Quantum Oscillator in 1D: How Can a Real Particle Have an Imaginary Velocity?

[RealKiller69]

I have got a simple qstion.
We have a particle in 1d oscillator with E0( fundamental level).We know that phi~ e^-x^2 for any x, so We can measure a position and get a value x=a, such that V(a)>E0 . In this case T<0 so the velocity of the particle is imaginary, how is this even possible?, (a real particle moving an imaginary velocity.)

 

[BvU]

We can measure a position and get a value x=a, such that V(a)>E0

Yes

T<0 so the velocity of the particle is imaginary

What is 'the velocity of the particle' in your context ?

 

[RealKiller69]

Yes
What is 'the velocity of the particle' in your context ?

Thats what I am trying to figure out, i can consider a wave package when the particle is in the permitted region for tht energy but if it gets out of that region ( the wave function doesn't restrict the particle in a specific region) i will get an imaginary value for the momentum. How do i interpret this thought experiment??.

 

[BvU]

In QM, the 1-D momentum operator is $\ \displaystyle { {\hbar\over i }{\partial \over \partial x }} \quad$ imaginary everywhere (*) -- no difference left or right of each of the classical turning points ...

As you found, the classically forbidden region past the turning points comes with a negative kinetic energy

Note that the expectation value for the momentum as well as for the position is zero for all eigenstates of the QM oscillator !(*) in the convention that we normalize to real amplitude coefficients. We can choose them purely imaginary, in which case the position operator $x\psi$ yields imaginary values !