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$\newcommand{\complexes}{\mathbb{C}}$I'm trying to make sense of Born's rule involving a single qubit. Probably, I'm mixing apples and oranges here, but I can't tell where or why. In "The Feynman Lectures on Physics", volume 3, chapter 1, Feynman describes the probability of finding an electron at a position $x$ to be dependent upon the numbers $\phi_1$ and $\phi_2$ (relative to the electron passing through the slit $1$ or $2$) as

\begin{align*} P_1 &= |\phi_1|^2, \text{where } \phi_1 \in \complexes \tag{1}\\ P_2 &= |\phi_2|^2, \text{where } \phi_2 \in \complexes\tag{2}\\ P_{12} &= |\phi_1 + \phi_2|^2\tag{3} \end{align*}

such that these numbers $\phi_1, \phi_2$ are to be defined by experiment.

When I put a qubit into uniform superposition, I write $$\frac{1}{\sqrt 2} |0\rangle + \frac{1}{\sqrt 2} |1\rangle\tag{4}$$ and then I'm taking the $1/\sqrt{2}$ to be $\phi_1$ and the other $1/\sqrt{2}$ to be $\phi_2$. When I ask myself --- what's the probability of finding the qubit in either state, I'm applying Born's rule and adding these amplitudes, computing the magnitude of the sum, getting $\sqrt{2}$, which must be wrong because when I square it I get $2$ and I should get $1$.

Where did I go wrong?

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2 Answers 2

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$\newcommand{\complexes}{\mathbb{C}}$I'm trying to make sense of Born's rule involving a single qubit. Probably, I'm mixing apples and oranges here, but I can't tell where or why. In "The Feynman Lectures on Physics", volume 3, chapter 1, Feynman describes the probability of finding an electron at a position $x$ to be dependent upon the numbers $\phi_1$ and $\phi_2$ (relative to the electron passing through the slit $1$ or $2$) as

\begin{align*} P_1 &= |\phi_1|^2, \text{where } \phi_1 \in \complexes \tag{1}\\ P_2 &= |\phi_2|^2, \text{where } \phi_2 \in \complexes\tag{2}\\ P_{12} &= |\phi_1 + \phi_2|^2\tag{3} \end{align*}

Yes, this is correct. Recall that the "1" and "2" refer to two different paths (via two different slits) by which the electron can make it to the detector. The amplitude $\phi_1+\phi_2$ is used when we can not distinguish which path the electron took (and there is thus interference).

When I put a qubit into uniform superposition, I write $$\frac{1}{\sqrt 2} |0\rangle + \frac{1}{\sqrt 2} |1\rangle\tag{4}$$ and then I'm taking the $1/\sqrt{2}$ to be $\phi_1$ and the other $1/\sqrt{2}$ to be $\phi_2$.

When I ask myself --- what's the probability of finding the qubit in either state, I'm applying Born's rule and adding these amplitudes,

No, you don't add the amplitudes here. In this case what is being measured is the spin and there are not multiple "paths" to interfere.

Where did I go wrong?

This is a good question. To understand the answer you should consider something in quantum computing that is analogous to having multiple interfering paths in the double slit experiment. But, in quantum computing we often ignore all the spatial degrees of freedom, so we are not usually looking at literal paths in real space.

So we want to look at a quantum computing analog to real-space propagation along a path. For a single qubit we do our "propagation" by acting with a 2x2 unitary matrix: $$ U = \left(\begin{matrix}a & b \\ -b^* & a^*\end{matrix} \right)\;, $$ where $|a|^2+|b|^2 = 1$

Suppose you start with the state: $$ |\psi\rangle = \psi_0|0\rangle + \psi_1|1\rangle\;, $$ (where $|\psi_0|^2+|\psi_1|^2=1$) and propagate it with $U$ and then measure spin-up (i.e., measure the bit value 0). What is the probability amplitude? It is: $$ P^U_{0} = |a\psi_0 + b\psi_1|^2\;, $$ where you now see some interference can happen.

But, as you noted, if you measure spin up before propagating with $U$ then the probability is just $P_0=|\psi_0|^2$.

Similarly, if you measure spin-down (i.e., measure the bit value 1) after propagation with $U$ the probability is: $$ P^U_1 = |-b^*\psi_0+a^*\psi_1|^2\;. $$

But if you measure spin down before you propagate with $U$ the probability is $P_1=|\psi_1|^2$.

