So for two quantum systems a and b in a joint state

$$\Psi = \lvert \psi_a \rangle\lvert \psi_b \rangle \tag{1}$$

With basis states $\{\lvert \phi_k \rangle\}$ for system a and $\{\lvert v_j \rangle\}$ for system b, and operator A acting on the Hilbert space of a with the spectral decomposition

$$ A = \lambda_1\lvert \phi_1 \rangle \langle \phi_1 \rvert + \lambda_2\sum_{n=2}^N \lvert \phi_n \rangle \langle \phi_n \rvert \ \tag{2}$$

I am trying to understand how I can write an expression for the projector acting on the total Hilbert space and which is associated to the measurement outcome $\lambda_2$ so I can then calculate the probability of obtaining $\lambda_2$ in a measurement of A on the state $\Psi$.

My understanding is that only the second term of equation 2 is relevant for this measurement as it contains the coefficient $\lambda_2$. Operator A only acts on the Hilbert space of a, so am I correct in writing that the projector acting on the total Hilbert space is the following? $$ (\lambda_2\sum_{n=2}^N \lvert \phi_n \rangle \langle \phi_n \rvert) \otimes I \tag{3} $$

And if the probability of measuring $\lambda_2$ is the following (because there are multiple projectors corresponding to the measurement $\lambda_2$) $$ \langle \Psi \rvert \ \sum_{n=2}^N \lvert \phi_n \rangle \langle \phi_n \rvert \Psi \rangle \tag{4} $$

If anyone could point out if I'm correct or wrong with my answers/reasoning and why, it would be much appreciated. I also would like to know if I can write my expressions in a simpler way. Finally, I would like to ask that since only the Hilbert space of a is concerned in this measurement, does that mean the basis states for system b is completely irrelevant even if I change them? Thank you.


1 Answer 1


Yes. You are correct. We can write operator $A$ as $$ A = \left[\lambda_1 |\phi_1 \rangle \langle \phi_1| + \lambda_2\sum |\phi_n\rangle \langle \phi_n| \right] \otimes I_b$$ as it acts as the identity operator on subsystem $b$. Consequently, it doesn't matter what is the basis in which we write subsystem $b$, as the identity operator is the same in every basis ($U I U^{\dagger} = I$). The state of subsystem $b$ will not be changed following the measurement of $A$.

As to your question about writing it more neatly. If (but only if) the sum over $n$ covers all the basis states other than $n=1$ then you can replace it with the projector that completes $|\phi_1\rangle\langle\phi_1|$ as $$ A = \left[\lambda_1 |\phi_1\rangle\langle\phi_1|+\lambda_2\left(I-|\phi_1\rangle\langle\phi_1|\right)\right]\otimes I_b $$

  • $\begingroup$ Thank you! I thought that was the case regarding the basis states of system b but I wasn't sure how to formally argue this. Does this mean with respect to calculating the probability of measuring $\lambda_2$, that equation 4 is correct? $\endgroup$
    – Coconut
    Commented Nov 5, 2019 at 11:36
  • $\begingroup$ @s.twenty yes. eq. 4 gives the probability to measure $\lambda_2$. $\endgroup$
    – user245141
    Commented Nov 5, 2019 at 12:02
  • $\begingroup$ Thank you. Lastly, I would like to ask because the product state is expressed in a separable form, does this mean that systems a and b are not entangled and therefore measurements of a and b are not correlated? $\endgroup$
    – Coconut
    Commented Nov 5, 2019 at 15:17

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