How to calculate the probabilities of observing quantum states using the "expectation_from_wavefunction" in cirq (and why)

How to calculate the probabilities of observing quantum states using the "expectation_from_wavefunction" in cirq (and why) Ask Question Asked 5 years, 11 months ago Modified yesterday Viewed 670 times 1 I am currently simulating some quantum circuits, and want to calculate the probabilities of observing each individual state. I am able to use Cirq for this, and calculate it using P 00 =|α | 2 𝑃 00 = | 𝛼 | 2 . Code: import cirq import sympy x0, x1 = sympy.symbols('x0 x1') q = cirq.GridQubit.rect(1, 2) circuit = cirq.Circuit( cirq.rx(x0).on(q[0]), cirq.rx(x1).on(q[1]), cirq.ry(3.14/4).on(q[0]), cirq.ry(3.14/4).on(q[1])) resolver = cirq.ParamResolver({x0: 0.2, x1: 0.3}) simulator = cirq.Simulator() results = simulator.simulate(program=circuit, param_resolver=resolver, qubit_order=q).final_state print("Internal quantum state:", results) print("Probabilities of observing each state:", [abs(x)**2 for x in results]) Output: internal quantum state: [0.8377083+0.08743566j 0.3529709-0.11249491j 0.35297093-0.06251574j 0.13120411-0.08743566j] probabilities of observing each state: [0.7094002059173512, 0.13724355108492148, 0.12849669756020887, 0.024859516013751914] However, in multiple tutorials (for instance from TFQ) I see the use of "expectation_from_wavefunction": z0 = cirq.Z(q[0]) qubit_map={q[0]: 1, q[1]: 1} z0.expectation_from_wavefunction(results, qubit_map).real output: 0.6757938265800476 My question: How can I use expectation_from_wavefunction to obtain the probabilities of observing the individual states ( P 00 , P 01 , P 10 , P 11 𝑃 00 , 𝑃 01 , 𝑃 10 , 𝑃 11 )? Bonus question: why would I favor this approach? simulationcirq Share Improve this question Follow asked Mar 22, 2020 at 9:24 Thomas Hubregtsen 6241 1 gold badge 4 4 silver badges 11 11 bronze badges Why would want to use "expectation_from_wavefunction()" to calculate the probabilities when you can just do what you did; square the values of your wavefunction? The former takes more work than the latter. –  Victory Omole Commented Mar 22, 2020 at 14:24 @VictoryOmole Because in the tutorial of TFQ that I refer to, they only use the "expectation_from_wavefunction". All following examples build upon it. I think it helps with batching, but I am not sure. –  Thomas Hubregtsen Commented Mar 22, 2020 at 21:06 sorry, forgot to mention: the top approach works in Cirq. I am trying to run Tensorflow Quantum, and all tutorials here rely on this method. –  Thomas Hubregtsen Commented Mar 22, 2020 at 21:23 Tensorflow Quantum combines Tensorflow with Cirq. If you can "import tfq" you can "import cirq" and thus use all the functionality in Cirq. –  Victory Omole Commented Mar 23, 2020 at 16:14 expectation_from_wavefunction is used when you don't want to write the logic for yourself. This is more useful in cases with multi-qubit observables involving the X and Y axies. –  Craig Gidney Commented Apr 21, 2020 at 22:56 Add a comment 2 Answers Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 0 why would I favor this approach? The Expectation value is defined as ⟨A⟩=⟨ψ|A|ψ⟩ ⟨ 𝐴 ⟩ = ⟨ 𝜓 | 𝐴 | 𝜓 ⟩ where ψ 𝜓 is the wavefunction and A 𝐴 is the operator. Use "expectation_from_wavefunction()" if you don't want to write code that calculates ⟨ψ|A|ψ⟩ ⟨ 𝜓 | 𝐴 | 𝜓 ⟩ . Share Improve this answer Follow answered Mar 22, 2020 at 14:28 Victory Omole 2,7061 1 gold badge 11 11 silver badges 28 28 bronze badges Thanks. I see now that the bottom approach calculates the expectation value, which is defined as ⟨𝐴⟩=⟨𝜓|𝐴|𝜓⟩. But how does this relate to the answer from the top approach? Because I see no relation between the numeric outcomes. For a 1-qubit system, would the expectation value not give me the probability of observing the qubit in the state 1? –  Thomas Hubregtsen Commented Mar 22, 2020 at 21:49 If you don't see the relation, what calculation did you perform? What do you get? –  Victory Omole Commented Mar 23, 2020 at 17:00 The code and its output should be shown in my question. I embed using Rx followed by an Ry. The top approach provides: internal quantum state: [0.8377083+0.08743566j 0.3529709-0.11249491j 0.35297093-0.06251574j 0.13120411-0.08743566j] probabilities of observing each state: [0.7094002059173512, 0.13724355108492148, 0.12849669756020887, 0.024859516013751914] The bottom approach (that seems to have an extra Rz, which I have also experimented with in the top approach to no affect) gives as output for the real component 0.6757938265800476. –  Thomas Hubregtsen Commented Mar 23, 2020 at 21:20 I meant: did you take the wavefunction and do a pen-and-paper calculation of that expectation value and get the same answer that z0.expectation_from_wavefunction(results, qubit_map).real gives you? –  Victory Omole Commented Mar 23, 2020 at 23:17 Add a comment 0 The expectation value that the system provided me with was 0.6757938265800476. This was in a range of [-1,1]. Mapping this to a range of [0,1]: (0.6757938265800476+1)/2=0.837896913290024. The expectation value was the expectation value for qubit 0. This implies that it observed qubit 0 in state 0 83.7896913290024% of the time, and 1-0.837896913290024 in state 1. The system also printed the following probabilities: P 00 =0.7094002059173512 𝑃 00 = 0.7094002059173512 P 01 =0.13724355108492148 𝑃 01 = 0.13724355108492148 P 10 =0.12849669756020887 𝑃 10 = 0.12849669756020887 P 11 =0.024859516013751914 𝑃 11 = 0.024859516013751914 The probability to observe qubit 0 (indexing right to left) in state 0: E q0 =0.8378969=0.7094002+0.1284966= P 00 + P 10 𝐸 𝑞 0 = 0.8378969 = 0.7094002 + 0.1284966 = 𝑃 00 + 𝑃 10 To answer the question "How can I use expectation_from_wavefunction to obtain the probabilities of observing the individual states": you can't. You can only relate this particular expectation value (with Pauli Z) to a set of state probabilities, but not to individual ones without extra information (such as observables in different computational basis) "Why would I favor this approach": access to the internal quantum state is not realistic with quantum system, and would require more computation. P.S. this is my current interpretation which I expect to be right on a high level. I think I still have details wrong, such as parts of the explanation. Very open to opinions. Share Improve this answer Follow edited Apr 22, 2020 at 11:08 answered Apr 22, 2020 at 11:02 Thomas Hubregtsen 6241 1 gold badge 4 4 silver badges 11 11 bronze badges Add a comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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