A Question while I study Modern Quantum physics by Sakurai

[BREAD]

In the book of Modern Quantum physics by Sakurai

I wonder how 1.6.26 can be 1.6.27.

 

[PeroK]

Actually, I'll put hats on the operators (something that perhaps Sakurai ought to have thought about):

If you set $\mathbf{dx'} = (dx_1, 0, 0)$ then $\mathbf{\hat{K}} \cdot \mathbf{dx'} = \hat{K_1} dx_1$ and (1.6.26) becomes:

$-i \mathbf{\hat{x}}\hat{K_1} dx_1 + i\hat{K_1}\mathbf{\hat{x}}dx_1 = (dx_1, 0, 0)$

Where $dx_1$ is an "infinitesimal" number and can be cancelled, and $\mathbf{\hat{x}} = (\hat{x_1}, \hat{x_2}, \hat{x_3})$; giving:

$-i (\hat{x_1}\hat{K_1} - \hat{K_1}\hat{x_1}, \ \hat{x_2}\hat{K_1} - \hat{K_1}\hat{x_2},\ \hat{x_3}\hat{K_1} - \hat{K_1}\hat{x_3}) = (1, 0, 0)$

Which gives the result for $\hat{K_1}$. Repeat for $\mathbf{dx'} = (0, dx_2, 0)$ etc.

 

[BvU]

Thanks, @PeroK. Out of curiosity: How did you know ${\bf x K} \cdot d{\bf x'}$ was to be interpreted as ${\bf x} \left ({\bf K} \cdot d{\bf x'} \right)$ and not as $\left ({\bf x K} \right ) \cdot d{\bf x'}$ ?

 

[PeroK]

Thanks, @PeroK. Out of curiosity: How did you know ${\bf x K} \cdot d{\bf x'}$ was to be interpreted as ${\bf x} \left ({\bf K} \cdot d{\bf x'} \right)$ and not as $\left ({\bf x K} \right ) \cdot d{\bf x'}$ ?

I looked at my notes and, unlike Sakurai, I tend to distinguish between operators and numbers/vectors. You have to do that at the beginning of the section/proof and keep track of it.

 

[whyohwhy]

I suspect I am being dumb here, but I can't work out how this squares with the line

"dx' is understood to be the number dx' multiplied by the identity operator in the ket space spanned by lx'>"

the |x'> ket is earlier defined as follows

Surely this means that the dx' is present in y directions and z directions as well.

Or am I misunderstanding something fundamental?

 

[PeroK]

The notation $\mathbf x = (x,y,z)$ is not uncommon.

 

[whyohwhy]

Ah, hang on, may have worked out my confusion

When it says "dx' is understood to be the number dx' multiplied by the identity operator in the ket space spanned by lx'>"

Doesn't this mean that is is an operator which can be written as a 3x3 matrix of the form:

(dx,0,0)
(0,dx,0)
(0,0,dx)

which means if we select a particular vector like (1,0,0)^t to multiply right, we get the equations you show?

 

[andresB]

dx is the vector dx=(dx,dy,dz), and $$\hat{\mathbf{K}}=(\hat{K}_{x},\hat{K}_{z},\hat{K}_{z})$$
such that
$$\hat{\mathbf{K}}\cdot d\mathbf{x}=\hat{K}_{x}dx+\hat{K}_{z}dy+\hat{K}_{z}dz$$

 

[whyohwhy]

So it sounds like what we are saying is that the following sentence is incorrect or at the very least misleading?

"dx' is understood to be the number dx' multiplied by the identity operator in the ket space spanned by lx'>"

either this or I am incorrect in viewing the identity operator in the ket space lx'> as:

(1,0,0)
(0,1,0)
(0,0,1)

or I am misunderstanding what is meant by "the number dx'" which I am interpreting to mean a scalar

 

[andresB]

when acting on kets (or bras), all numbers can be understood as the number multiplying the identity operator. For example
$$3:=3\hat{I}$$,
means
$$(3\hat{I})\left|A\right\rangle =3\left(\hat{I}\left|A\right\rangle \right)=3\left|A\right\rangle $$

 

[andresB]

or I am misunderstanding what is meant by "the number dx'" which I am interpreting to mean a scalar

In here the word vector has two meanings. dx is a vector in the ordinary sense of vector calculus(more precisely, it's a differential form).

The other meaning of the word vectors refers to the elements belonging to the Hilbert space spanned by the kets $$\left|\mathbf{x}\right\rangle$$.

 

[whyohwhy]

Yes but a vector cannot be right? This is a vector times a identity operator?

 

[whyohwhy]

In here the word vector has two meanings. dx is a vector in the ordinary sense of vector calculus(more precisely, it's a differential form).

The other meaning of the word vectors refers to the elements belonging to the Hilbert space spanned by the kets $\left|\mathbf{x}\right\rangle$ .

So this sounds like it could be right to me. But I don't get the distinction between elements belonging to Hilbert space and vectors in the ordinary sense. Looks like I have some more reading to do! Thanks so much

 

[andresB]

The position operator is a vector operator, loosely this means that it has components in each axis
$$\hat{\mathbf{X}}=\hat{x}\mathbf{i}+\hat{y}\mathbf{j}+\hat{z}\mathbf{k}$$
its actions on a 3d position ket is
$$\hat{\mathbf{X}}\left|\mathbf{x}\right\rangle =\left(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\right)\left|\mathbf{x}\right\rangle $$

 

[andresB]

So this sounds like it could be right to me. But I don't get the distinction between elements belonging to Hilbert space and vectors in the ordinary sense. Looks like I have some more reading to do! Thanks so much

I recommend you to first get used to the derivation of the commutation relations with the translation operator for the 1d case (1 spatial dimension, the Hilbert space of square-integrable functions is still infinite-dimensional).

 

[whyohwhy]

The position operator is a vector operator, loosely this means that it has components in each axis
$$\hat{\mathbf{X}}=\hat{x}\mathbf{i}+\hat{y}\mathbf{j}+\hat{z}\mathbf{k}$$
its actions on a 3d position ket is
$$\hat{\mathbf{X}}\left|\mathbf{x}\right\rangle =\left(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\right)\left|\mathbf{x}\right\rangle $$

Ah! This makes sense! But isn't mentioned in the textbook up to this point. Or at least not explicitly.

 

[andresB]

Ah! This makes sense! But isn't mentioned in the textbook up to this point. Or at least not explicitly.

The 3d vector notation is unusual, generally speaking, books write the action of each component individually

$$\hat{x}\left|\mathbf{x}\right\rangle =x\left|\mathbf{x}\right\rangle ,\;\hat{y}\left|\mathbf{x}\right\rangle =y\left|\mathbf{x}\right\rangle ,\;\hat{z}\left|\mathbf{x}\right\rangle =z\left|\mathbf{x}\right\rangle $$