Coordinate representation of the momentum operator

[MSLion]

The position operator in coordinate representation is:
X_ab=aδ(a-b)
this is diagonal as expected

The momentum operator turns out to look like
P_ab=-ih∂_aδ(a-b)

Now, this is not supposed to be diagonal because it does not commute with X.
However it looks pretty diagonal to me.

What am I missing?

 

[haael]

The momentum operator is not what you wrote here. It is equal to:
$$P_{ab} = e^{i(a-b)}$$
with some scaling factors.

 

[MSLion]

The momentum operator is not what you wrote here. It is equal to:
$$P_{ab} = e^{i(a-b)}$$
with some scaling factors.

What exactly are you saying?

Anyone familiar with quantum mechanics in dirac notation?

 

[geoduck]

The position operator in coordinate representation is:
X_ab=aδ(a-b)
this is diagonal as expected

The momentum operator turns out to look like
P_ab=-ih∂_aδ(a-b)

Now, this is not supposed to be diagonal because it does not commute with X.
However it looks pretty diagonal to me.

What am I missing?

I don't see how it looks diagonal. For something to be diagonal, it needs to be multiplied by a delta function. An example of a diagonal matrix is:

O_ab=δ(a-b) f(a,b)

for arbitrary function f(a,b).

The delta function is essentially the identity matrix, and f(a,b) are the diagonal components.

 

[MSLion]

From what I understood on distributions the support of the derivative of a distribution is contained within the support of the original distribution.

Therefore ∂_aδ(a-b) should be zero wherever δ(a-b) is.

 

[fzero]

From what I understood on distributions the support of the derivative of a distribution is contained within the support of the original distribution.

Therefore ∂_aδ(a-b) should be zero wherever δ(a-b) is.

Distributions are supported on functions $f(a)$, not on the domain of the functions themselves. Given a function $f(a)$, we have $\delta[f]=f(0)$ and $\delta'[f]=f'(0)$. These are generally not equal. The same argument can be made for delta functions which are shifted away from the origin.

$\delta'$ is as independent of $\delta$ as an ordinary function is from its derivative. There's no unitary operator that can map one to another (for all test functions $f$).