J.J. Sakurai: Error of 2nd Order in d x' Eq (1.6.24)

[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?

 

[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.

 

[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.

 

[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?

 

[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?

 

[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 don't need to project the ket into function.