10 views

Skip to first unread message

Aug 23, 2006, 5:23:31 PM8/23/06

to

Does anyone have any ideas about what the commutation

relation of [x_i, P*P] should be for a single,

relativistic particle (e.g. an electron)?

relation of [x_i, P*P] should be for a single,

relativistic particle (e.g. an electron)?

I've been arriving at an apparent contradiction, and

I'm wondering if anyone has any thoughts on this.

P is an operator for the momentum 4-vector

x_i is the i_th component of the position operator.

All vectors are 4-vectors, but I'm using Latin indices

instead of the conventional Greek here.

According to the relativity, P*P = (mc)^2, where m is

the particle's mass. Since m is a scalar constant,

[x_i, P*P] = [x_i, (mc)^2] = 0.

However, if I try to expand the commutation relation

by its components, I get the following:

[x_i, P*P] = [x_i, P_j*P_j]

= [x_i, P_j]P_j + P_j [x_i, P_j]

= i*h-bar (d_ij * P_j + d_ij * P_j)

= i*h-bar * P_i

I have taken the canonical commutation relation

[x_i,p_j] = i*h-bar*d_ij from non-relativistic quantum

mechanics and assumed it still applies in the

relativistic case.

d_ij is the Kronecker-delta.

Anyways, the components approach does not give 0.

Does anyone have know why the two approaches do not

agree?

It seems to me that either mass is indeed a

non-commuting operator in relativistic quantum

mechanics, or I have to use a different canonical

commutation relation.

Thanks.

__________________________________________________

Do You Yahoo!?

Tired of spam? Yahoo! Mail has the best spam protection around

http://mail.yahoo.com

Aug 23, 2006, 6:44:28 PM8/23/06

to

On 2006-08-23, Physics Student <phy_s...@yahoo.com> wrote:

> Does anyone have any ideas about what the commutation

> relation of [x_i, P*P] should be for a single,

> relativistic particle (e.g. an electron)?

>

> I've been arriving at an apparent contradiction, and

> I'm wondering if anyone has any thoughts on this.

>

> P is an operator for the momentum 4-vector

>

> x_i is the i_th component of the position operator.

>

> All vectors are 4-vectors, but I'm using Latin indices

> instead of the conventional Greek here.

> Does anyone have any ideas about what the commutation

> relation of [x_i, P*P] should be for a single,

> relativistic particle (e.g. an electron)?

>

> I've been arriving at an apparent contradiction, and

> I'm wondering if anyone has any thoughts on this.

>

> P is an operator for the momentum 4-vector

>

> x_i is the i_th component of the position operator.

>

> All vectors are 4-vectors, but I'm using Latin indices

> instead of the conventional Greek here.

> According to the relativity, P*P = (mc)^2, where m is

> the particle's mass. Since m is a scalar constant,

>

> [x_i, P*P] = [x_i, (mc)^2] = 0.

>

> However, if I try to expand the commutation relation

> by its components, I get the following:

>

> [x_i, P*P] = [x_i, P_j*P_j]

>= [x_i, P_j]P_j + P_j [x_i, P_j]

>= i*h-bar (d_ij * P_j + d_ij * P_j)

>= i*h-bar * P_i

>

> I have taken the canonical commutation relation

> [x_i,p_j] = i*h-bar*d_ij from non-relativistic quantum

> mechanics and assumed it still applies in the

> relativistic case.

>

> d_ij is the Kronecker-delta.

The 4-vector x_i is made up of three operators x_1, x_2, x_3, and one

scalar variable t. While, the 4-vector P_i is made up of the three

operators p_1, p_2, p_3, and the operator H/c = sqrt((mc)^2 + p^2),

where p^2 = p_1^2 + p_2^2 + p_3^2. The x_i and p_i operators, for

i=1,2,3, even in the relativistic case are still canonically conjugate

pairs so the still have the commutation relations [x_i,p_j] = i*hbar*d_ij.

