# How to calculate the projector for a joint product space? (Quantum Mechanics)

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.

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 $$A = \left[\lambda_1 |\phi_1\rangle\langle\phi_1|+\lambda_2\left(I-|\phi_1\rangle\langle\phi_1|\right)\right]\otimes I_b$$
• 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? Nov 5 '19 at 11:36
• @s.twenty yes. eq. 4 gives the probability to measure $\lambda_2$.