Solving the Representation of d/dq: Mistake in Reasoning?

In summary, Dirac deduced that the representation of d/dq in "matrix" form for a system with one degree of freedom is a scalar matrix in the q representation. However, this contradicts the quantum condition pq-qp=iħ. The mistake in this line of reasoning was assuming that the derivative of the delta function is of the same form as the delta function itself. In reality, the derivative has different properties and its representation as a "matrix" would have only two diagonals with non-zero values.
  • #1
Spin One
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0
When considering the position representation of a system with one degree of freedom endowed with canonical co-ordinates and momenta q and p, Dirac deduces that:
upload_2018-8-22_9-43-41.png

which is the representation of d/dq in "matrix" form. But the derivative of a delta function is, I assume (from the definition of the delta function), of the form
upload_2018-8-22_9-45-32.png
, with ξ' and ξ" interchanged for q' and q", since the gradient of the delta function will be 0 everywhere except at q'=q", where it will be undefined.
So the representation of d/dq is a scalar "matrix" in the q representation(from the definition)
But the momentum operator is just iħd/dq, meaning it is a scalar matrix too.
q is of course also a scalar matrix (since the representation is built upon it), and since scalar matrices commute, we get that:
pq-qp=0, which contradicts the quantum condition pq-qp=iħ.

There must be a mistake in my line of reasoning, and I would much appreciate it if someone can point it out.
 

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  • #2
Spin One said:
But the derivative of a delta function is, I assume (from the definition of the delta function), of the form
upload_2018-8-22_9-45-32-png.png
, with ξ' and ξ" interchanged for q' and q", since the gradient of the delta function will be 0 everywhere except at q'=q", where it will be undefined.
This is an incorrect assumption. The derivative of the delta distribution is different from the delta distribution itself. Instead of having the property
$$
\int f(x) \delta(x) dx = f(0)
$$
as the delta distribution, it has the property
$$
\int f(x) \delta'(x) dx = - \int f'(x) \delta(x) dx = -f'(0).
$$
 

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  • #3
Orodruin said:
This is an incorrect assumption. The derivative of the delta distribution is different from the delta distribution itself. Instead of having the property
$$
\int f(x) \delta(x) dx = f(0)
$$
as the delta distribution, it has the property
$$
\int f(x) \delta'(x) dx = - \int f'(x) \delta(x) dx = -f'(0).
$$
Oh ok, thanks! I see now how my intuition was wrong, since taking the δ function in the sense of a limit of a proper function, the derivative must be 0 at δ(0), +∞ at δ(-ε) and -∞ at δ(+ε). In that case, will the representation of d/dq be a "matrix" (of the generalized kind) with only two diagonals just above and just below the main diagonal having a non-zero value?
 
Last edited:

1. What is the representation of d/dq and why is it important in solving problems?

The representation of d/dq is a mathematical notation used to represent the derivative of a function with respect to a variable q. It is important in problem solving because it allows us to find the rate at which a function is changing at a specific point, which is crucial in many scientific and mathematical applications.

2. What is the most common mistake in reasoning when dealing with d/dq?

The most common mistake is forgetting to apply the chain rule when differentiating composite functions. This can lead to incorrect results and should be carefully checked when solving problems involving d/dq.

3. How can one avoid making mistakes when working with d/dq?

Practice and careful attention to detail are key in avoiding mistakes when working with d/dq. It is also important to double check all steps and make sure the correct rules and formulas are being applied.

4. Are there any specific techniques or tips for solving problems involving d/dq?

Yes, it is helpful to break down the problem into smaller steps and use the chain rule and other derivative rules as needed. It can also be useful to draw diagrams or graphs to visualize the problem and better understand the problem and its solution.

5. Can d/dq be applied to any type of function?

Yes, d/dq can be applied to any type of function as long as it is continuous and differentiable. However, the methods and techniques used may vary depending on the type of function being differentiated.

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