However, neither t nor H are independent observables. The former is just

an independent scalar variable, while the latter is a function of the

p_i. If it makes sense to you to consider the commutation relation

[t,P*P], then it's safe to say that it vanishes, since t is not an

operator. To complete the calculation you are interested in, you should

compute [x_i,P*P] = [x_i, H^2-p^2] with the standard canonical

commutation relations. You should find 0, as you expected.

Hope this helps.

Igor

Aug 23, 2006, 6:44:28 PM8/23/06

to

Hi Physics Student,

Your mistake is in the assumption

[x_i, P_0] = d_i0 = 0

In fact, P_0 is operator of energy (Hamiltonian), and the

correct commutator is

[x_i, P_0] = P_i/P_0 = i*h-bar V_i (velocity)

(note that here i=1,2,3, because there is no such thing

as operator of time t = x_0)

Then the commutator of x_i with P*P is, as expected,

[x_i, P*P] = [x_i, P_0^2 - P_i^2]

= 2*i*h-bar P_i - 2*i*h-bar P_i

= 0

Eugene.

Aug 27, 2006, 4:22:34 PM8/27/06

to

Thanks guys for responding. The two responses I've

received so far seem to say essentially the same

thing: a) t is an independent scalar variable and so

should commute with anything; and b) the P_0 component

is a function of spatial momenta and should therefore

NOT commute with spatial x_i. Therefore, the

commutation relations for [x_i, P_j] in space-time

should consist of three parts:

1) [x_i, P_j] = i*h-bar*d_ij, for i and j = 1,2,3

2) [x_0, P_j] = 0

3) [x_i, P_0] = not zero, where i = 1,2,3

received so far seem to say essentially the same

thing: a) t is an independent scalar variable and so

should commute with anything; and b) the P_0 component

is a function of spatial momenta and should therefore

NOT commute with spatial x_i. Therefore, the

commutation relations for [x_i, P_j] in space-time

should consist of three parts:

1) [x_i, P_j] = i*h-bar*d_ij, for i and j = 1,2,3

2) [x_0, P_j] = 0

3) [x_i, P_0] = not zero, where i = 1,2,3

By setting part (3) appropriately, I should get

[x_i, P*P] = 0.

This makes perfect sense to me, but there's one small

issue I have with the [x_i, P_j] when defined in this

way. Namely, it's no longer an isotropic rank-(1,1)

tensor, and when Lorentz-transformed to a new frame,

the [x'_i, P'_j] no longer obeys any of the three

properties listed above. Am I correct then in

concluding that the commutator is frame-dependent and

is therefore not a tensor?

The commutator appears to be composed of the outer

product of two 4-vectors, and mathematically, this

should make it a tensor. Or is it that x_i and P_j

are not tensors when in their "operator form"?

Aug 28, 2006, 4:16:47 AM8/28/06

to

Physics Student wrote:

> Thanks guys for responding. The two responses I've

> received so far seem to say essentially the same

> thing: a) t is an independent scalar variable and so

> should commute with anything; and b) the P_0 component

> is a function of spatial momenta and should therefore

> NOT commute with spatial x_i. Therefore, the

> commutation relations for [x_i, P_j] in space-time

> should consist of three parts:

> 1) [x_i, P_j] = i*h-bar*d_ij, for i and j = 1,2,3

> 2) [x_0, P_j] = 0

> 3) [x_i, P_0] = not zero, where i = 1,2,3

>

> By setting part (3) appropriately, I should get

> [x_i, P*P] = 0.

>

> This makes perfect sense to me, but there's one small

> issue I have with the [x_i, P_j] when defined in this

> way. Namely, it's no longer an isotropic rank-(1,1)

> tensor, and when Lorentz-transformed to a new frame,

> the [x'_i, P'_j] no longer obeys any of the three

> properties listed above. Am I correct then in

> concluding that the commutator is frame-dependent and

> is therefore not a tensor?

