I am currently working with an exercise set about discriminating between two different quantum states. We consider a 2-dimensional Hilbert space expanded by $|\phi_1\rangle$, $|\phi_2\rangle$. The system is secretly prepared into one of the following states at random:

$|\psi_1\rangle=|{\phi_1}\rangle \tag{1}\\$ $|\psi_2\rangle=\cos\theta|\phi_1\rangle+\sin\theta |\phi_2\rangle \tag{2}$

Where angle $\theta$ is a real number between $0$ and $\frac{\pi}{2}$.

We are handed one of these states and want to identify if we receive $|\psi_1\rangle$ or $|\psi_2\rangle$.

To start with we consider another basis $|e_1\rangle$, $|e_2\rangle$ which is rotated with respect to $|\phi_1\rangle$ and $|\phi_2\rangle$:

$$|e_1\rangle=\cos\alpha|\phi_1\rangle+\sin\alpha|\phi_2\rangle\\$$ $$|e_2\rangle=-\sin\alpha|\phi_1\rangle+\cos\alpha|\phi_2\rangle$$

Where angle $\alpha$ originates from the fact that basis $|e_1\rangle$, $|e_2\rangle$ is rotated relative to $|\phi_1\rangle$, $|\phi_2\rangle$.

I am asked to write down general expressions using $|\phi_1\rangle$ and $|\phi_2\rangle$ for the projectors corresponding to a measurement in the $|e_1\rangle$, $|e_2\rangle$ basis.

Attempted solution:

I have previously constructed the basis $|\phi_1\rangle$ and $|\phi_2\rangle$ in terms of $|e_1\rangle$ and $|e_2\rangle$ as:

$$|\phi_1\rangle=-\sin\alpha |e_2\rangle$$

$$|\phi_2\rangle=\sin\alpha |e_1\rangle+\cos\alpha |e_2\rangle$$

But I see now that they want me to construct the projectors instead of the vectors themselves. What do they mean exactly, and how do I determine these projectors?


Following the first answer, i get:

$P=\sum_i|\phi_i\rangle\langle\phi_i|\delta_{a,a_i}=|e_1\rangle\langle e_1|+|e_2\rangle\langle e_2|$

$|e_1\rangle\langle e_1| = [\text{cos}\alpha | \phi_1 \rangle + \text{sin}\alpha |\phi_2 \rangle]\;[\text{cos}\alpha \langle \phi_1 |+\text{sin}\alpha \langle \phi_2 |]=\text{cos}^2\alpha |\phi_1\rangle \langle \phi_1 | + \text{sin}^2\alpha |\phi_2\rangle\langle \phi_2 |$

$|e_2\rangle\langle e_2| = [-\text{sin}\alpha | \phi_1 \rangle + \text{cos}\alpha |\phi_2 \rangle]\;[-\text{sin}\alpha \langle \phi_1 |+\text{cos}\alpha \langle \phi_2 |]=\text{sin}^2\alpha |\phi_1\rangle \langle \phi_1 | + \text{cos}^2\alpha |\phi_2\rangle\langle \psi_2 |$

So $P=|\phi_1\rangle\langle \phi_1|+|\phi_2\rangle\langle \phi_2|$

Is it correct to only look at the non-perpencular states for $|e_1\rangle\langle e_1|$, i.e. does the kronecker delta 'survive' for this expression when considering the states $|\phi_1\rangle$, $|\phi_2\rangle$?


1 Answer 1


If "the projectors corresponding to a measurement in the |e1⟩, |e2⟩ basis" means operators that project out the |e1> and |e2> components from a state, aren't they just |e1><e1| and |e2><e2|? You just need to plug in the expression of |e1> and |e2> in terms of |ϕ1⟩ and |ϕ2⟩.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jul 3, 2023 at 4:41
  • 1
    $\begingroup$ I have considered your answer with an update to my post. It is correct that the kronecker delta is 'alive' when considering the terms $|e_1\rangle\langle e_2$, so only the non-perpendicular projections survive? $\endgroup$ Commented Jul 3, 2023 at 8:12
  • 1
    $\begingroup$ If I understand the question correctly, there should be 2 projectors, |e1><e1| for projecting out the |e1> component, and |e2><e2| for projecting out the |e2> component, not the sum of the two. $\endgroup$
    – Sgia
    Commented Jul 4, 2023 at 15:47

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