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$$\langle x| M|x'\rangle=M(x)\langle x|x'\rangle=M(x)\,\delta(x-x')$$

I know this is true for if $M$ is a momentum operator or position operator, is this is true for a general operator $M $?

$\langle x'|M|\psi\rangle=\langle x'|\alpha\rangle=\alpha(x')$

this is equivalent to

$$\int \langle x'|M|x\rangle\langle x|\psi\rangle\,dx.$$

If equation 1 is true we can write this as

$$\int \langle x'|M|x\rangle\langle x|\psi\rangle\,dx=\int M(x)\,\delta(x-x')\,\psi(x)\,dx=M(x')\,\psi(x')=\alpha(x')$$

Equation one is not true implies equation 2 is also not true, but actually equation 2 is true in fact this implies equation one is also true.

Hear by $M(x')$ I mean operator operates on position representation of wave function; please clarify me where I am wrong, please help me to clear my concept.

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  • $\begingroup$ You are claiming $\langle x\mid$ is an eigenstate of $M$ with eigenvalue $M(x)$. This is not necessarily true. $\endgroup$ Jun 3 '19 at 4:48
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Yes, $$\langle\ x\rvert \hat M\lvert\ x'\rangle=M(x)\langle\ x\lvert\ x'\rangle=M(x)\delta(x-x'),$$ provided you understand what you claim you do, namely $$\langle\ x\rvert \hat p\lvert\ x'\rangle=-i\hbar \partial_x \langle\ x\lvert\ x'\rangle=-i\hbar \partial_x\delta(x-x'),$$ so then that M(x) represents an operator consisting of functions of x and derivatives w.r.t. x in a specific associative order.

Proceed to verify that, for instance, indeed, $$\langle\ x\rvert f(\hat p) g(\hat x) h(\hat p) k(\hat x) \lvert\ x'\rangle \\ = k( x) h(-i\hbar \partial_x) g(x) f(-i\hbar \partial_x) \langle\ x\lvert\ x'\rangle \\ = k( x) h(-i\hbar \partial_x) g(x) f(-i\hbar \partial_x) \delta(x-x')~.$$

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  • $\begingroup$ I asked same question on physics stalk exchange they answered differently its link is physics.stackexchange.com/questions/483891/… @Cosmas Zachos $\endgroup$ Jun 4 '19 at 3:46
  • $\begingroup$ thanks for your valuable answer @Cosmas Zachos $\endgroup$ Jun 4 '19 at 3:46
  • $\begingroup$ They are reading your question too narrowly. They misconstrue M(x) as a mere function of x and not as an operator-valued function indexed by x, as is done here. The case of a single power of momentum, indeed, illustrates their reading, and, here, mine! Are we being clear? $\endgroup$ Jun 4 '19 at 13:20
  • $\begingroup$ Ur answer is very clear, But I have 2 doubts Can all operators can be written as a function of momentum operator and position operator ? What happens if 'M' is an integral operator and also why operators arranged in a specific associative order?, @ Cosmas Zachos $\endgroup$ Jun 4 '19 at 16:41
  • $\begingroup$ Yes, most operators are presumed to be strings of p s and q s. Integral operators are normally represented as inverses of derivative operators, so, then, Integral Green function kernels. recall the coordinate representation of the identity is the delta function! Coordinates and derivatives multiply each other associatively, and since they do not commute, their ordering matters. $\endgroup$ Jun 6 '19 at 14:45

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