Relativistic Quantum Mechanics

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Physics Student

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

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.

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Igor Khavkine

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

> 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

Eugene Stefanovich

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

Physics Student

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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

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"?

Igor Khavkine

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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?

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

sr

unread,
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.

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