# Expectation value of a quantum circuit [closed]

The expectation value of an operator $$A$$ is defined by this equation $$\langle A \rangle_\psi = \sum_j a_j |\langle \psi | \phi_j \rangle|^2$$.

My first question is does it mean that the expectation can be a negative value? If we want to calculate the observable $$Z$$ in the state $$|1\rangle$$, the expectation value is -1, right?

Suppose we have a four-qubit circuit, and we measure it using qiskit. As the measurement is done on $$Z$$ basis, it corresponds to calculate the expectation value of observable $$ZZZZ$$. Let's say the output state is $$|0001\rangle$$ using a simulator.

Since the eigenvalue of $$|0001\rangle$$ is -1, the expectation value, in this case, is -1. (Is it correct?)

When I execute the circuit on a real quantum chip with shots 1170, I get the result like: $$|0000\rangle : 100$$, $$|0001\rangle :900$$, $$|1001\rangle : 50$$, $$|0011\rangle : 120$$.

How can I calculate the expectation value based on the results?

Is it equal to the sum of each state times its probability ($$0\times\frac{100}{1170} + 1\times\frac{900}{1170}+ 9\times\frac{50}{1170} + 3\times\frac{120}{1170}$$)

or the sum of state probability times its eigenvalue $$(1\times100 + (-1)\times900 + 1\times50 + 1\times120)/1170$$?

• The last one. And yes, expectation values can be negative. May 19 '21 at 14:19
• @peachnuts Note that in general, to get the exact expectation, you will need to do infinite number of experiments/shots. May 19 '21 at 15:43