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Hi. I will give you a question I have looked at and then tell you where I am confused.
The wavefunction for a particle of mass m is ψ(x) = sin(kx)exp(-iωt) where k is a constant.
(i) Is this particle in a state of defined momentum ? If so , determine its momentum.
(ii) Is this particle in a state of defined energy ? If so , determine its energy.
For (i) i thought if k is a constant then p=[itex]\hbar[/itex]k must also be a constant ie a definite momentum. The solution says to apply the momentum operator to ψ which shows that ψ is not an eigenfunction of momentum. Does p=[itex]\hbar[/itex]k not always apply ? If ψ is not an eigenfunction of momentum what happens when momentum is measured ?
For (ii) i just assumed E=[itex]\hbar[/itex]ω which in this case turns out to be right although according to the solution i should have used the energy operator i[itex]\hbar[/itex]dψ/dt which gives an eigenvalue of E=[itex]\hbar[/itex]ω. Could i instead apply the Hamiltonian operator to give an energy eigenvalue of [itex]\hbar[/itex]^2k^2/2m ? Are these 2 methods equivalent ?
The wavefunction for a particle of mass m is ψ(x) = sin(kx)exp(-iωt) where k is a constant.
(i) Is this particle in a state of defined momentum ? If so , determine its momentum.
(ii) Is this particle in a state of defined energy ? If so , determine its energy.
For (i) i thought if k is a constant then p=[itex]\hbar[/itex]k must also be a constant ie a definite momentum. The solution says to apply the momentum operator to ψ which shows that ψ is not an eigenfunction of momentum. Does p=[itex]\hbar[/itex]k not always apply ? If ψ is not an eigenfunction of momentum what happens when momentum is measured ?
For (ii) i just assumed E=[itex]\hbar[/itex]ω which in this case turns out to be right although according to the solution i should have used the energy operator i[itex]\hbar[/itex]dψ/dt which gives an eigenvalue of E=[itex]\hbar[/itex]ω. Could i instead apply the Hamiltonian operator to give an energy eigenvalue of [itex]\hbar[/itex]^2k^2/2m ? Are these 2 methods equivalent ?