0
$\begingroup$

I want to ask an easy question from Problem 4.2.1, Quantum Mechanics(2nd) by Shankar.

Let's say Operators, $L_{x}$, $L_{y}$, $L_{z}$ are

$$L_{x}$ = $1/2^{1/2}$ $\begin{pmatrix} 0& 1 &0 \\ 1& 0&1 \\ 0& 1& 0 \end{pmatrix}$$

$$L_{y}$ = $1/2^{1/2}$ $\begin{pmatrix} 0& -i &0 \\ i& 0&-i \\ 0& i& 0 \end{pmatrix}$$

$$L_{z}$ = $\begin{pmatrix} 1& 0 &0 \\ 0& 0& 0 \\ 0& 0& -1 \end{pmatrix}$$

We consider a state described as below, in the $L_{z}$ basis,

$$\mid\Psi\rangle = \left(\begin{smallmatrix} 1/2\\ 1/2\\ 1/2^{1/2} \end{smallmatrix}\right)$$

If I measure $L_{z}^{2}$ and get a result +1, what would be the state of this vector?

It bugs me because it's degenerated. It should be linear combination of the two degenerate states, but the answer says

$$\frac{1}{(1/4 +1/2)^{1/2}}\left (\begin{smallmatrix} 1/2\\ 0\\ 1/2^{1/2} \end{smallmatrix}\right )$$

$\endgroup$
  • 1
    $\begingroup$ Why “but”? What is your projection operator? What is your proposal? $\endgroup$ – Cosmas Zachos Jan 20 at 11:21
  • $\begingroup$ I just got the answer. I skipped some part ahead of this section. The explanation was presented in some previous paragraphs. $\endgroup$ – ComplexNumber Jan 20 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.