I Solving Dirac Algebra Arithmetic

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The discussion focuses on simplifying Dirac notation arithmetic in quantum mechanics, specifically transitioning from one equation to another without converting to matrices and vectors. The measurement operator M is defined as |0><0| ⊗ I, projecting onto states of the form |0x>. The participant seeks a more efficient method to solve the equations directly in Dirac notation. The solution involves recognizing that the projection simplifies the state to only include components |00> and |01>, leading to the final result. This approach avoids cumbersome conversions and highlights the utility of Dirac notation in quantum calculations.
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I have a question regarding the Dirac notation arithmetic. Below is a measurement of a general 2 qubit state with the measurement operator M=|0><0| ⊗ I , where I is the identity operator. To go from equation (2) to equation (3), I've been converting all the Dirac notation to matrices and column vectors and then carrying out the arithmetic, but this is very cumbersome. Is there a more simple way to go through the arithmetic with the Dirac notation to directly solve (2) for (3) instead of first converting? Thanks for looking.

Prob = <ψ|M† M |ψ > (1)
= (α ∗ <00| + β ∗ <01| + γ ∗ <10| + δ ∗ <11|) (|0><0| ⊗ I) (α |00>+ β |01>+ γ |10>+ δ |11>) (2)
= (α ∗ <00| + β ∗ <01|) (α |00>+ β |01>) (3)
= |α|^2 + |β|^2 (4)
 
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##M## is clearly a projection onto the space spanned by states on the form ##|0x\rangle = |0\rangle \otimes |x\rangle##. As such, it should be clear that
$$
|\phi\rangle = M|\psi\rangle = \alpha |00\rangle + \beta |01\rangle
$$
and therefore ##\langle \phi| = \langle \psi| M^\dagger = \alpha^* \langle 00| + \beta^* \langle 01|##, leading directly to your end result.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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