Ladder operators to find Hamiltonian of harmonic oscillator

In summary, the conversation discusses the derivation of the equation for the Hamiltonian for the harmonic oscillator. The process involves defining the operators ##\hat {a}_{\pm}## and using the commutator ##[\hat {x}, \hat {p}] = i \hbar##. The conversation also mentions differences between the explanations in a YouTube video and Griffiths' book. The final equations are corrected and the Hamiltonian is determined to be ##H=\dfrac{p^2}{2 m} + \dfrac{1}{2} m \omega^2 x^2##.
  • #1
gfd43tg
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Hello, I was just watching a youtube video deriving the equation for the Hamiltonian for the harmonic oscillator, and I am also following Griffiths explanation. I just got stuck at a part here, and was wondering if I could get some help understanding the next step (both the video and book glanced over this part)

Video Link

The derivation starts at about 8:00

First defining
$$\hat {a}_{\pm} = \frac {1}{\sqrt{2 \hbar m \omega}}( \mp i \hat {p} + m \omega \hat {x})$$

$$ \hat {a}_{-} \hat {a}_{+} = \frac {1}{2 \hbar m \omega}(i \hat {p} + m \omega \hat {x})(-i \hat {p} + m \omega \hat {x}) $$
$$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} + i m \omega \hat {p} \hat {x} - i m \omega \hat {x} \hat {p} + m^{2} \omega^{2} \hat {x}^{2}) $$
$$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} - i m \omega (\hat {x} \hat {p} - \hat {p} \hat {x}) + m^{2} \omega^{2} \hat {x}^{2})$$
And I know the commutator ##[\hat {x}, \hat {p}] = i \hbar##
$$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} - i m \omega ( i \hbar) + m^{2} \omega^{2} \hat {x}^{2} )$$
At this point both the video and Griffiths stop, although each do something different
I have no idea how Griffiths goes from this
upload_2015-3-5_0-11-14.png

To this
upload_2015-3-5_0-14-12.png

I will continue with what I was doing,
$$ = \frac {1}{2 \hbar m \omega}( \hat {p}^{2} + m \omega \hbar + m^{2} \omega^{2} \hat {x}^{2})$$
$$ = \frac {1}{2 \hbar m \omega}\hat {p}^{2} + \frac {1}{2 \hbar^{2} m^{2} \omega^{2}} + \frac {m \omega \hat {x}^{2}}{2 \hbar} $$
And from here I am not sure how both the video and Griffiths conclude
upload_2015-3-5_0-14-34.png

How do I get to the answer from my steps? Since I am not skipping..
 

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  • #2
Maylis said:
I will continue with what I was doing,
=12(p̂ 2+(i)+m2ω2x̂ 2)
= \frac {1}{2 \hbar m \omega}( \hat {p}^{2} + m \omega ( i \hbar) + m^{2} \omega^{2} \hat {x}^{2})

There is an i too much in the term in the middle
Maylis said:
=12mωp̂ 2+122m2ω2+mωx̂ 22

The term in the middle is again wrong...
Basically the last two equations you wrote before the Griffiths insert are wrong.
Once you have them right, you know that the Hamiltonian is
[tex]H=\dfrac{p^2}{2 m} + \dfrac{1}{2} m \omega^2 x^2[/tex]
 
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  • #3
Woops, I caught my error and now I am able to see how it was done. Thanks.
 

1. What are ladder operators in quantum mechanics?

Ladder operators are mathematical operators used in quantum mechanics to describe the energy levels of a system. They are used to raise or lower the energy of a quantum state by a fixed amount.

2. How do ladder operators relate to the Hamiltonian of a harmonic oscillator?

The Hamiltonian of a harmonic oscillator can be expressed in terms of the ladder operators. The raising and lowering operators, along with the energy of the ground state, make up the Hamiltonian of a harmonic oscillator.

3. How do ladder operators help find the energy levels of a harmonic oscillator?

By applying the ladder operators to the ground state energy level, the energy levels of a harmonic oscillator can be determined. Each application of the raising or lowering operator changes the energy level by a fixed amount.

4. Can ladder operators be used to find the wavefunction of a harmonic oscillator?

Yes, the ladder operators can be used to find the wavefunction of a harmonic oscillator. The wavefunction can be expressed in terms of the ladder operators and the ground state energy.

5. Are ladder operators only applicable to harmonic oscillators?

No, ladder operators can be applied to other quantum systems as well. However, they are commonly used in the context of harmonic oscillators due to their simple and elegant mathematical properties.

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