Velocity of qm particle in a box

  • Thread starter fk08
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  • #1
fk08
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Can I draw conclusions from the curvature of a wavefunction on the kin. energy of the particle? For instance, the lowest bound solution of a particle in the box is a half of sinus:
Psi(x) = sin(pi*x/L)

Since the second deriv. of a wavefunction is proportional to the kin. energy, this would imply that the highest kinetic energy of the particle is exactly in the middle of the box, and zero at the nodes:

d^2Psi/dx^2 = -sin(pi*x/L)
e.g for x = L---> d^2Psi/dx^2 = 0 = Ekin = p^2
 

Answers and Replies

  • #2
eaglelake
131
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Can I draw conclusions from the curvature of a wavefunction on the kin. energy of the particle? For instance, the lowest bound solution of a particle in the box is a half of sinus:
Psi(x) = sin(pi*x/L)

Since the second deriv. of a wavefunction is proportional to the kin. energy, this would imply that the highest kinetic energy of the particle is exactly in the middle of the box, and zero at the nodes:

d^2Psi/dx^2 = -sin(pi*x/L)
e.g for x = L---> d^2Psi/dx^2 = 0 = Ekin = p^2

In the infinitely deep square well (a box) the total energy is the kinetic energy and, therefore, [tex]\psi (x) = \sin \frac{{\pi x}}{L}[/tex] is an eigenfunction of the kinetic energy operator corresponding to the kinetic energy eigenvalue [tex]E_1 = (KE)_1 = \frac{{\hbar ^2 \pi ^2 }}{{2mL^2 }}[/tex]. In the ground state, then, a measurement of the kinetic energy always yields the value [tex](KE)_1 = \frac{{\hbar ^2 \pi ^2 }}{{2mL^2 }}[/tex]. The ground state energy is also the ground state kinetic energy, in this case. You assumed the kinetic energy to be proportional to the eigenfunction which is not correct.
 
  • #3
tom.stoer
Science Advisor
5,778
170
In order to talk about a velocity one has to define it via an operator. It seems natural to define

[tex]{\bf v}=\frac{\bf p}{m}[/tex]

or

[tex]{\bf v^2}=\frac{2{\bf E}_\text{kin}}{m}[/tex]

Then one can calculate the velocity for an arbitrary state via

[tex]\langle v \rangle = \langle\psi|{\bf v}|\psi\rangle[/tex]

or

[tex]\langle v^2 \rangle = \langle\psi|{\bf v^2}|\psi\rangle[/tex]

respectively.

This shows that one cannot associate one piece of the wave function with velocity ore something like that. Classical entities are encoded in the whole state, not only in a single "piece" of its wave function.
 

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