9
A quick and dirty list:
$T_{1}$ and $T_{2}$ - colloquially known as decoherence times, but slightly more precisely also as the (qubit) relaxation time ($T_{1}$) and the (qubit) dephasing time ($T_{2}$).
$T_{1}$ is a measure of how quickly a qubit in the excited ($|1\rangle$) state spontaneously relaxes to the ground ($|0\rangle$) state.
$T_{2}$ is slightly ...
6
The IBM circuit composer used to be also known as IBM Quantum Experience, or IQX for short. For those historical reasons, the name of the style is iqx:
from qiskit import *
circuit = QuantumCircuit(2)
circuit.h(0)
circuit.cx(0, 1)
circuit.t(1)
circuit.draw('mpl', style='iqx')
Compare with the composer look:
6
That’s possible, just do
circuit.draw('mpl', style='iqx')
6
Just adding some stuff to the already good answer :
The gate time actually is related to the connexion between qubits, so is related to the CNOT, not the single-qubit gates.
The frequency is defined as the difference in energy between the ground and excited states, i.e. the |0⟩ and |1⟩ states, respectively.
for more info about how to measure all this (T1, ...
4
There will be a lot of swapping and additional gates during the execution of your circuit since CCX is not a native gate. The actual circuit that is being executed is something like:
This circuit has 11 qubits and depth of 79 and close to 100 CNOT gates, this is way too much for current hardware. Below is the noise level for Melbourne.
4
The number of qubits is part of the backend configuration:
FakeManhattan().configuration().n_qubits
65
If you need to filter the list of mocked backends based on the amount of qubits:
from qiskit.test.mock import FakeProvider
provider = FakeProvider()
[ b.name() for b in provider.backends() if b.configuration().n_qubits > 20]
['fake_cambridge',
'...
4
All the jobs are sent in a non-blocking way. You can send all your jobs with job = backend.run and recover their results in a fully different session.
In order to fully recover your Job object from a backend, you need the job id (given you are running Qiskit 0.24 or later. See @jyu00's comment)
You can save your job ids like this:
jobs = []
for circuit in ...
3
The $R_{YY}$ gate is not available within the Qiskit composer. This is because you can build it from other available gates within the composer. For instance, there is a nice way to represent $R_{YY}$ using the set $\{ R_{XX}, R_X, R_Y, P \}$. That is, suppose you want to implement $R_{YY}(\theta)$ then you can do it as:
Now, there is an option to add custom ...
3
This post contains a few questions, so let's first recap the steps the circuit sampler goes through (you can also check out the source code here), before answering your questions.
How the Circuit Sampler works
Provided with a backend/quantum instance and an operator expression, the job of the circuit sampler is to execute all circuits in the operator ...
3
This is just a quirk of how complex numbers are implemented in Python/Numpy, etc. At the end of the day, these are represented as floating-point numbers within the target simulator. These are then transformed via various mathematical operations to implement the simulation and this eventually leads to an accumulation of round off error. For all intents and ...
3
To follow up on hizqial's answer, these two spheres represent two different things.
The Bloch sphere is a way to visualize the state of a single qubit in every way possible. On the image below you see that the North Pole of the sphere represents the $|0\rangle$ state, whereas the South Pole represents the $|1\rangle$ state. Every point between the two ...
2
If you want, everything is detailed here about the system configuration, it will explain everything you can find in the configuration.
Just a quick resume about how to see the info : personally, I sometimes use backend.configuration().to_dict(), with this all of the information will be printed in a dict, and it's pretty easy to use. Now, you can also do a &...
2
A single qubit can be written down as $\cos(\theta/2)|0\rangle + \mathrm{e}^{i\varphi}\sin(\theta/2)|1\rangle$. So, the qubit is described by two angles ($\theta$ and $\varphi$). Hence, it can be mapped to a unit sphere (or with any other radius but then unit is used as a convention). This is a Bloch sphere, a visualization of a sigle qubit.
A Q-spehere is a ...
2
I think you underestimated how long your circuit really is....
When running your circuit on the hardware, it has to be transpile into the set of gates that is known to the hardware. For IBM machines, these are $\{ CX, ID, RZ, SX, X \}$ . Furthermore, there is a constraint on the qubit layout of the hardware as well. Not all qubits are connected. Thus there ...
1
If you want to get the final qubit layout (regarding to your first question), you can look at the circuit data. Here is an example:
circuit = QuantumCircuit(4, 4)
for i in range(4):
circuit.x(i)
circuit.measure([0,1,2,3], [0,1,2,3])
Circuit_Transpile = transpile(circuit, provider.get_backend('ibmq_16_melbourne') ,
...
1
Here is the qubit layout and error map for `ibmq_quito':
And when you try to execute your original circuit on 'ibmq_quito', what you are executing is something like the circuit below given ibmq_quito's qubits connectivity* and its native gate set ${CX, ID, RZ, SX, X}$ . This circuit has 196 gates altogether with about 50 CNOT gates.
You can use the error ...
1
On the Qiskit slack there have been others who have the same issue connecting to Quantum Lab. A 503 Service Unavailable Error may be due to service maintenance or being overloaded. You may have to just wait for the server to return.
1
Yes, you can do that without a problem. They only control the total number of experiments or jobs that you have submitted simultaneously. As long as the total number of submitted jobs (currently 5) are met with your account, regardless of the specifications from where they are submitted, it is ok.
Good luck!
1
A Bloch Sphere is the state space of all possible points to which a state vector can point to. A Bloch sphere can only demonstrate the state of one qubit while a Q-sphere can demonstrate multiple qubits. We use Q-sphere more often than not because quantum circuits are generally more than one qubit.
1
Without the OpenQASM code to import this is hard to verify what went wrong. However, in the case of unexpected results, it is important to note that every SWAP gate gets decomposed into 3 CX gates, and every CCX gate becomes six CX gates and a handful of single-qubit gates. So your circuit is already quite long when decomposed. Add to this the need for ...
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