> Thanks guys for responding. The two responses I've

> received so far seem to say essentially the same

> thing: a) t is an independent scalar variable and so

> should commute with anything; and b) the P_0 component

> is a function of spatial momenta and should therefore

> NOT commute with spatial x_i. Therefore, the

> commutation relations for [x_i, P_j] in space-time

> should consist of three parts:

> 1) [x_i, P_j] = i*h-bar*d_ij, for i and j = 1,2,3

> 2) [x_0, P_j] = 0

> 3) [x_i, P_0] = not zero, where i = 1,2,3

>

> By setting part (3) appropriately, I should get

> [x_i, P*P] = 0.

>

> This makes perfect sense to me, but there's one small

> issue I have with the [x_i, P_j] when defined in this

> way. Namely, it's no longer an isotropic rank-(1,1)

> tensor, and when Lorentz-transformed to a new frame,

> the [x'_i, P'_j] no longer obeys any of the three

> properties listed above. Am I correct then in

> concluding that the commutator is frame-dependent and

> is therefore not a tensor?

Yes.

> The commutator appears to be composed of the outer

> product of two 4-vectors, and mathematically, this

> should make it a tensor. Or is it that x_i and P_j

> are not tensors when in their "operator form"?

While the commutator does have this appearance superficially, it really

is not a product of two 4-vectors. You have to remember that the x_i

are not space-time coordinates, they are canonical coordinates on the

phase space of a relativistic particle. The same can be said about P_j.

Fortunately, the quadruple (P_0,P_j), where P_0 = sqrt((mc)^2+P_j*P_J),

does transform like a Minkowski 4-vector under Lorentz transformations.

Unfortunately, the same is not true for the quadruple (t,x_i). So the

answer to your question is that x_i is not a tensor, neither as a

classical canonical observable nor as a quantum operator.

To be more explicit, the canonical observables x_i represent the

spatial coordinates (in a chosen frame) of the particle at t=0 (with

the time coordinate t taken from the same frame). Intuitively, it

already seems wrong to put together a certain time t and the x_i, which

are associated to t=0. But we can also explicitly check the

transformation properties of x_i and see that they are not what we'd

expect for the spatial part of a 4-vector.

First, suppose that (t,x_i) does form a Minkowski 4-vector. Then, under

an infinitesimal boost of magnitude b along the z-axis, its components

transform as

dt = b z,

dz = b t, (*)

dx = dy = 0.

On the other hand, treating x_i as the intersection point of the

particle's trajectory with the t=0 hyperplane, applying the same

infinitesimal boost to the space-time coordinates (hence changing the

t=0 hyperplane and the intersection point), we find that (derivation

left as an instructive exercise)

dt = 0,

dx_i = -b v_i z. (**)

Here v_i is the particle's velocity at t=0 (measured in the same

frame). Also, dt = 0 since t, being just a number, is proportional to

the constant function on the phase space, which obviously transforms

trivially as the phase space is mapped onto itself by the boost. Quite

clearly, the transformation properties (**) are not the same as (*),

which we would expect for a Minkowski 4-vector. Similarly, you can also

check that the quadruple (t, x_i + t v_i), which gives the space-time

coordinates of the particle at time t (again, in the chosen frame),

does not transform like a 4-vector either.

Hope this helps.

Igor

Aug 28, 2006, 4:16:46 AM8/28/06

to

PhysicsStudent asked:

> Am I correct then in concluding that the

> commutator is frame-dependent and is

> therefore not a tensor?

You need the invariant form of the commutation

relations, involving (time derivative of) the Pauli-Jordan fn.

An introduction to this can be found in Greiner & Reinhardt's

"Field Quantization", starting on p100. It's not precisely what you

were asking since they talk in the context of QFT,

rather than single-particle RQM, but it might give

you the general idea.

HTH.

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu