0
$\begingroup$

I don't know how to write the equation form. Assuming my notation as Dirac notation, what do I get from

$$ ( 0 | 0 | 0 ) ~?$$

$\endgroup$
  • $\begingroup$ There is no such thing as a $0$ eigenvector $\endgroup$ – Hugo V Dec 17 '18 at 15:46
  • $\begingroup$ @HugoV I am working with a ground state simple harmonic oscillator. Should I denote it differently? $\endgroup$ – user121392 Dec 17 '18 at 15:47
  • 1
    $\begingroup$ Yes, 0 as a vector cannot be an eigenvector, the ground state's wavefunction, $\psi_0$ may be what you are referring to and is also not zero for the case of the harmonic oscillator. It would be helpful if you add some more context to your question. $\endgroup$ – ohneVal Dec 17 '18 at 15:49
  • $\begingroup$ @ohneVal I am working with quantum simple harmonic oscillators and obtain the result of position and momentum operators. When expanding the position-squared operator, one of the resulting terms is this. $\endgroup$ – user121392 Dec 17 '18 at 16:05
0
$\begingroup$

It is not quite clear what you mean with "0 eigenvector". However, acting with the "0 operator" on any vector gives zero (i.e., the zero vector). So the answer to your question is definitely "zero", $$ \langle \psi | 0 | \psi \rangle = 0 . $$

$\endgroup$

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.