I Solution of Quantum differential equation

Edge5
Messages
22
Reaction score
0
pZgfb3s

HPKZ6KD.jpg

(I think I couldn't add the image)
you can see my answer in link

https://pasteboard.co/HPKZ6KD.jpg

(Please first see my answer in the link)
But in answer it is φ= Asin(kx) + Bcos(kx)

I know that euler formula is eix = cosx +isinx

But I can't get this answer can you help me?
 
Physics news on Phys.org
In the answer key does it say if the constants A,B are real or complex?

My opinion is that if ##\phi## is complex valued then the general solution is as your answer says (and in your answer the constants A,B can be complex constants).
However if ##\phi## is real valued then the correct answer is as the answer key says that is ##\phi=A\sin(kx)+B\cos(kx)## where A,B are real constants here.
 
Delta2 said:
In the answer key does it say if the constants A,B are real or complex?

My opinion is that if ##\phi## is complex valued then the general solution is as your answer says (and in your answer the constants A,B can be complex constants).
However if ##\phi## is real valued then the correct answer is as the answer key says that is ##\phi=A\sin(kx)+B\cos(kx)## where A,B are real constants here.
Assuming answer is real. How do I get from this general answer to ##\phi=A\sin(kx)+B\cos(kx)##
 
if we assume ##\phi## is real valued then from your general answer (for which i ll use ##A'## and ##B'## to denote the complex constants as well as ##\phi'## for the complex valued function) we ll have (i use Euler's formula to rewrite your general answer).

$$\phi'=A'\cos(kx)+B'\cos(-kx)+i(A'\sin(kx)+B'\sin(-kx))=(A'+B')\cos(kx)+i(A'-B')\sin(kx) \text{(1)}$$

So we just looking for complex constants ##A',B'## such that ##(A'+B')=B (2) ## and ##(A'-B')=-iA (3)## and then for these constants it would be ##\phi=\phi'##. The system of equations (2),(3) has unique solutions for ##A',B'## given that ##A,B## are real.
 
Last edited:
  • Like
Likes Edge5
Now that i think of it again, if we allowed for ##\phi## to be complex valued (since it is a wave function it would be complex valued in general case), and also allow for constants A,B to be complex in the answer key (the answer that your book says), then the answer key and your answer are equivalent.
 
  • Like
Likes Edge5
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...
Back
Top