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Kashmiri
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Cohen Tannoudji pp 215

Third postulate :The only possible result of the measurement of a physical quantity $\mathscr A$ is one of the eigenvalues of the corresponding observable $\mathbf A$.

pp 225

...there of course exists an operator associated with the velocity of the particle...

pp 223

The observable $\mathbf A$ which describes a classically defined physicallyphysical quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively

From above we can say that there exists a velocity operator $\mathbf v=\frac{\mathbf p}{m}$ ,whose eigenvalues are the observed values of velocity.

  1. I've seen multiple times that we can't define velocity in quantum mechanics, but here I find that the eigenvalues of $\mathbf v$ can be the values of velocity

  2. I've also seen that the velocity operator in Schrodinger picture is :

the $j$th velocity operator is defined as : $\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$ But we also have another expression for the velocity operator as $\mathbf v=\frac{\mathbf p}{m}$, so how do we reconcile the two.

Please help me, Thank you.

Cohen Tannoudji pp 215

Third postulate :The only possible result of the measurement of a physical quantity $\mathscr A$ is one of the eigenvalues of the corresponding observable $\mathbf A$.

pp 225

...there of course exists an operator associated with the velocity of the particle...

pp 223

The observable $\mathbf A$ which describes a classically defined physically quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively

From above we can say that there exists a velocity operator $\mathbf v=\frac{\mathbf p}{m}$ ,whose eigenvalues are the observed values of velocity.

  1. I've seen multiple times that we can't define velocity in quantum mechanics, but here I find that the eigenvalues of $\mathbf v$ can be the values of velocity

  2. I've also seen that the velocity operator in Schrodinger picture is :

the $j$th velocity operator is defined as : $\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$ But we also have another expression for the velocity operator as $\mathbf v=\frac{\mathbf p}{m}$, so how do we reconcile the two.

Please help me, Thank you.

Cohen Tannoudji pp 215

Third postulate :The only possible result of the measurement of a physical quantity $\mathscr A$ is one of the eigenvalues of the corresponding observable $\mathbf A$.

pp 225

...there of course exists an operator associated with the velocity of the particle...

pp 223

The observable $\mathbf A$ which describes a classically defined physical quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively

From above we can say that there exists a velocity operator $\mathbf v=\frac{\mathbf p}{m}$ ,whose eigenvalues are the observed values of velocity.

  1. I've seen multiple times that we can't define velocity in quantum mechanics, but here I find that the eigenvalues of $\mathbf v$ can be the values of velocity

  2. I've also seen that the velocity operator in Schrodinger picture is :

the $j$th velocity operator is defined as : $\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$ But we also have another expression for the velocity operator as $\mathbf v=\frac{\mathbf p}{m}$, so how do we reconcile the two.

Please help me, Thank you.

Source Link
Kashmiri
Kashmiri
  • 1.3k
  • 14
  • 39

Velocity operator, its expression and eigenvalues

Cohen Tannoudji pp 215

Third postulate :The only possible result of the measurement of a physical quantity $\mathscr A$ is one of the eigenvalues of the corresponding observable $\mathbf A$.

pp 225

...there of course exists an operator associated with the velocity of the particle...

pp 223

The observable $\mathbf A$ which describes a classically defined physically quantity $\mathscr A$ is obtained by replacing, in the suitably symmetrized expression for $\mathscr A, \mathbf r$ and $\mathbf p$ by the observables $\mathbf R$ and $\mathbf P$ respectively

From above we can say that there exists a velocity operator $\mathbf v=\frac{\mathbf p}{m}$ ,whose eigenvalues are the observed values of velocity.

  1. I've seen multiple times that we can't define velocity in quantum mechanics, but here I find that the eigenvalues of $\mathbf v$ can be the values of velocity

  2. I've also seen that the velocity operator in Schrodinger picture is :

the $j$th velocity operator is defined as : $\frac{1}{i\hbar}[\hat{q}^j, \hat{H}]$ But we also have another expression for the velocity operator as $\mathbf v=\frac{\mathbf p}{m}$, so how do we reconcile the two.

Please help me, Thank you.