2
$\begingroup$

I'm reading Griffiths's Introduction to Quantum Mechanics 3rd ed textbook [1]. On p.43, the author explains:

What if I apply the lowering operator repeatedly? Eventually I’m going to reach a state with energy less than zero, which (according to the general theorem in Problem 2.3) does not exist! At some point the machine must fail. How can that happen?

We know that $a_-ψ$ is a new solution to the Schrödinger equation, but there is no guarantee that it will be normalizable—it might be zero, or its square-integral might be infinite. In practice it is the former: There occurs a “lowest rung” (call it $ψ_0$) such that $$a_−ψ_0 = 0 $$

I understood why $a_−ψ_0$ should not be normalized. But why should it be non-normalized like $a_−ψ_0 = 0$? As the author mentioned in the book, the possibility of its square-integral value being infinite may also exist (satisfying with the non-normalizable condition). The author went over this point, and I wonder what happens to the case I mentioned.

Reference

Griffiths, D. J.; Schroeter, D. F. Introduction to Quantum Mechanics 3rd ed; Cambridge University Press, 2018. ISBN 978-1107189638.

$\endgroup$
1
0
$\begingroup$

$a_-\psi_0$ results in the zero-vector. Call this vector $\vert\hbox{0 vector}\rangle$. Then in any computation $$ \langle \psi_n|\hat T \vert\hbox{0 vector}\rangle =0 $$ for any operator. In particular the length of $\vert \hbox{0 vector}\rangle$ is $0$ and in this sense it cannot be normalized.

$\endgroup$
0
$\begingroup$

I'll be appealing to you physical intuition, so that I don't have dive deep into functional analysis.

Wavefunctions are mathematical objects which are defined in a Hilbert space, which is square integrable. Now, the operators you see in QM are defined on this Hilbert space. Roughly, there are kind of linear functions which map from one Hilbert space to another. Typically, Hamiltonian operators are semi-bounded: there is a lowest energy.

Now, here is the intuition: In the same or previous page, you have the relation $(a_+a_- +\frac{1}{2}\hbar\omega)\psi= E\psi$. If you were to plug $\psi_0$ here and conjecture that $a_-\psi_0$ is infinity, you immediately see that $E$ would also be infinite for the ground state, and thus also for higher states in a quantum harmonic oscillator irrespective of $\omega$. This would also affect the correspondence principle as clearly there exists classical harmonic oscillator, and you won't recover that for any $\omega$ for larger number of states or if $\hbar \rightarrow 0$.

Thus the other possibility of $a_-\psi_0$ being $0$ is reasonable.

Hope that helps.

$\endgroup$
4
  • $\begingroup$ This is incorrect. If $\psi_n$ is normalized and has eigenvalue $E_n$, then $\lambda\psi_n$, with $\vert\lambda\vert\ne 1$, is also an eigenfunction with the same eigenvalue, i.e. the size of the eigenvalue does not depend on the norm of the eigenvector, except if this eigenvector is the zero vector. Thus even if $a_-\psi_0$ is infinite, this does NOT imply the eigenvalue is infinite. $\endgroup$ Aug 10 '20 at 12:53
  • $\begingroup$ @ZeroTheHero I didn't intend to say that the eigenvalue is $\infty$ for $a_-$. As a matter of fact, that is not even an eigenvalue equation for any $\psi$. Yes, the language was sloppy (at the brink of incorrectness). But that sloppiness was at the level of the book. For reference you can refer to the next page as to the passage the OP mentioned. $\endgroup$
    – Rounak
    Aug 11 '20 at 16:45
  • $\begingroup$ @ZeroTheHero The correct answer would indeed be the semi boundedness of the spectrum of the operator. $\endgroup$
    – Rounak
    Aug 11 '20 at 16:48
  • $\begingroup$ Please clarify your answer and I will be happy to remove my downvote. $\endgroup$ Aug 11 '20 at 20:58

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.