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슈레딩거 방정식에서의 i의 의미?

익명(1.235) 2025-03-29 06:54:00 추천 0

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https://youtu.be/uVKMY-WTrVo?si=RiV5iCPEF4LICo9G

What is the i really doing in Schrödinger's equation?

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The reason why multiplying j means a differentiation? because we have introduced the phase concept. We do not use cos(x) but cos(x) +jsin(x). and jsin(x) is not natural and added for nothing?. According to Euler cos(x) +jsin(x) is e^(jx) and. The real value is only cos(x) . but multiplying j makes the value into jcos(x)-sin(x). real number part has switched from cos(x) into -sin(x) and it means a differentiation. if we make a differentiation on e^(jx) it is je^(jx).. So multiplying j is a one time differntiation in respect to x. So multiplying complex unit means a one-time-more differentiation. and it is same to the classical wave equation that the ratio of two time differentiation in respect to time with the Laplacian is wave velocity square.

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