Sakurai: Page 46

1. Apr 21, 2013

omoplata

From page 46 of "Modern Quantum Mechanics, revised edition", by J.J. Sakurai.

In equation (1.6.24),
$$\left[\mathbf{x}, \mathcal{T}(d\mathbf{x'}) \right] = d \mathbf{x'} \mid \mathbf{x'} + d \mathbf{x'} \rangle \approx d \mathbf{x'} \mid \mathbf{x'} \rangle$$
It is written: "where the error made in writing the last line of (1.6.24) is of second order in $d \mathbf{x'}$". How does that happen? From a Taylor expansion of $\mid \mathbf{x'} + d \mathbf{x'} \rangle$ ? If so, how do you Taylor expand a ket?

Last edited: Apr 21, 2013
2. Apr 21, 2013

MisterX

You may just use the definition of the transformation operator $\mathcal{T}$.
$$\mathcal{T}(d\mathbf{x'})\mid \mathbf{x'} \rangle = \mid \mathbf{x'} +d\mathbf{x'} \rangle$$
$$\left[\mathbf{x}, \mathcal{T}(d\mathbf{x'}) \right]\mid \mathbf{x'} \rangle = d \mathbf{x'} \mid \mathbf{x'} + d \mathbf{x'} \rangle = d \mathbf{x'}\mathcal{T}(d\mathbf{x'})\mid\mathbf{x'} \rangle = d \mathbf{x'} \left(1 - i\mathbf{K} \cdot d\mathbf{x'}\right) \mid\mathbf{x'} \rangle$$

You may notice that $1 - i\mathbf{K} \cdot d\mathbf{x'}$ does look like the first two terms of a Taylor expansion.

Last edited: Apr 21, 2013
3. Apr 21, 2013

omoplata

Oh, yeah. That never occurred to me. Thanks!

I also found out just now that I had written my equation wrong. I just corrected it. Sorry about that and thanks for looking at Sakurai to get through that.

4. Apr 10, 2014

bassam

You can’t use $(1-\mathbf{K}\cdot dx')$ because you need to prove this later. I believe ket can be expanded as $$|x'+dx'\rangle=|x'\rangle+\frac{|x'\rangle}{dx'}dx'+\dots$$ and then ignore $O(dx')^2$. What do you think?

Last edited: Apr 10, 2014
5. Apr 10, 2014

ChrisVer

The problem is how you can expand the ket?
I guess a good way would be to use this:
$<f|x+dx>= f(x+dx) \approx f(x)+ f'(x) dx = <f|x>+ \frac{<f| x+dx>-<f|x>}{dx} dx= <f| (|x>+\frac{| x+dx>-<f|x>}{dx} dx)$
or
$|x+dx> \approx |x>+\frac{| x+dx>-|x>}{dx} dx$

so multiplying with dx:
$dx |x+dx> \approx dx |x>+\frac{| x+dx>-|x>}{dx} (dx)^{2}$

Am I somewhere wrong?

6. Apr 10, 2014

bassam

Exactly ... the second term $$\frac{|x+dx\rangle-|x\rangle}{dx}$$ is nothing but $d|x\rangle/dx$ since $dx$ is infinitesimal. It is totally legal to apply dervative to ket. Remember SHE $\hat{H} |\psi\rangle=i\hbar d|\psi\rangle/dt$. So you dont need to project the ket into function.

Last edited: Apr 10, 2014