# What is the difference between the quantum states $\left|01\right> + \left|10\right>$ and $\left|01\right> - \left|10\right>$? [duplicate]

I am finding it difficult to understand the four states of two entangled qubits, $$\left|00\right>$$ $$\left|01\right> + \left|10\right>$$ $$\left|01\right> - \left|10\right>$$ $$\left|11\right>$$

This Veritasium video states that two qubits have these four states, and calls the $\left|01\right> + \left|10\right>$ the $\left|T_0\right>$ state, and $\left|01\right> - \left|10\right>$ the singlet state, $\left|S\right>$.
However, what does the difference between the minus and plus signify? I understand that in both states the qubits have the property of being opposite to one another, but that's it.

It would be greatly appreciated if someone could help me understand this difference!

Before I mark an answer as correct, the main understanding I now have, is that, although when squared having an identical probability, the states differ when transformations are applied to it?
Is this basically like this image, taken from here? Note; I think my question differs from this post. The OP of that post already understands the presence and meaning of the minus sign, but rather asks why the minus sign is associated with $S=0$, which is a different question. Yet this question has been marked as duplicate, without any argument why it is still the same. I would like to see an argument against.

• A nice way to get a better understanding is to have a look at a system of two spin 1/2 particles: wikiwand.com/en/Triplet_state It is the difference between the singlet and one of the triplet states which you would like to understand. Nov 6, 2016 at 18:41
• @pawel_winzig thank you for your reference. I see that the answer is in there, but I still fail to understand the text; how these states are derived. Nov 6, 2016 at 19:18
• @knzhou I would argue against. The OP of that post already understands the presence and meaning of the minus sign, but rather asks " why is the combination with the minus sign is the one for S=0?", which is a different question. Nov 6, 2016 at 19:22
• Veritasium is an idiot. Don't watch his videos. Aug 16, 2022 at 17:42

The name of the concept you are looking for is probability amplitude.

The two states $\lvert T\rangle = \frac{1}{\sqrt2}(\lvert 10\rangle + \lvert 01\rangle)$ and $\lvert S\rangle = \frac{1}{\sqrt2}(\lvert 10\rangle - \lvert 01\rangle)$ you mention differ in the probability amplitude of finding the system in the state $\lvert 01 \rangle.$

In both cases the probability of finding the system in the state $\lvert 01\rangle$ is $1/2$, but nonetheless, the two states are very different.

One way to see this is to ask how do they evolve under some transformation (some gate, in the common language of quantum information you may be reading about). Take for example the unitary evolution described by the 4x4 Hadamard matrix: $$U = \begin{pmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&-1&1\\1&-1&1&-1\end{pmatrix}.$$ This is To do this, you have to multiply $U$ by $\lvert T\rangle$ and $\lvert S \rangle$. If you do this, you will find that $\lvert T \rangle$ evolves into the state $$\frac{1}{\sqrt2}(\lvert 00 \rangle - \lvert 10 \rangle),$$ while $\lvert S \rangle$ evolves into the state: $$\frac{1}{\sqrt2}(\lvert 01 \rangle - \lvert 11 \rangle).$$ As you can see, these final states are completely different, which is a consequence of the fact the initial states were also very different.

• Hi, please see my edit Nov 6, 2016 at 20:10
• @Isaiah yes it is like what you said in your edit, with the only difference that in that picture you have a single qubit, while your states are two-qubit states, which is why I had to use a 4x4 Hadamard matrix instead of the simpler 2x2 one.
– glS
Nov 6, 2016 at 20:49
• Thank you. One more question, why is 10-01 no basic state? Nov 7, 2016 at 11:39
• @Isaiah what do you mean by "basic state"?
– glS
Nov 7, 2016 at 12:15
• yes (but we call it "superposition" of states, not "mixed" states, as the word "mixed" in this context has an entirely different meaning)
– glS
Nov 7, 2016 at 16:40

The Bell states are the states $$|-,~+\rangle~+~|+,~-\rangle$$ $$|-,~+\rangle~-~|+,~-\rangle$$ $$|-,~-\rangle~+~|+,~+\rangle$$ $$|-,~-\rangle~-~|+,~+\rangle.$$ The combinations of $|s_1,~m_1\rangle$ and $|s_2,~s_2\rangle$ form a total spin state by $$|s,~m\rangle~=~\sum_{m_1+m_2=m}C^{ss_1s_2}_{mm_1m_2}|s_1,m_1\rangle|s_2,m_2\rangle,$$ for $C^{ss_1s_2}_{mm_1m_2}$ the Clebsch-Gordon coefficient. We then have the following sums: $$|0,~0\rangle~=~\frac{1}{\sqrt{2}}(|+,~-\rangle|-,~+\rangle~-~|-,~+\rangle|+,~-\rangle)$$ $$|1,~0\rangle~=~\frac{1}{\sqrt{2}}(|+,~-\rangle|-,~+\rangle~+~|-,~+\rangle|+,~-\rangle)$$ $$|1,~1\rangle~=~|+,~+\rangle$$ $$|1,~-1\rangle~=~|-,~-\rangle$$ Because the first of these has $s~=~0$ it is called a singlet state. The other three with $s~=~1$ are triplet states. The last two of these form two linearly independent combinations to form the Bell states.

It is not hard to show these are linearly independent. Computations between the triplet states $\langle 1,~s|1,~m'\rangle~=~0$ for $m~\ne~m'$. Similarly, inner products between the singlet state and any of the triplet states is zero.

I thougt, those states are just some functions, and we give them a name, $\left| 1 \right>$ for example. So if the choice of these functions is arbitrary, how can the $+$ or $-$ sign make a difference?
The statement is correct in some cases. But now you add $\left| 10 \right> + \left| 01 \right>$, and at this point it is too late to make arbitrary choices: both $\left| 1 \right>$s in each of the summands mean the same! Well, they mean the same single-particle-state, and in the first summand the first particle is in it, and in the second summand the secont particle.
If you accept this, then there is an obvious difference between the states $\left| 10 \right> + \left| 10 \right>$ and $\left| 10 \right> - \left| 10 \right>$ with respect to what happens, if you interchange the particles. The first expression doesn't change - that's why we call this state (two-particle state, constructed in a clever fashion of arbitrary single-particle-states) symmetric - whereas the second state is called antisymmetric, since it changes its sign.