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I was recently studying Quantum mechanics from R.Shankar's Principles of Quantum Mechanics. I recently encountered improper vectors, and function and infinte-dimensional vectors. But I got confused at a point:
The book introduced a hermitian operator that takes the derivate of a ket $|{\psi} \rangle$ from the Hermitian operator K which is defined as: $$K_{xx'}=-i\delta'(x-x')$$ Then it solved for the eigenfunction of K:$$\psi_k(x)=Ae^{ikx}$$ where k is the eigenvalue and $\psi_k$ is the corresponding eigenfunction. Then the book proved: $$\langle k|k'\rangle=\delta(k-k')$$ here k depicts eigenfunction in vector form. The book then Introduces the X operator and does some maths(I didn't understood a thing,except that it multiplies a function by x.). Then the book says:
In the k basis, K operator just multiplies with the function with k while the X operator becomes $i\frac{d}{dx}$.
I wanted to give it a try,$$K_{kk'}=\langle k|K|k'\rangle$$ $$K_{kk'}=\int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\langle k|x\rangle\langle x|K|x'\rangle\langle x'|k'\rangle dxdx'}}$$ $$K_{kk'}=C\int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\delta(x-x')e^{-ikx}e^{ik'x'}dx'dx}}$$ $$K_{kk'}=C\int_{-\infty}^{\infty}{e^{-ikx}e^{ik'x}dx}$$ which corresponds to:$$K_{kk'}=\langle p|p'\rangle$$ Which is not accurate(The above integral was taken from the book only).The Calculation for X in k basis was even more terrible. So here is my question:

  • What is the X operator actually?
  • What is wrong with my calculations above? I hope you guys will help me, I want to know Where I am getting it wrong, I hope you ideas will help. Thanks in advance.
Quantum Mechanic
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Charu _Bamble
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    You wrote down $K_{xx'}$ wrong. It should be $K_{xx'} = -i\delta'(x-x')$. – march Dec 14 '23 at 16:30
  • What is the $\hat{X}$ operator? It is the operator that acts as multiplication by $x$ on position-space wave functions, i.e., $$ \hat{X}: \lvert{\psi}\rangle\mapsto \hat{X}\lvert\psi\rangle \Longleftrightarrow \hat{X}: \psi(x) \mapsto x\psi(x),. $$ In other words if $\psi(x)$ represents $\lvert \psi\rangle$ in position space, then $x\psi(x)$ represents $\hat{X}\lvert\psi\rangle$ in position space.

    In QM, $\hat{X}$ is an operator that corresponds to the position of a quantum particle. All of this will be spelled out in detail (you're only on Ch 1, a mathematical introduction!).

     – march Dec 14 '23 at 16:37
  • Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis. – Frobenius Dec 14 '23 at 16:57
  • @hft go easy, I edited it, my fault can revert! I was assuming the $\langle x|K|x\rangle$ used in the derivation had the correct form – Quantum Mechanic Dec 14 '23 at 19:14
  • @march it was my fault for editing – Quantum Mechanic Dec 14 '23 at 19:14
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    @QuantumMechanic Yeah, roll back the edit. My apologies to OP for assuming it was OP's fault. – hft Dec 14 '23 at 19:15
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    @QuantumMechanic In your defense, you did not introduce the typo in OP's sixth equation. That does seem to be OP's fault... – hft Dec 14 '23 at 19:19

2 Answers2

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The integral in question is exactly a definition of the Dirac delta function $$\delta(z-z^\prime)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i q(z-z^\prime)}dq.$$ I have specifically used different variables $(z, z^\prime, q)$ than the ones in the question because this is true for any variables.

When taking integrals from the book, make sure that the variables match. On one side you have $k$ and $k^\prime$ while on the other there is $p$ and $p^\prime$ - either they are the same, or you must use the definitions for how they are related to each other (and the factor of $2\pi$ might work out nicely).

Quantum Mechanic
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The last integral is equal to a Dirac delta function of k prime minus k. In position space the X operator is the multiply by x operator, while in k space it is the pure imaginary differentiate by k operator. Similarly, in k space the K operator is the multiply by k operator and in position space it is the purely imaginary differentiate by p operator. As a consequence of Fourrier mathematics, the eigenfunctions of the momentum in position space have the same form as the eigenfunctions of position in momentum space.

Albertus Magnus
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