# Measurement probability of a state from three hilbert spaces

I'm curious how to find the probability measurement of a state when one qubit is measured. For example this state:

$$|\gamma\rangle = \frac{1}{\sqrt{2}}(| 010 \rangle + | 101 \rangle )$$

Assuming these are qubits and if we measure the last qubit in the $$| + \rangle ,| - \rangle$$ basis, what is the probability measurement of the state after measuring the third qubit?

A probability usually means something like $$|\langle\phi|\psi \rangle|^2$$ but if we take the results from after measuring the last qubit in the $$+,-$$ basis we get:

$$|\psi_+\rangle$$ = $$\frac{1}{2}(| 01 \rangle + | 10 \rangle )$$

$$|\psi_-\rangle$$ = $$\frac{1}{2}(| 01 \rangle - | 10 \rangle )$$

How would we calculate the corresponding measurement probabilities of these two states? Is it really just $$|\langle\psi_+|\psi_+ \rangle|^2$$ Sounds too easy/doesn't make much sense.

• What does "the probability measurement of the state" mean? – Rammus Feb 26 at 20:19
• That's what I am trying to figure out. This was a question I saw somewhere. – mikanim Feb 26 at 21:43

A measurement can be associated with a decomposition of the identity $$I$$ into a sum of projectors. For example, the measurement of a qubit in the $$| + \rangle ,| - \rangle$$ basis is associated with the decomposition $$I = P_+ + P_- = |+\rangle\langle+| ~+~ |-\rangle\langle-|.$$ Then the probabilities are $$\text{Tr}(P_+|\phi\rangle\langle\phi|) = |\langle\phi|+ \rangle|^2$$ and $$\text{Tr}(P_-|\phi\rangle\langle\phi|) = |\langle\phi|- \rangle|^2$$ if we measure a qubit state $$|\phi\rangle$$.

When we measure only one subsystem this is equivalent to a measurement of the whole system with a naturally extended decomposition. For example, in your case it will be $$I_{123} = I_{12} \otimes P_+ + I_{12} \otimes P_-,$$ where $$I_{123}$$ is the identity operator on the whole system of three qubits, and $$I_{12}$$ is identity on just first two qubits.

Then the probabilities are $$p_+ = \text{Tr}((I_{12} \otimes P_+)|\gamma\rangle\langle\gamma|)$$ with the post-measurement state $$\frac{1}{\sqrt{p_+} } (I_{12} \otimes P_+) |\gamma\rangle$$ and $$p_- = \text{Tr}((I_{12} \otimes P_-)|\gamma\rangle\langle\gamma|)$$ with the post-measurement state $$\frac{1}{\sqrt{p_-} } (I_{12} \otimes P_-) |\gamma\rangle$$.

If you are not interested in post-measurement states then there is another approach of calculating probabilities $$p_+,p_-$$. You can calculate the density matrix of the third qubit $$\rho_3 = \text{Tr}_{12} (|\gamma\rangle\langle\gamma|) = \text{Tr}_{12}\big(\frac{1}{2}(| 010 \rangle + | 101 \rangle ) (\langle 010 | + \langle 101 | )\big)$$ $$= \text{Tr}_{12}\big(\frac{1}{2}(|010\rangle\langle 010| + |010\rangle\langle 101| + |101\rangle\langle 010| + |101\rangle\langle 101| )\big) =$$ $$= \frac{1}{2} (|0\rangle\langle 0| + |1\rangle\langle 1|) = \frac{1}{2} I.$$

Then the probabilities are $$p_+ = \text{Tr}(P_+ \rho_3) = \text{Tr}(P_+ \frac{1}{2} I) = \frac{1}{2}$$ and $$p_- = \text{Tr}(P_- \rho_3) = \text{Tr}(P_- \frac{1}{2} I) = \frac{1}{2}$$.

The post-measurement states will actually be $$\frac{1}{\sqrt{p_+} } (I_{12} \otimes P_+) |\gamma\rangle = (I_{12} \otimes |+\rangle\langle+|)(|010\rangle + |101\rangle) = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)|+\rangle$$ and $$\frac{1}{\sqrt{p_-} } (I_{12} \otimes P_-) |\gamma\rangle = (I_{12} \otimes |-\rangle\langle-|)(|010\rangle + |101\rangle) = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)|-\rangle$$

• Thanks for the insightful comment. The post-measurement states are wrong with the prefactor p_+,- are in the denominator and they equal 1/2. Also, could you point me to the derivation why the probability is 1/sqrt(p) for the post-measurement state? – mikanim Feb 27 at 20:09
• Better put: The post-measurement states are wrong with the prefactor p_+,- are in the denominator. With how you now have it, the state would have sqrt(2) as the prefactor which wouldn't give us a probability bigger than 1 for each state. – mikanim Feb 27 at 21:37
• $|\gamma\rangle$ has a factor $1/\sqrt{2}$, which I removed in a couple with $1/\sqrt{p} = \sqrt{2}$ – Danylo Y Feb 28 at 9:41
• The last $1/\sqrt{2}$ factor is due to projection. $\langle - | 0 \rangle = 1/\sqrt{2}$. – Danylo Y Feb 28 at 9:43
• It's just a postulate, classical Nielsen&Chuang book has it. – Danylo Y Feb 28 at 11:34