I Simple harmonic oscillator

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  • In the series solution of simple harmonic oscillator,why do we have a factor of 1/2 in the trial solution?
 

BvU

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There's a lot more factors there. Could you be a bit more specific and post the trial solution you are referring to ?
 
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In the series solution of simple harmonic oscillator,why do we have a factor of 1/2 in the trial solution?
the factor is chosen from the asymptotic behavior of the solution of differential equation ...in the above situation as the 2nd order partial derivative should yield square of the variable times the function the factor 1/2 is chosen...
see for details
<http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/lecture-notes/MIT8_04S13_Lec08.pdf> [Broken]

however the trial functions are always chosen by anticipating the behaviour of the solution as r going to zero and infinity as it is easy to estimate. for bound states at both these extremities the wave function should go to zero.
 
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the factor is chosen from the asymptotic behavior of the solution of differential equation ...in the above situation as the 2nd order partial derivative should yield square of the variable times the function the factor 1/2 is chosen...
see for details
<http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/lecture-notes/MIT8_04S13_Lec08.pdf> [Broken]

however the trial functions are always chosen by anticipating the behaviour of the solution as r going to zero and infinity as it is easy to estimate. for bound states at both these extremities the wave function should go to zero.
But still we are left with the term e^-q^2/2
 
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PeroK

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You take ##\psi = e^{-q^2 /2}## because that's the solution. A better question is how do you get to that solution?

If you try ##\psi = e^{-q}## then ##\psi'' = \psi## so that doesn't work.

If you try ##\psi = e^{-q^2}## then ##\psi'' = (4q^2 - 2)\psi## so that doesn't quite work, but it should give you the idea:

If you try ##\psi = e^{-q^2 /2}## then ##\psi'' = (q^2 - 1)\psi## and, if you ignore the ##-1## for large ##q##, then you have a solution.
 
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You take ##\psi = e^{-q^2 /2}## because that's the solution. A better question is how do you get to that solution?

If you try ##\psi = e^{-q}## then ##\psi'' = \psi## so that doesn't work.

If you try ##\psi = e^{-q^2}## then ##\psi'' = (4q^2 - 2)\psi## so that doesn't quite work, but it should give you the idea:

If you try ##\psi = e^{-q^2 /2}## then ##\psi'' = (q^2 - 1)\psi## and, if you ignore the ##-1## for large ##q##, then you have a solution.
But if we take the 2nd case still our assumption is correct for large value of q
 

PeroK

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But if we take the 2nd case still our assumption is correct for large value of q
Really? Like ##4q^2 - 2 \approx q^2## for large ##q##?
 
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BvU

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So basically the trial solution is ##\ e^{-\alpha q^2} \ ## and the 'characteristic equation' yields ##\ 4\alpha^2q^2-2\alpha q - q^2 = 0 \ ##, to expand a little on PeroK post #8.
 

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