Schrodinger equation and free particles

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SUMMARY

The discussion centers on whether the functions psi_I = A cos(kx - wt) and psi_II = A sin(kx - wt) are solutions to the Schrödinger equation for a free particle. It is established that neither function satisfies the equation when substituted, as shown through the calculations involving second derivatives and time derivatives. The participants suggest that a linear combination of the two states, expressed as psi = Cpsi_I + Dpsi_II, may yield a valid solution, prompting further exploration of the coefficients C and D.

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kehler
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Homework Statement


Show whether the functions
psi_I = A cos(kx - wt)
psi_II = A sin(kx - wt)
are solutions of Schrödinger equation for a free particle

Homework Equations


Schrödinger equation

The Attempt at a Solution


For psi_I = A cos(kx - wt),
d2psi_I/dx2 = -Ak2psi[/SUB]I[/SUB]
dpsi_I/dt = Awsin(kx - wt)
Substituting into S.E,
ih_bar Awsin(kx - wt) = ((h_bar)2/2m) Ak2psi_I + 0 psi (free particle so V=O)
So it doesn't satisfy the S.E.

For psi_II = A sin(kx - wt),
d2psi_II/dx2 = -Ak2psi[/SUB]II[/SUB]
dpsi_II/dt = -Awcos(kx - wt)
Substituting into S.E,
ih_bar -Awcos(kx - wt) = ((h_bar)2/2m) Ak2psi_II + 0 psi
So it doesn't satisfy the S.E either.

Is this correct? Or is there some way that I'm supposed to manipulate both sides to equal each other? :S
 
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hi kehler

as you've shown, they will not be a solution of the Time Independent SE for a free particle by themselves...

How about a linear combination of the 2 states? ie
\psi = C\psi_1 + D\psi_2

what do C & D have to staisfy to be a solution...?
 

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