# Is $|A\rangle = \frac{1}{\sqrt2} |00\rangle + \frac{1}{\sqrt2} |01\rangle$ a valid quantum state?

Is $$|A\rangle = \frac{1}{\sqrt2} |00\rangle + \frac{1}{\sqrt2} |01\rangle$$ a valid quantum state? Or does a quantum state need to be a superposition of the entire basis, i.e., $$|A\rangle = \frac{1}{2} |00\rangle + \frac{1}{2} |01\rangle + \frac{1}{2} |10\rangle + \frac{1}{2} |11\rangle$$

• One way to check your work computationally is with Qiskit's Statevector.is_valid() docs.quantum.ibm.com/api/qiskit/… Mar 22 at 3:58

Yes, the state $$|A\rangle = \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |01\rangle$$ is indeed a valid quantum state. In quantum mechanics, a valid quantum state can be any normalized linear combination of basis states.

The two states you've mentioned, $$|A\rangle = \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |01\rangle$$ and $$|A\rangle = \frac{1}{2} |00\rangle + \frac{1}{2} |01\rangle + \frac{1}{2} |10\rangle + \frac{1}{2} |11\rangle$$ are both valid quantum states, and they represent different superpositions of the basis states $$|00\rangle$$ and $$|01\rangle$$.

The state $$|A\rangle = \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |01\rangle$$ represents a superposition where the system is in a state that is a linear combination of $$|00\rangle$$ and $$|01\rangle$$, with equal probability amplitudes $$\frac{1}{\sqrt{2}}$$.

On the other hand, the state $$|A\rangle = \frac{1}{2} |00\rangle + \frac{1}{2} |01\rangle + \frac{1}{2} |10\rangle + \frac{1}{2} |11\rangle$$ represents a superposition where the system is in a state that is a linear combination of all four basis states $$|00\rangle$$, $$|01\rangle$$, $$|10\rangle$$, and $$|11\rangle$$, with equal probability amplitudes $$\frac{1}{2}$$.

Both states are valid quantum states.

• Very nice answer! And just to add a little bit: a very easy way to check if a state is valid is, if all the terms contain orthogonal states (e.g. taking the $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle$ example, we see that $|00\rangle$ and $|01\rangle$ are indeed orthogonal), then you just check whether the sum of the square of the amplitudes is equal to 1 (e.g. $|\frac{1}{\sqrt{2}}|^2+|\frac{1}{\sqrt{2}}||^2=1)$ Mar 26 at 15:21

To add to the answer from Yet another Random Guy, you can see for yourself by recreating the state programmatically (here using the Amazon Braket SDK):

from braket.circuits import Circuit
from braket.devices import LocalSimulator

circuit = Circuit().i(0).h(1).state_vector()
LocalSimulator().run(circuit).result().values[0]
$$$$
`

Given that we are talking about the state of 2 qubits, the canonical basis of the state space is $$(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$$

A physically valid quantum state is just a vector in that vector space whose Hermitian norm is 1 - normalized-, ie $$|\psi\rangle=\alpha_1|00\rangle + \alpha_2 |01\rangle + \alpha_3|10\rangle + \alpha_4 |11\rangle$$ $$\forall i,\quad\alpha_i \in\mathbb{C}$$ with $$\langle \psi | \psi \rangle = |\alpha_1|^2+|\alpha_2|^2+|\alpha_3|^2+|\alpha_4|^2 = 1$$

• Is Hermitian Norm some well-defined and regularly used terminology/quantity/name? I have never heard the term Hermitian Norm in any literature before. Mar 26 at 16:01
• If you just meant the regular Euclidean norm using the inner product, technically, you are missing a square root. We define it as $$|| |\psi\rangle || = \sqrt{\langle \psi | \psi \rangle}\,.$$ Mar 26 at 16:05
• Since the condition is to be equal to 1 it is completely equivalent to use the square of the norm or the norm. Much more readable without the square root. And Euclidean for me refers to real vector space not complex. I did not make a study of the use of the word Hermitian, but obviously you understood the meaning this is only what is important to me. Mar 26 at 16:14
• Since you used it, I was just trying to learn something new, a quantity that I had not come across before called the hermitian norm. So, that's why I just asked for clarification. Mar 26 at 16:26
• This is a word used in French maths textbooks more than English ones, I think. Mar 26 at 16:30