And, as you can check, in either case: $$ P^U_0+P^U_1 = 1 $$ and $$ P_0+P_1=1\;. $$

In quantum computing, the stuff happening "inside the computer" is all the unitary propagation stuff (all your $H$ gates and $CNOT$ gates and whatever other unitary gates you want to apply). This "internal to the computer" stuff can cause interference to happen.

The stuff happening at the end (measurement, e.g., readout of the classical bit values) does not cause interference. At the end we just get whatever classical bit string we get and we calculate the probability of that classical bit string by the Born rule.

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    $\begingroup$ What is meant by "measure spin up"? $\endgroup$ Commented Nov 17, 2023 at 1:10
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    $\begingroup$ A single qubit is a two-state quantum system. It is common, at least pedagogically, to take those two states to be the spin-z eigenstates for a spin-1/2 particle (e.g., electron). The spin-z operator $\hat S_z$ for a single-qubit system is a quantum observable. This means it is a Hermitian operator whose eigenvectors span the space. The eigenvalues of $S_z$ are the possible measurement results (this is a basic rule of quantum mechanics). The eigenvector denoted $|0\rangle$ conventionally corresponds to spin-up. The eigenvector $|1\rangle$ conventionally corresponds to spin-down. $\endgroup$
    – hft
    Commented Nov 17, 2023 at 1:26
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    $\begingroup$ So, "measure spin up" is the same as "measure the bit value to be zero." $\endgroup$
    – hft
    Commented Nov 17, 2023 at 1:28
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    $\begingroup$ And "measure spin down" is the same as "measure the bit value to be 1." $\endgroup$
    – hft
    Commented Nov 17, 2023 at 1:28
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The $\phi_1$ and $\phi_2$ are not just "numbers" but "wavefunctions". For the two-slit experiment, they depend on a position, $x$, so the "functions" look like $\phi_1(x)$ and $\phi_2(x)$, and can even be written as $|\phi_1\rangle$ and $|\phi_2\rangle$ (not only is this allowed, but it also can be very useful when writing things like inner products). Since we can essentially use whatever symbol for $\phi_1$ and $\phi_2$ that we want, we could use any of the following types of notation (I'm just picking arbitrary symbols, just like you can call the x-axis the m-axis or q-axis or whatever else is appropriate):

$$\tag{1} | \phi_1 \rangle = |A_1\rangle = |K\rangle = |0\rangle$$ $$\tag{2} | \phi_2 \rangle = |A_2\rangle = |F\rangle = |1\rangle$$

Qubits like the spin of an electron do not depend on a continuous function such as the position $x$. Instead of wavefunctions like $|\phi_1(x)\rangle$ and $|\phi_2(x)\rangle$, in which we have dependence on $x$, we simply have $|0\rangle$ and $|1\rangle$.

For the wavefunction:

$$ \tag{3} \frac{1}{\sqrt{2}} | 0\rangle + \frac{1}{\sqrt{2}} | 1\rangle, $$

I disagree with your choice of making $\frac{1}{\sqrt{2}} = \phi_1$ and $\frac{1}{\sqrt{2}} = \phi_2$. If anything, I would say $|0\rangle = \phi_1$ and $|1\rangle = \phi_2$ as I already did in Eqs. 1 and 2.

If you want to get the number 1, you can get it in the following way:

\begin{align}\tag{4} &\left(\frac{1}{\sqrt{2}} | 0\rangle + \frac{1}{\sqrt{2}} | 1\rangle\right)^\dagger\left(\frac{1}{\sqrt{2}} | 0\rangle + \frac{1}{\sqrt{2}} | 1\rangle\right) \\ \tag{5}&= \left(\frac{1}{\sqrt{2}} \langle 0| + \frac{1}{\sqrt{2}} \langle 1| \right)\left(\frac{1}{\sqrt{2}} | 0\rangle + \frac{1}{\sqrt{2}} | 1\rangle\right)\\\tag{6} &=\frac{1}{2}\langle 0 | 0 \rangle + 0 + 0 + \frac{1}{2}\langle1 | 1 \rangle\\ &=\frac{1}{2} + \frac{1}{2}\tag{7}\\ &=1.\tag{8} \end{align}

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  • $\begingroup$ Did you mean to write the second dagger or you left it accidentally? $\endgroup$ Commented Nov 16, 2023 at 10:27
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    $\begingroup$ The dagger in the second line was an accident, feel free to edit it because I'm on a phone! $\endgroup$ Commented Nov 16, 2023 at 13:09
  • $\begingroup$ It looks like someone did it. (Special thanks to the ``Community.'') $\endgroup$ Commented Nov 16, 2023 at 17:10

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