Finding the Value of [A,B] When Dependent on x,y,z,t

[Karlisbad]

let be A and B 2 operator so their commutator is:

$$[A,B]=1$$

then my question is if A and B are dependant of x,y,z,t then what would be the value of commutator:

$$[\partial_{c} A , \partial_{b} B]=?$$

where c and b can be x,y,z or t..thanks...

 

[cristo]

Well, if A is independent of x^b, then $$\partial_b A = \frac{\partial A}{\partial x^b}=0 \forall b$$. Similarly for B, and so the commutator is zero.

 

[dextercioby]

One doesn't know. Iff one specifies the domain and ranges of the partially differentiated operators, can one compute the value of the commutator.

Daniel.

 

[Marco_84]

I'm not so sure, but after a fast calculus i found out that it should be zero the answer.
Using the propertie:

$$[AB,C]=A[B,C]+[A,C]B$$

and remebering that:

$$[\partial_{c}, \partial_{b}]=0$$
since b and c are indip. variables.
you reach this condition:

$$[\partial_{c}A, \partial_{b}B]=[\partial_{b}B, \partial_{c}A]=$$

wich means that:

$$[A,B]=[B,A]=AB-BA=BA-AB$$

so AB=-AB ==> i think zero is the only solution.

Bye MARCO

 

[StatMechGuy]

$$[\partial_{c}A, \partial_{b}B]=[\partial_{b}B, \partial_{c}A]$$

The commutator product is antisymmetric under interchange, no symmetric, so everything following this point is incorrect. I worked out the algebra (took about a page) and found that

$$\left [ \partial_c A, \partial_b A \right ] = \partial_c \left ( \left [ A, \partial_b \right ] B \right ) + \partial_b \left ( \left [ B, \partial_c \right ] A \right )$$

although I might have made a computational error.

 

[Marco_84]

The commutator product is antisymmetric under interchange, no symmetric, so everything following this point is incorrect. I worked out the algebra (took about a page) and found that

$$\left [ \partial_c A, \partial_b A \right ] = \partial_c \left ( \left [ A, \partial_b \right ] B \right ) + \partial_b \left ( \left [ B, \partial_c \right ] A \right )$$

although I might have made a computational error.

i know is antysemmetric but if you keep expanding your calculus you find
a symmetric relation which tells you that the commutator must be zero since:
[A,B]=-[B,A] and also [A,B]=[B,A]

 

[dextercioby]

The commutator product is antisymmetric under interchange, no symmetric, so everything following this point is incorrect. I worked out the algebra (took about a page) and found that

$$\left [ \partial_c A, \partial_b A \right ] = \partial_c \left ( \left [ A, \partial_b \right ] B \right ) + \partial_b \left ( \left [ B, \partial_c \right ] A \right )$$

although I might have made a computational error.

I see you imply that

$$\partial_{\mu}[A,B]=\left[\partial_{\mu}A, B\right] +\left[A,\partial_{\mu}B\right]$$

,which is completely incorrect.

Daniel.

 

[StatMechGuy]

I see you imply that

$$\partial_{\mu}[A,B]=\left[\partial_{\mu}A, B\right] +\left[A,\partial_{\mu}B\right]$$

,which is completely incorrect.

Daniel.

I'm not sure how I imply that at all. Lemme go through my steps, maybe you can find my error.

$$[ \partial_c A, \partial_b B ] & = & \partial_c [A, \partial_b B] + [\partial_c, \partial_b B] A$$
$$& = & - \left \{ \partial_c[\partial_b B, A] + [\partial_b B, \partial_c] A \right \} + \left \{ \partial_b [B, \partial_c] + [\partial_b, \partial_c] B \right \} A$$
$$& = & - \partial_c \partial_b [B, A] - \partial_c([\partial_b, A] B) + \partial_b ([B, \partial_c ] A)$$
$$& = & - \partial_c ([\partial_b, A] B) + \partial_b([B, \partial_c] A)$$

 

[dextercioby]

Let's see whether i got the problem right. It happened to me a few days to get a problem wrong and therefore make erroneous statements.

1.What is $[,]$ supposed to mean ?
2.Who is A/B ?
3. Why can you take out the (possibly) space(-time) derivative from the bracket structure ?

Daniel.

