -infinity!
So why do you claim that something will go wrong with time,
where it clearly is no problem with the space coordinates?!
> Moreover, I do not see, how you could get any real dynamical description
> of quantum systems, since it is not clear, which equation you would
> solve. The equation of motion i\partial_t psi(t,x)=H psi(t,x) would be a
> tautology (despite the fact that with our sign choice of (4) I'd have to
> write -i\partial_t on the left-hand side).
For instance a "z-Hamitonian", H_z, which describes the
evolution in the spatial z-direction. It would be like
using the Dirac equation but not isolating the d/dt term
(by factoring out a gamma_0) but instead isolating a d/dz.
Obviously, the result would have positive and negative p_z
solutions (as with the original Dirac energy spectrum!) but
this need not bother us now, so this choice is much better!
Conceptually, it comes down to the question whether we can
describe an "evolution" mechanism in any direction of space-
time, or only in the t-direction. The latter is of course
special because of the signature of the metric, but it is
not immediately clear (to me) what goes wrong if you take
the other choice!
--
Jos
Hendrik van Hees
May 25, 2012, 3:07:28 AM5/25/12
to
On 24/05/12 23:09, Jos Bergervoet wrote:
> On 5/23/2012 6:01 PM, Hendrik van Hees wrote:
>> On 23/05/12 03:05, Jos Bergervoet wrote:
> ...
>> If you could describe time as an observable in quantum theory, then it
>> must fulfill Eq. (4) since by definition the Hamiltonian generates time
>> translations. Of course, then you could define H as -i \partial_t and T
>> as the multiplication with t,
>
> Well, we could call H just p_0. No need to give it a
> special name in that case!
It doesn't help to rename things. You can denote your Hamiltonian with
> On 5/23/2012 6:01 PM, Hendrik van Hees wrote:
>> On 23/05/12 03:05, Jos Bergervoet wrote:
> ...
>> If you could describe time as an observable in quantum theory, then it
>> must fulfill Eq. (4) since by definition the Hamiltonian generates time
>> translations. Of course, then you could define H as -i \partial_t and T
>> as the multiplication with t,
>
> Well, we could call H just p_0. No need to give it a
> special name in that case!
whatever letter you like.
>
>> .. but it is bad for (at
>> least) two reasons.
>>
>> If you assume time to be an observable and obeying the above commutation
>> relation then by the same arguments as for position and momentum
>> operators in the Heisenberg algebra [x_i,p_j]=i delta_{ij} one has whole
>> R as the spectrum of H, and thus energy wouldn't be bounded from below,
>> and there was no stable matter, contrary to everyday experience.
>
> That is not completely obvious.. The assumption seems to be
> that the energy would tend to go to -infinity. But (since
> we're comparing the coordinates) we do not see that all
> particles in our universe tend to get p_x -> -infinity!
> So why do you claim that something will go wrong with time,
> where it clearly is no problem with the space coordinates?!
The reason is that H is the generator for time translations, i.e., it
>> .. but it is bad for (at
>> least) two reasons.
>>
>> If you assume time to be an observable and obeying the above commutation
>> relation then by the same arguments as for position and momentum
>> operators in the Heisenberg algebra [x_i,p_j]=i delta_{ij} one has whole
>> R as the spectrum of H, and thus energy wouldn't be bounded from below,
>> and there was no stable matter, contrary to everyday experience.
>
> That is not completely obvious.. The assumption seems to be
> that the energy would tend to go to -infinity. But (since
> we're comparing the coordinates) we do not see that all
> particles in our universe tend to get p_x -> -infinity!
> So why do you claim that something will go wrong with time,
> where it clearly is no problem with the space coordinates?!
governs the time evolution of the system. If the energy is not bounded
from below, there is no stable ground state. Applied to the description
of particles this would mean that matter wouldn't be stable, contrary to
everyday experience.
Momenta generate translations in space, and the Cartesian coordinates of
the position vector are indeed unbounded, and there is no problem with
stability whatsoever.
>
>> Moreover, I do not see, how you could get any real dynamical description
>> of quantum systems, since it is not clear, which equation you would
>> solve. The equation of motion i\partial_t psi(t,x)=H psi(t,x) would be a
>> tautology (despite the fact that with our sign choice of (4) I'd have to
>> write -i\partial_t on the left-hand side).
>
> We could make *another* choice for the evolution operator!
> For instance a "z-Hamitonian", H_z, which describes the
> evolution in the spatial z-direction. It would be like
> using the Dirac equation but not isolating the d/dt term
> (by factoring out a gamma_0) but instead isolating a d/dz.
Let's discuss non-relativistic quantum mechanics here. In relativistic
>> Moreover, I do not see, how you could get any real dynamical description
>> of quantum systems, since it is not clear, which equation you would
>> solve. The equation of motion i\partial_t psi(t,x)=H psi(t,x) would be a
>> tautology (despite the fact that with our sign choice of (4) I'd have to
>> write -i\partial_t on the left-hand side).
>
> We could make *another* choice for the evolution operator!
> For instance a "z-Hamitonian", H_z, which describes the
> evolution in the spatial z-direction. It would be like
> using the Dirac equation but not isolating the d/dt term
> (by factoring out a gamma_0) but instead isolating a d/dz.
quantum theory things are more complicated, particularly with respect to
the definition of position operators. Of course also there, the above
argument against the interpretation of time as an observable holds. The
existence of a stable ground state is an important constraint for model
building also in relativistic quantum theory in terms of a local causal
quantum field theory. Together with locality it determines the general
structure of physically sensible QFTs, leading to quite general
predictions as the connection between spin and statistics and CPT
invariance.
>
> Obviously, the result would have positive and negative p_z
> solutions (as with the original Dirac energy spectrum!) but
> this need not bother us now, so this choice is much better!
