To the point:

How should I think about the equation $$\hat{\mathbf{x}}\mid\mathbf{x'}\rangle = \mathbf{x'}\mid\mathbf{x'}\rangle~?$$

Is it a triple of equations

$\hat{x}\mid x'\rangle = x'\mid x'\rangle$

$\hat{y}\mid y'\rangle = y'\mid y'\rangle$

$\hat{z}\mid z'\rangle = z'\mid z'\rangle$

or can the 1-D operators act directly on the (vector) state $\mid\mathbf{x'}\rangle$ such as

$\hat{x}\mid\mathbf{x'}\rangle = x'\mid\mathbf{x'}\rangle$?

In the latter case can we interpret $\mid\mathbf{x'}\rangle$ as a simultaneous eigenket of the operators $x, y$ and $z$, given they all commute? ( i.e $\mid\mathbf{x'}\rangle = \mid x',y',z' \rangle$ )

I have tried thinking about this problem in terms of discrete states as sometimes I have an easier time with matrix notation but in this case I am again stuck. As above should I interpret the vector equations as a triple of matrix equations (or a matrix of matrices)?

Finally, how do I interpret operator functions of $x, y$ and $z$ such as $F(r)\mid\mathbf{x'}\rangle$, where $r = \sqrt{x^2+y^2+z^2}$. See, for example Q1.27 Sakurai, Modern Quantum Mechanics.

  • $\begingroup$ think of $|\mathbf v\rangle$ as $|v_1\rangle\otimes\cdots\otimes|v_n\rangle$ for any $\mathbf v\in\mathbb R^n$, assuming that the operators associated to each component generate a commutative algebra. This is certainly the case for ordinary coordinates. $\endgroup$ – Phoenix87 Jun 15 '15 at 12:49
  • 1
    $\begingroup$ Thank you! So for example, where I wrote, $\hat{x} \mid \mathbf{x'} \rangle = x' \mid \mathbf{x'} \rangle$, to make it clear, I should write $[ \hat{x} \otimes \hat{1_y} \otimes \hat{1_z} ] ( \mid x' \rangle \otimes \mid y' \rangle \otimes \mid z' \rangle ) = x' \mid x' \rangle \otimes \mid y' \rangle \otimes \mid z' \rangle$ $\endgroup$ – mauiaw1 Jun 15 '15 at 13:39
  • $\begingroup$ that's correct $\ $ $\endgroup$ – Phoenix87 Jun 15 '15 at 13:58

Your second option is correct.

It is a simultaneous eigenket of all three commuting operators. And that is how it should be interpreted. To be explicit, the equation $$\hat{\mathbf{x}}\mid\mathbf{x'}\rangle = \mathbf{x'}\mid\mathbf{x'}\rangle~?$$

Is a triple of equations

$\hat{x}\mid \mathbf x'\rangle = x'\mid \mathbf x'\rangle$

$\hat{y}\mid \mathbf x'\rangle = y'\mid \mathbf x'\rangle$

$\hat{z}\mid \mathbf x'\rangle = z'\mid \mathbf x'\rangle$

And you can use a product to have all three components as part of one object.

As for operators inside functions, if you consider a simultaneous eigenket then it isn't strange to define the operator to act by scalar multiplication of the scalar value determined by putting the eigenvalues into the function.

You can work out the entire functional calculus but that is not what most everyday physicists do and probably not what is expected in a first course. For instance in the position representation pointwise multiplication of a function of the coordinates does exactly this and usually a fuss is not made.


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