2
$\begingroup$

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., $$ |A\rangle = \frac{1}{2} |00\rangle + \frac{1}{2} |01\rangle + \frac{1}{2} |10\rangle + \frac{1}{2} |11\rangle$$

$\endgroup$
1

3 Answers 3

6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $|\frac{1}{\sqrt{2}}|^2+|\frac{1}{\sqrt{2}}||^2=1)$ $\endgroup$ Mar 26 at 15:21
0
$\begingroup$

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]
```
$\endgroup$
0
$\begingroup$

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 $$\langle \psi | \psi \rangle = |\alpha_1|^2+|\alpha_2|^2+|\alpha_3|^2+|\alpha_4|^2 = 1$$

$\endgroup$
6
  • $\begingroup$ Is Hermitian Norm some well-defined and regularly used terminology/quantity/name? I have never heard the term Hermitian Norm in any literature before. $\endgroup$
    – FDGod
    Mar 26 at 16:01
  • $\begingroup$ If you just meant the regular Euclidean norm using the inner product, technically, you are missing a square root. We define it as $$|| |\psi\rangle || = \sqrt{\langle \psi | \psi \rangle}\,.$$ $\endgroup$
    – FDGod
    Mar 26 at 16:05
  • $\begingroup$ 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. $\endgroup$ Mar 26 at 16:14
  • $\begingroup$ 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. $\endgroup$
    – FDGod
    Mar 26 at 16:26
  • $\begingroup$ This is a word used in French maths textbooks more than English ones, I think. $\endgroup$ Mar 26 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.