Sure, also the Cartesian components of momentum have an unbounded
> Obviously, the result would have positive and negative p_z
> solutions (as with the original Dirac energy spectrum!) but
> this need not bother us now, so this choice is much better!
spectrum, which is no problem either.
>
> Conceptually, it comes down to the question whether we can
> describe an "evolution" mechanism in any direction of space-
> time, or only in the t-direction. The latter is of course
> special because of the signature of the metric, but it is
> not immediately clear (to me) what goes wrong if you take
> the other choice!
Of course, time is fundamentally different from space since it
> Conceptually, it comes down to the question whether we can
> describe an "evolution" mechanism in any direction of space-
> time, or only in the t-direction. The latter is of course
> special because of the signature of the metric, but it is
> not immediately clear (to me) what goes wrong if you take
> the other choice!
parametrizes the "causality flow" of events, and that's why the metric
in relativity has its specific structure (1,3) or (3,1) in the
west-coast or east-cost notation, respectively.
Jos Bergervoet
May 31, 2012, 8:46:38 AM5/31/12
to
On 5/25/2012 9:07 AM, Hendrik van Hees wrote:
..
about the difference between x^0 and the 1,2,3-
components. Why is there a special operator for
the time direction and not for the other three?
Why can't we, alternatively, use a Hamiltonian
that governs the z-evolution?
> If the energy is not bounded
> from below, there is no stable ground state. Applied to the description
> of particles this would mean that matter wouldn't be stable, contrary to
> everyday experience.
Why? The positions of particles are not bounded
to the left or right, or any other direction. By
the same argument that would still mean that
matter isn't stable. Unless you explain why you
use different criteria for time than for space..
> Momenta generate translations in space, and the Cartesian coordinates of
> the position vector are indeed unbounded, and there is no problem with
> stability whatsoever.
So why *would* there be a problem with time?
..
just an arbitrary choice we make, picking one of the 4
coordinates..
> and that's why the metric
> in relativity has its specific structure (1,3) or (3,1) in the
> west-coast or east-cost notation, respectively.
Take the simple case of 1+1 dimensions, then you could
just as well pick the "space" coordinate to be the one
for which you construct an evolution operator. Why would
that fail? Where is the specialness of time coming from,
in that case?
And what would happen in 2+2 dimensions?
--
Jos
..
>> So why do you claim that something will go wrong with time,
>> where it clearly is no problem with the space coordinates?!
>
> The reason is that H is the generator for time translations,
> i.e., it governs the time evolution of the system.
That is circular argument, since this thread asks
>> where it clearly is no problem with the space coordinates?!
>
> The reason is that H is the generator for time translations,
> i.e., it governs the time evolution of the system.
about the difference between x^0 and the 1,2,3-
components. Why is there a special operator for
the time direction and not for the other three?
Why can't we, alternatively, use a Hamiltonian
that governs the z-evolution?
> If the energy is not bounded
> from below, there is no stable ground state. Applied to the description
> of particles this would mean that matter wouldn't be stable, contrary to
> everyday experience.
to the left or right, or any other direction. By
the same argument that would still mean that
matter isn't stable. Unless you explain why you
use different criteria for time than for space..
> Momenta generate translations in space, and the Cartesian coordinates of
> the position vector are indeed unbounded, and there is no problem with
> stability whatsoever.
..
>> Conceptually, it comes down to the question whether we can
>> describe an "evolution" mechanism in any direction of space-
>> time, or only in the t-direction. The latter is of course
>> special because of the signature of the metric, but it is
>> not immediately clear (to me) what goes wrong if you take
>> the other choice!
>
> Of course, time is fundamentally different from space since it
> parametrizes the "causality flow" of events,
But why does it? In the explanation here it seems to be
>> describe an "evolution" mechanism in any direction of space-
>> time, or only in the t-direction. The latter is of course
>> special because of the signature of the metric, but it is
>> not immediately clear (to me) what goes wrong if you take
>> the other choice!
>
> Of course, time is fundamentally different from space since it
> parametrizes the "causality flow" of events,
just an arbitrary choice we make, picking one of the 4
coordinates..
> and that's why the metric
> in relativity has its specific structure (1,3) or (3,1) in the
> west-coast or east-cost notation, respectively.
just as well pick the "space" coordinate to be the one
for which you construct an evolution operator. Why would
that fail? Where is the specialness of time coming from,
in that case?
And what would happen in 2+2 dimensions?
--
Jos
Jay R. Yablon
May 31, 2012, 2:10:25 PM5/31/12
to
"Hendrik van Hees" wrote"
>"The equation of motion i\partial_t psi(t,x)=H psi(t,x) . . ."
I now ask:
I have a separate but related question. If we create a four-vector:
H^u == (H, p_x, p_y, p_z) (1)
then is the equation:
i\partial_u psi(t,x)=H^u psi(t,x) (2)
a correct four-dimensional covariant equation?
Similarly, for an observable O? Is
i\partial_u O=[H^u, O]
a correct equation, and are there any limits to its applicability?
Jay
"Hendrik van Hees" wrote in message news:a23hce...@mid.dfncis.de...
>"The equation of motion i\partial_t psi(t,x)=H psi(t,x) . . ."
I now ask:
I have a separate but related question. If we create a four-vector:
H^u == (H, p_x, p_y, p_z) (1)
then is the equation:
i\partial_u psi(t,x)=H^u psi(t,x) (2)
a correct four-dimensional covariant equation?
Similarly, for an observable O? Is
i\partial_u O=[H^u, O]
a correct equation, and are there any limits to its applicability?
Jay
"Hendrik van Hees" wrote in message news:a23hce...@mid.dfncis.de...
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