Undergrad Understanding the Raising and Lowering Operator: A Scientific Analysis

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SUMMARY

The discussion focuses on the mathematical proof involving the raising operator $$\hat a$$ and lowering operator $$\hat a^{\dagger}$$ in quantum mechanics. It establishes that $$\hat a|n\rangle = k|n-1\rangle$$, where $$k$$ is a positive real constant. The proof utilizes the normalization of eigenstates and the identity $$\langle \psi |\hat X |\phi \rangle = \langle \phi |\hat X^{\dagger} |\psi \rangle^*$$ to derive the equation $$k^2 = n$$, confirming the relationship between the operators and their eigenstates. The discussion encourages further exploration of the proof for the raising operator $$\hat a^{\dagger}$$.

PREREQUISITES
  • Understanding of quantum mechanics and operator theory
  • Familiarity with eigenstates and eigenvalues
  • Knowledge of normalization in quantum states
  • Proficiency in mathematical proofs involving complex numbers
NEXT STEPS
  • Study the properties of quantum mechanical operators, specifically $$\hat a$$ and $$\hat a^{\dagger}$$
  • Learn about the implications of the identity $$\langle \psi |\hat X |\phi \rangle = \langle \phi |\hat X^{\dagger} |\psi \rangle^*$$ in quantum mechanics
  • Explore the normalization conditions for quantum states and their significance
  • Practice deriving proofs involving raising and lowering operators in quantum harmonic oscillators
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Students and professionals in physics, particularly those specializing in quantum mechanics, as well as educators looking to enhance their understanding of operator theory and eigenstate relationships.

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TL;DR
I was reading up on linear harmonic oscillator and here when they define 2 new operators for solving the problem.
I still am finding it difficult how they ended up with the relation mentioned in the grey box, describing how the above mentioned operators act on given eigen state
Thanks in advance
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The proof is not immediately obvious. We start by noting that $$\hat a|n\rangle = k|n-1\rangle$$for some constant ##k##, where we take ##k## to be real and positive by convention. Note that eigenstates are defined only up to a complex phase factor, so we need this convention.

Next, using $$\hat a^{\dagger}\hat a |n\rangle = n|n\rangle$$ and the normalisation of ##|n\rangle## we see that$$\langle n|\hat a^{\dagger}\hat a |n\rangle = \langle n| n|n\rangle = n$$Finally, we evaluate the LHS a different way: $$\langle n|\hat a^{\dagger}\hat a |n\rangle = \langle n|\hat a^{\dagger}k|n-1\rangle = k\langle n-1|\hat a|n\rangle^* = k\langle n-1|k|n-1\rangle^* = k^2$$using the normalisation of ##|n-1\rangle## and the important identity: $$\langle \psi |\hat X |\phi \rangle = \langle \phi |\hat X^{\dagger} |\psi \rangle^*$$which applies for all vectors and operators. That leads to the required equation $$k^2 = n$$
 
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PS it would be a good exercise to do the proof for ##\hat a^{\dagger}## yourself.
 
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PeroK said:
PS it would be a good exercise to do the proof for ##\hat a^{\dagger}## yourself.
Yeah I will try it for sure,
Thanks!
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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