1
$\begingroup$

I do not understand why $a_-a_+\psi_n=(n+1)\psi_n$ and not $\sqrt n(n+1))\psi_n$ or how the Energy formula can help me understand this (I was told that it would).

In the introduction to quantum mechanics by Griffiths in the proof section of the hermitian conjugate Griffiths goes on to prove that $a_+\psi_n=c_n\psi_{n+1}$ and $a_p\sin=d_n\psi_{n-1}$. At one point he states that by equations $\hbar\omega(a_{+/-}a_{+/-} +/- 1/2)\psi=E\psi$ and $\psi_n(x)=A_n(a_+)^n\psi_0(x)$ [with $E_n=(n+1/2)\hbar\omega$] the statements $a_+a_\psi=n\psi_n$ and $a_-a_+\psi_n=(n+1)\psi_n$. With all this jumble of proofs and equations I can't seem to find a simple way to understand this.

$\endgroup$
  • 1
    $\begingroup$ Please use MathJax to write your equations. $\endgroup$ – garyp Nov 26 '15 at 3:01
  • $\begingroup$ It's very hard to answer this because it's hard to find a specific question in your question. The inconsistent equation formatting isn't helping, but the real problem is that I think your question is too broad. Can you point to a specific proof that has you puzzled, for example? $\endgroup$ – elifino Nov 26 '15 at 3:48
  • $\begingroup$ Assuming $\psi_n$ is a quantum state, there is no difference between $\psi_n$, $(n+1)\psi_n$ and $\sqrt{n}(n+1)\psi_n$. $\endgroup$ – WillO Nov 26 '15 at 3:51
  • $\begingroup$ I don't understand how I can prove $a_-a_+\psi_n=(n+1)\psi_n$ $\endgroup$ – user96828 Nov 26 '15 at 3:53
  • $\begingroup$ Okay. So kηives' answer covers the case where you already know the formulas for $c_n$ and $d_n$. Is that working out for you, or do you want to understand how those coefficients are derived? $\endgroup$ – elifino Nov 26 '15 at 4:12
2
$\begingroup$

using $$a^{\dagger} |n\rangle = \sqrt{n+1}| n+1 \rangle$$ and $$ a |n\rangle = \sqrt{n} |n-1\rangle, $$ apply these rules in order: $$ a(a^{\dagger}|n\rangle) = a\sqrt{n+1}|n+1\rangle = \sqrt{n+1}a|n+1\rangle = (\sqrt{n+1})^2 |n\rangle = (n+1)|n\rangle $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.