 

[Marco_84]

Maybe I've made some mistake during the calculus.
Well i bet that the problem is that we don't know in which space A and B are defined. But if the space is L^2(RN) i hope u understand what i mean. We can use Von-Neumann theorem which say:
if two self-adjoint oerator A and B have the Heisenberg commutation rules, there exists u:
q=uAù and p=uBù.

so we can treat the problem using all we know about the position and the momentum operator.
Sorry i forgot to tell you that u is a unitary trasformation so that
uù=ùu=1.

 

[StatMechGuy]

Let's see whether i got the problem right. It happened to me a few days to get a problem wrong and therefore make erroneous statements.

1.What is $[,]$ supposed to mean ?
2.Who is A/B ?
3. Why can you take out the (possibly) space(-time) derivative from the bracket structure ?

Daniel.

1. Commutator brackets
2. I suppose two arbitrary operators
3. I'm using commutator identities to "pull those things out".

 

[dextercioby]

1. Commutator brackets
2. I suppose two arbitrary operators
3. I'm using commutator identities to "pull those things out".

1. Thought so myself.
2. Acting on what kinda space ?
3. If you make an assumption about 2., then i can argue against 3.

Daniel.

 

[dextercioby]

Maybe I've made some mistake during the calculus.
Well i bet that the problem is that we don't know in which space A and B are defined. But if the space is L^2(RN) i hope u understand what i mean. We can use Von-Neumann theorem which say:
if two self-adjoint oerator A and B have the Heisenberg commutation rules, there exists u:
q=uAù and p=uBù.

so we can treat the problem using all we know about the position and the momentum operator.
Sorry i forgot to tell you that u is a unitary trasformation so that
uù=ùu=1.

In the way it's formulated, i'd say the problem is in the realms of quantum field theory, in which A and B are field operators depending on $x^{\mu}$ and the derivative $\partial_{a}$ is actually the more familiar $\partial_{\mu}$.

The sad thing is that the problem is too abstractly (read "vaguely") formulated and all we can do is make convenient assumptions, as the OP never bothered to answer to our (possible) solutions.

Daniel.

 

[Marco_84]

In the way it's formulated, i'd say the problem is in the realms of quantum field theory, in which A and B are field operators depending on $x^{\mu}$ and the derivative $\partial_{a}$ is actually the more familiar $\partial_{\mu}$.

The sad thing is that the problem is too abstractly (read "vaguely") formulated and all we can do is make convenient assumptions, as the OP never bothered to answer to our (possible) solutions.

Daniel.

Yes i know what you mean, but, you know for sure that the fock space still an Hilbert space (IT'S JUST A COUNTOUBLY ADDITION OF SUCH THOSE SPACES) so i think the theorem works good anyway.
I'm pretty sure also the question was about the relativistic wiew. It doesn't change the algebraic properties of the operators a lot. P (the momentum) still a "illimitato" operator, not bounded it's right?
what ever you say. You can pick a sequence of function (fn) in the Fock space that: (||Pfn||)/(||fn||)=infinity when n goes too infinity. where the norm/scalar product its the appropriate of the Hilbert space in exam. for example on L2(0,+infinity) fn=exp(-nx).

Bye Marco

 

[dextercioby]

Veramente giusto. Unfortunately though, using unitary transformations to bring the problem back to canonical (anti)commutation relations for the fundamental fields of the theory (i.e. analogues of the x and p in qm of systens with finite # of degrees of freedom) is not a great simplification, since the fock space is not unique and this reflects itself in different physical models for basically the same comm.relations.

Either way, this is already philosophy and not mathematics.

Daniel.

 

[marlon]

Veramente giusto. Unfortunately though, using unitary transformations to bring the problem back to canonical (anti)commutation relations for the fundamental fields of the theory (i.e. analogues of the x and p in qm of systens with finite # of degrees of freedom) is not a great simplification, since the fock space is not unique and this reflects itself in different physical models for basically the same comm.relations.

Either way, this is already philosophy and not mathematics.

Daniel.

Dexter, beware that you do not become too theoretical a physicist.
People will get scared of you. :rofl:


marlon

 

[StatMechGuy]

1. Thought so myself.
2. Acting on what kinda space ?
3. If you make an assumption about 2., then i can argue against 3.

Daniel.

What kind of space do you like? I viewed A and B as operators that were functions of x, p and t in your standard classical quantum mechanics. What else could they be?