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kengrimwood
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(If the equation below do not appear correctly, you can read all of the question in the attached file.)
Solving the time dependent 1D Schrödinger equation, one can show that in all points (x,t),
[tex] i\bar{h}\frac{\partial}{\partial t}\Psi(x,t)=-\frac{\bar{h}^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)=E\Psi(x,t)[/tex]
for a certain value of the energy E. This means that the wave function is necessarily an eigenfunction of the hamiltonian operator.
On the other hand, we may specify an initial condition for this problem. My question is: what if the initial condition DOES NOT verify the above equations. Namely, if [tex]-\frac{\bar{h}^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)[/tex] is not proportional to [tex]\Psi(x,t)[/tex].
Does it mean that not all initial conditions will give a solution or am I wrong somewhere? Thank you for your help!
Solving the time dependent 1D Schrödinger equation, one can show that in all points (x,t),
[tex] i\bar{h}\frac{\partial}{\partial t}\Psi(x,t)=-\frac{\bar{h}^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)=E\Psi(x,t)[/tex]
for a certain value of the energy E. This means that the wave function is necessarily an eigenfunction of the hamiltonian operator.
On the other hand, we may specify an initial condition for this problem. My question is: what if the initial condition DOES NOT verify the above equations. Namely, if [tex]-\frac{\bar{h}^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)[/tex] is not proportional to [tex]\Psi(x,t)[/tex].
Does it mean that not all initial conditions will give a solution or am I wrong somewhere? Thank you for your help!
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