Harmonic oscillator – raising and lowering operator calculations

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.12.

In the study of the harmonic oscillator, we can express {x} and {p} in terms of the raising and lowering operators:

\displaystyle   x \displaystyle  = \displaystyle  \sqrt{\frac{\hbar}{2m\omega}}(a_{+}+a_{-})\ \ \ \ \ (1)
\displaystyle  p \displaystyle  = \displaystyle  i\sqrt{\frac{\hbar m\omega}{2}}(a_{+}-a_{-}) \ \ \ \ \ (2)

We now have

\displaystyle   \langle x\rangle \displaystyle  = \displaystyle  \sqrt{\frac{\hbar}{2m\omega}}\int\psi_{n}^*(a_{+}+a_{-})\psi_{n}dx\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  0 \ \ \ \ \ (4)

The reason this is zero is that, as we saw when working out the normalization of the stationary states,

\displaystyle   a_{+}\psi_{n} \displaystyle  = \displaystyle  \sqrt{n+1}\psi_{n+1}\ \ \ \ \ (5)
\displaystyle  a_{-}\psi_{n} \displaystyle  = \displaystyle  \sqrt{n}\psi_{n-1} \ \ \ \ \ (6)
\displaystyle   a_{+}a_{-}\psi_{n} \displaystyle  = \displaystyle  n\psi_{n}\ \ \ \ \ (7)
\displaystyle  a_{-}a_{+}\psi_{n} \displaystyle  = \displaystyle  \left(n+1\right)\psi_{n} \ \ \ \ \ (8)

and since the wave functions are orthogonal, we get

\displaystyle  \int\psi_{n}^*\psi_{n+1}dx=\int\psi_{n}^*\psi_{n-1}dx=0 \ \ \ \ \ (9)

Similarly:

\displaystyle   \langle p\rangle \displaystyle  = \displaystyle  i\sqrt{\frac{\hbar m\omega}{2}}\int\psi_{n}^*(a_{+}-a_{-})\psi_{n}dx\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  0 \ \ \ \ \ (11)

for the same reason.

For the mean squares:

\displaystyle   \langle x^{2}\rangle \displaystyle  = \displaystyle  \left(\frac{\hbar}{2m\omega}\right)\int\psi_{n}^*(a_{+}+a_{-})(a_{+}+a_{-})\psi_{n}dx\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{\hbar}{2m\omega}\right)\int\psi_{n}^*(a_{+}a_{-}+a_{-}a_{+})\psi_{n}dx\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{\hbar}{2m\omega}\right)(2n+1)\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right) \ \ \ \ \ (15)

In going from the first to the second line, we’ve thrown out terms where we integrate two orthogonal functions. For example,

\displaystyle   \int\psi_{n}^*a_{+}a_{+}\psi_{n}dx \displaystyle  = \displaystyle  \int\psi_{n}^*\sqrt{\left(n+1\right)\left(n+2\right)}\psi_{n+2}dx\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  0 \ \ \ \ \ (17)

We have used the relations above and the fact that {\psi_{n}} is normalized to get the third line.

Similarly:

\displaystyle   \langle p^{2}\rangle \displaystyle  = \displaystyle  -\frac{\hbar m\omega}{2}\int\psi_{n}^*(-a_{+}a_{-}-a_{-}a_{+})\psi_{n}dx\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \hbar m\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (19)

The uncertainty principle then becomes

\displaystyle   \sigma_{p}\sigma_{x} \displaystyle  = \displaystyle  \sqrt{\left\langle x^{2}\right\rangle \left\langle p^{2}\right\rangle }\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \hbar\left(n+\frac{1}{2}\right) \ \ \ \ \ (21)

and the kinetic energy is

\displaystyle  \langle T\rangle=\frac{\langle p^{2}\rangle}{2m}=\frac{1}{2}\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (22)

which is half the total energy, as it should be.

11 thoughts on “Harmonic oscillator – raising and lowering operator calculations

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  7. apashanka das

    sir my question is,the ground state wave function which is here is actually the representation of the ground state wave function in the basis of eigenvectors of position operator,how can it be total wave function?please explain ,thank you

    Reply
  8. apashanka das

    sir what will be the representation of position and momentum operator in the basis of eigenvectors of position operator?please reply ,thanks

    Reply

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