Hello, I have been confusing about the problems for a long time.

My code is illustrated as below:

```
import pennylane as qml
import numpy as np
qubit_num = 2
dev = qml.device("default.qubit", wires=qubit_num)
@qml.qnode(dev)
def circuit(x):
qml.RZ(x[:, 0], wires=0)
qml.RZ(x[:, 1], wires=1)
return qml.state()
x = np.random.randn(5, 2)/10
print(x)
print(circuit(x))
'''
terminal:
[[-0.00343289 0.07952211]
[-0.11131135 -0.0997912 ]
[-0.11075909 -0.22640184]
[ 0.15814072 -0.11102991]
[ 0.00028335 -0.10640625]]
0. +0.j ]
[0.99443463+0.10535539j 0. +0.j 0. +0.j
0. +0.j ]
[0.98582393+0.1677831j 0. +0.j 0. +0.j
0. +0.j ]
[0.99972258-0.02355323j 0. +0.j 0. +0.j
0. +0.j ]
[0.99859257+0.05303655j 0. +0.j 0. +0.j
0. +0.j ]]
'''
```

As you can see, the parameters of `qml.RZ`

are 5-dim vectors.

Then the circuit returns five different states.

However, if I replace `qml.state()`

with expectations, the circuit will return identical values:

```
@qml.qnode(dev)
def circuit(x):
qml.RZ(x[:, 0], wires=0)
qml.RZ(x[:, 1], wires=1)
exp_vals_z = [qml.expval(qml.PauliZ(wires=w)) for w in [0, 1]]
return exp_vals_z
x = np.random.randn(5, 2)/10
print(x)
print(circuit(x))
'''
terminal:
[[-0.09886417 -0.08193338]
[-0.07531188 0.0685733 ]
[-0.04081865 0.06245962]
[ 0.13144776 -0.09798962]
[-0.16170616 0.07845339]]
[[1. 1. 1. 1. 1.]
[1. 1. 1. 1. 1.]]
'''
```

The measurement values for each wire are identical.

Could I say that the values are identical since the values are averaged (i.e. expectation values)?

Thanks for your reply in advance!

Sincerely,

Jeff

1 Like

Hi @q36111095! The reason why you are getting different state vectors but the same expectation values is that you are only applying RZ gates. Qubits are initialized in the |0\rangle state and therefore performing a qml.RZ gate only introduces a phase. This is not reflected when computing the expectation value, as the probability of being in the |0\rangle Z eigenstate (with eingenvalue 1) is still one.

You can try a different circuit and you will observe different expectation values, for example

```
@qml.qnode(dev)
def circuit(x):
qml.RX(x[:, 0], wires=0)
qml.Hadamard(0)
qml.RX(x[:, 1], wires=1)
exp_vals_z = [qml.expval(qml.PauliZ(wires=w)) for w in [0, 1]]
return exp_vals_z
print(qml.draw(circuit)(x))
x = np.random.randn(5, 2)
print(x)
print(circuit(x))
```

```
0: ββRX(M1)ββHββ€ <Z>
1: ββRX(M0)βββββ€ <Z>
[[ 0.13833687 -0.60442826]
[ 0.05236643 0.8771775 ]
[-0.58109614 -0.26983257]
[ 0.94531585 0.9756361 ]
[-0.28200791 0.45957891]]
[[0. 0. 0. 0. 0. ]
[0.82282715 0.63932402 0.96381554 0.56064144 0.89623936]]
```

Hope this can help!

2 Likes

Hey everyone, and welcome to the forum @q36111095!

@Pablo_Vinasβs answer here is correct . Letβs see the math:

\begin{align*}
\vert \psi \rangle = R_z(\theta) \vert 0 \rangle &= e^{-i \theta / 2} \vert 0 \rangle \\
\langle \psi \vert Z \vert \psi \rangle &= \langle 0 \vert e^{i \theta / 2} Z e^{-i \theta / 2} \vert 0 \rangle \\
&= \langle 0 \vert Z \cancelto{1}{e^{-i \theta / 2} e^{i \theta / 2}} \vert 0 \rangle \\
& = \langle 0 \vert Z \vert 0 \rangle \\
& = 1.
\end{align*}

But, regarding the 5-dim vectors point, PennyLane will automatically broadcast parameters! This was a feature introduced in v0.24. See our blog post for more details!

2 Likes

Thanks for your reply again!!!

Okay, I will try some different gates to introduce features into quantum circuits.

It really helps me!

Sincerely,

Jeff

2 Likes

Oh, I got it!!

The phase will disappear when we measure expectation values of states!

Thanks for your derivation @isaacdevlugt , it really helps me!!

Sincerely,

Jeff

1 Like

Awesome! Glad to hear this worked for you

1 Like