Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
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The key to understanding many quantum protocols and circuits is in the following circuit: This is especially true in the case where $U^2=I$, such that $U$ has eigenvalues $\pm1$. You can readily calculate that if the input, $|\psi\rangle$, of the second qubit has an amplitude $\alpha_+$ for being supported on the $+1$ eigenspace, then at the end of the ...


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Mathematically it is a relationship between a bipartite linear operator vector space $L(X\otimes Y)$ and a superoperator vector space $C(X): L(X)\to L(Y)$ (maps of linear operators to linear operators). Bipartite density matrices are contained in the former, and quantum channels in the latter. The real "physical" meaning of the isomorphism for quantum ...


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Write down the two reduced density matrices of the single qubits that you have access to. Apply the Helstrom measurement (there are several descriptions of this on the site already). The problem is that, in this case, the two reduced density matrices are the same. That means that you cannot tell them apart. More explicitly, $$ |\varphi_2\rangle=(I\otimes X)|...


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What that cSWAP test does (and doesn't) do The important thing about the controlled-SWAP test is that what it does isn't just to SWAP, or to not SWAP, the two inputs. The controlled-SWAP test involves a control qubit which is in a superposition of $\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\ket{0}$ and $\ket{1}$: that is, we measure the first ...


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Apparently $\vert W \rangle$ was first reported (and the naming convention first adopted) by Dür, Vidal and Cirac in this preprint on May 26, 2000 (version 1 of 2). This is supported by the footnote on page 4 of this preprint on June 25, 2000 (version 3 of 3, this footnote did not appear in the earlier versions), which states (in part) Very recently Dürr ...


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$\mathrm{X}$ is not equivalent to a $\mathrm{CNOT}$ gate. The former is a 1-qubit gate whereas the 2nd is a 2-qubit gate (in essence, a controlled-$\mathrm{X}$). The $\mathrm{X}$ basically flips the state of qubit B i.e., $|0\rangle_B\to|1\rangle_B$ and $|1\rangle\to|0\rangle_B$, and does not depend on the state of qubit A.


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I think I understand where you're getting tripped up. When considering this portion of a circuit, it seems contradictory that both $\vert x \rangle$ is unchanged by $\text{CNOT}$, and $\text{CNOT}$ maps $\vert \psi_1 \rangle = \vert +- \rangle \rightarrow \vert -- \rangle = \vert \psi_2 \rangle$. It turns out that this is not a contradiction, but it is ...


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To prepare this state specifically, start with $|000\rangle$ and apply H to the 0th and 1st qubits, yielding $$ \frac{1}{4} (|000\rangle + |010\rangle + |100\rangle + |110\rangle) $$ Then, we can apply the Toffoli gate to the 0th and 1st qubit with the 2nd qubit as the target, yielding: $$ \frac{1}{4} (|000\rangle + |010\rangle + |100\rangle + |111\...


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Upon some reflection, the answer is that no, they most definitely do not. A first way to see this is to notice that a hypersphere has a very "regular" structure. Suppose there were some $2(d-1)$ basis (normalised) vectors $\{\mathbf v_1,...,\mathbf v_{2d-2}\}\subset\mathbb R^{d^2-1}$ forming a sphere containing the pure states. This obviously cannot be the ...


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This depends on precisely what you mean by "combinations". Let's go back to just one qubit to be clear. For one qubit, there are two classical states, $|0\rangle$ and $|1\rangle$. These are really the only things you can count. Of course, as a qubit, you are allowed any superposition of the form $$ \alpha|0\rangle+\beta|1\rangle,\qquad |\alpha|^2+|\beta|^2=1....


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Your understanding is correct. In the theory of photon polarization, the parametrization of the Bloch sphere (or its surfave) has traditionally another name. On the wikipedia page for the Jones calculus (the parametrization of the Bloch sphere surface), you'll find a table for the correspondence between kets and polarizations. To summarize, eigenstates ...


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Initially $|x\rangle|x\oplus y\rangle$ is a perfectly valid state. The first qubit is in $|x\rangle$ and the second qubit is in $|x\oplus y\rangle$ - that is, the second qubit is entangled with the first qubit. As you study other algorithms you might grok that we can have a state $|x\rangle|f(x)\rangle$ for some function $f(x)$ - the first register might ...


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IBM Q Experience and Qiskit both use the little-endian format. The multi-qubit state $|xy\rangle$ can be represented in a quantum circuit by putting the $|x\rangle$ qubit (left) on top, and the $|y\rangle$ one below, or by putting the $|y\rangle$ qubit (right) on top, and the $|x\rangle$ one below. Is that correct? Yes. The IBM Q ...


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I see the heart of your question. I'd like to clarify a bit, before answering your question though. Matrices (aka operators) do not measure quantum states--they operate on them. Specifically, they project the state into the matrix's eigenvectors. We can then measure that projected state in a particular basis that may be the same or different than what the ...


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Well, it's still $2^4 = 16$ I believe. The point of superposition is that these 16 combinations can encode 16 inputs, and with something called "quantum parallelism" they can be calculated simutaneously. Even though they don't give out all $16$ results simultaneouly, the result can be manipulated in various ways to give you what you need while drastically ...


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Pulling from another question (Big Endian vs Little Endian in Qiskit) and their documentation (https://community.qiskit.org/textbook/ch-prerequisites/qiskit.html), it appears that the register is intentionally labeled right-to-left so that integers are intuitively represented. Consider your example, where you applied X(qr[0]): $ |00\rangle \rightarrow |01\...


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Let's say $|a\rangle$ is the pure state qubit and $|b\rangle$ is the bell state. Putting $|a\rangle$ through the Pauli channel depends on which of the Pauli gates (which of the $\sigma_j$) you want acting on it. Once you choose this (call it $\sigma_a$), apply $2\times2$ Identity operators to it on either side until it's the same size as $\rho$, then just ...


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In principle, yes, you can always do it. The Bloch representation can be generalised to arbitrary dimensions, and you can always parametrise states in it by their "angle coordinates". For example, you can write an arbitrary 3-modes pure state as $$|\psi\rangle=\cos\alpha|0\rangle + e^{i\theta}\sin\alpha\cos\beta|1\rangle+e^{i\phi}\sin\alpha\sin\beta|2\...


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The state of single qubit can be described as a point on the Bloch sphere. All the allowed transformations of a single qubit can then be described as rotations on the Bloch sphere. Unfortunately, bigger quantum systems can no longer be described as fitting on a sphere like geometry. As a result, this idea of considering transformations as rotations does not ...


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So you want to calculate $\left<\beta |D\left(\alpha\right)|\beta\right>$ where $\left|\beta\right>$ is a coherent state and $D\left(\alpha\right)$ is the displacement operator. The easiest way of doing this is to take $\left<\beta |D\left(\alpha\right)|\beta\right> = \left<0|D^\dagger\left(\beta\right) D\left(\alpha\right)D\left(\beta\...


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In addition to @D. Tran's answer and for anyone out there (just in case), this did it for me. The first mistake I made was that I kept on thinking superimposed qubits could only hold 0s and 1s as pairs (2 states) i.e. say for $n=3$ qubits I thought it was equivalent of ##-##-## (e.g. 01-11-01 which is just $2^{n+n}$ combinations at max) but it turned out a ...


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As other have said, you can see 4 qubits as having either 16 different combinations, or as continuum-many superpositions of these combinations. (Continuum-many means "the cardinality of the real numbers") However you can also ask about the dimension of the space, whcich is a different notion of size. A single qbit is described by a vector in a 2-dimensional ...


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