Momentum operator in Schrodinger equation

In summary, the momentum operator is used in place of momentum because it is an operator and all observable quantities in quantum mechanics are replaced by operators. This is why the Heisenberg Uncertainty Principle is considered a basic postulate of QM.
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Bose
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Why momentum is replaced by momentum operator in Schrodinger equation ?
 
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  • #2
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A key basic idea in QM is that the commutator of p(momentum) and q(position),

qp - pq = i (The tradition in high energy physics is to use units in which c=1, and h bar=1)

For his equation to hold, both q and p must be operators. Note also that this commutator is responsible for the Heisenberg Uncertainty Principle.

In fact this is an assumption, but a good one, and as basic as it gets. You would benefit from a bit of homework, as suggested by malawi_glenn above.
Regards,
Reilly Atkinson
 
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So we have a set of quantum rule or postulate that can not be derived. That will be a bit strange, because then where did Schrodinger get his equation?
 
  • #5
Bose said:
So we have a set of quantum rule or postulate that can not be derived. That will be a bit strange, because then where did Schrodinger get his equation?

He took one part de Broglie and one part Einstein and mixed it together... on a serious note, [tex]p \mapsto -i \hbar \tfrac{\partial}{\partial x}[/tex] follows from de Broglie's work and is based on analogies between classical mechanics of particles and waves.
 
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  • #6
Bose said:
So we have a set of quantum rule or postulate that can not be derived. That will be a bit strange, because then where did Schrodinger get his equation?

That is not strange, every field has its postulates (axioms) that can't be derived nor be prooved.

However, you may postulate something, then from those axioms derive formulas and relations which can be verified and thus one can tell if you can build up a coherent system (a paradigm) or not.

Here you can read Schrodingers first publication about his new equation:
http://home.tiscali.nl/physis/HistoricPaper/Schroedinger/Schroedinger1926c.pdf

Hanve fun :-)
 
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1. What is the momentum operator in the Schrodinger equation?

The momentum operator in the Schrodinger equation is a mathematical operator that represents the momentum of a quantum particle. It is denoted by p and is given by p = -iħ∇, where i is the imaginary number and ħ is the reduced Planck's constant.

2. How is the momentum operator used in the Schrodinger equation?

The momentum operator is used in the Schrodinger equation to find the momentum of a quantum particle at a specific point in space. It is applied to the wave function, a mathematical representation of the particle's quantum state, to calculate its momentum.

3. What is the physical significance of the momentum operator?

The physical significance of the momentum operator is that it represents the observable quantity of momentum in quantum mechanics. It is a fundamental operator that helps us understand the behavior and properties of quantum particles.

4. How does the momentum operator relate to Heisenberg's uncertainty principle?

The momentum operator is related to Heisenberg's uncertainty principle, which states that it is impossible to simultaneously know the exact position and momentum of a particle. The momentum operator is used to calculate the uncertainty in a particle's momentum.

5. Can the momentum operator be applied to all quantum systems?

Yes, the momentum operator can be applied to all quantum systems, as it is a fundamental operator in quantum mechanics. However, it may require different mathematical forms depending on the specific system and its properties.

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