- #1
jstrunk
- 55
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I am completely baffled by bit of notation in Quantum Mechanics Concepts and Applications by Zitteli. He is trying to get the differential equation for the ground state of a harmonic oscillator using the algebraic method as opposed to Schrodinger's equation. I suspect he is compressing a lot of steps into one and probably also abusing the notation. Can someone clarify this, or point me to a source that develops Dirac notation slowly and clearly?
The annihilation operator is [itex]\hat a = \frac{1}{{\sqrt 2 {x_0}}}\left( {\hat X + x_0^2\frac{d}{{dx}}} \right)[/itex]
The eigenstates are given by [itex]\hat a\left. {\left| n \right.} \right\rangle = \sqrt n \left. {\left| {n - 1} \right.} \right\rangle[/itex]
The ground state is given by [itex]\hat a\left. {\left| 0 \right.} \right\rangle = 0[/itex]
He deduces the differential equation like this
[itex]
\begin{array}{l}
\left\langle {\left. x \right|} \right.\hat a\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right) = 0\\
x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right) = 0
\end{array}
[/itex]
The part I don't understand is
[itex]\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right)[/itex]
I would expect this [itex]\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \int_{ - \infty }^\infty {{x^ * }\left( {x + x_0^2\frac{d}{{dx}}} \right)} 0dx = 0[/itex]
The book went to a lot of trouble to show that n has to be a non-negative integer, but even if we somehow allow n=0 to morph into a continuous variable [itex]{\psi _0}\left( x \right)[/itex], I don't get his answer, I get the following
[itex]\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \int_{ - \infty }^\infty {{x^ * }\left( {x{\psi _0}\left( x \right) + x_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx = \int_{ - \infty }^\infty {\left( {{x^2}{\psi _0}\left( x \right) + xx_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx[/itex]
On the other hand, if we start with [itex]\hat a\left. {\left| 0 \right.} \right\rangle = 0[/itex] instead of [itex]\left\langle {\left. x \right|} \right.\hat a\left. {\left| 0 \right.} \right\rangle = 0[/itex] and allow n=0 to morph to [itex]{\psi _0}\left( x \right)[/itex] I can easily get the desired answer
[itex]
\begin{array}{l}
\hat a\left. {\left| 0 \right.} \right\rangle = 0\\
\frac{1}{{\sqrt 2 {x_0}}}\left( {\hat X + x_0^2\frac{d}{{dx}}} \right)\left. {\left| {{\psi _0}\left( x \right)} \right.} \right\rangle = 0\\
\frac{1}{{\sqrt 2 {x_0}}}\left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right) = 0\\
x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right) = 0
\end{array}
[/itex]
The annihilation operator is [itex]\hat a = \frac{1}{{\sqrt 2 {x_0}}}\left( {\hat X + x_0^2\frac{d}{{dx}}} \right)[/itex]
The eigenstates are given by [itex]\hat a\left. {\left| n \right.} \right\rangle = \sqrt n \left. {\left| {n - 1} \right.} \right\rangle[/itex]
The ground state is given by [itex]\hat a\left. {\left| 0 \right.} \right\rangle = 0[/itex]
He deduces the differential equation like this
[itex]
\begin{array}{l}
\left\langle {\left. x \right|} \right.\hat a\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \frac{1}{{\sqrt 2 {x_0}}}\left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right) = 0\\
x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right) = 0
\end{array}
[/itex]
The part I don't understand is
[itex]\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right)[/itex]
I would expect this [itex]\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \int_{ - \infty }^\infty {{x^ * }\left( {x + x_0^2\frac{d}{{dx}}} \right)} 0dx = 0[/itex]
The book went to a lot of trouble to show that n has to be a non-negative integer, but even if we somehow allow n=0 to morph into a continuous variable [itex]{\psi _0}\left( x \right)[/itex], I don't get his answer, I get the following
[itex]\left\langle {\left. x \right|} \right.\hat X + x_0^2\frac{d}{{dx}}\left. {\left| 0 \right.} \right\rangle = \int_{ - \infty }^\infty {{x^ * }\left( {x{\psi _0}\left( x \right) + x_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx = \int_{ - \infty }^\infty {\left( {{x^2}{\psi _0}\left( x \right) + xx_0^2\frac{{d{\psi _0}\left( x \right)}}{{dx}}} \right)} dx[/itex]
On the other hand, if we start with [itex]\hat a\left. {\left| 0 \right.} \right\rangle = 0[/itex] instead of [itex]\left\langle {\left. x \right|} \right.\hat a\left. {\left| 0 \right.} \right\rangle = 0[/itex] and allow n=0 to morph to [itex]{\psi _0}\left( x \right)[/itex] I can easily get the desired answer
[itex]
\begin{array}{l}
\hat a\left. {\left| 0 \right.} \right\rangle = 0\\
\frac{1}{{\sqrt 2 {x_0}}}\left( {\hat X + x_0^2\frac{d}{{dx}}} \right)\left. {\left| {{\psi _0}\left( x \right)} \right.} \right\rangle = 0\\
\frac{1}{{\sqrt 2 {x_0}}}\left( {x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right)} \right) = 0\\
x{\psi _0}\left( x \right) + x_0^2\frac{d}{{dx}}{\psi _0}\left( x \right) = 0
\end{array}
[/itex]