Mixing
Optimal strategy to a quantum state game
$begingroup$
Consider the following game:
I flip a fair coin, and depending on the outcome (either heads/tails), I'll give you one of the following states:
$$|0rangle text{ or } cos(x)|0rangle + sin(x)|1rangle.$$
Here, $x$ is a known constant angle. But, I don't tell you which state I give you.
How can I describe a measurement procedure (i.e. an orthonormal qubit basis) to guess which state I'm given, while maximizing the chance of being right? Is there an optimal solution?
I've been self-studying quantum computing, and I came across this exercise. I don't really know how to even start, and I would really appreciate some help.
I think that a good strategy would be to perform an orthogonal transformation with
$$begin{bmatrix}
cos(x) & -sin(theta)\
sin(x) & cos(theta)
end{bmatrix}.$$
Can't make much progress...
quantum-state measurement games
edited Jan 28 at 10:49
Blue♦
6,57541555
asked Jan 27 at 22:55
jjbidjjbid
513
$endgroup$
add a comment |
$begingroup$
Consider the following game:
I flip a fair coin, and depending on the outcome (either heads/tails), I'll give you one of the following states:
$$|0rangle text{ or } cos(x)|0rangle + sin(x)|1rangle.$$
Here, $x$ is a known constant angle. But, I don't tell you which state I give you.
How can I describe a measurement procedure (i.e. an orthonormal qubit basis) to guess which state I'm given, while maximizing the chance of being right? Is there an optimal solution?
I've been self-studying quantum computing, and I came across this exercise. I don't really know how to even start, and I would really appreciate some help.
I think that a good strategy would be to perform an orthogonal transformation with
$$begin{bmatrix}
cos(x) & -sin(theta)\
sin(x) & cos(theta)
end{bmatrix}.$$
Can't make much progress...
quantum-state measurement games
edited Jan 28 at 10:49
Blue♦
6,57541555
asked Jan 27 at 22:55
jjbidjjbid
513
$endgroup$
$begingroup$
Intuitively the answer is to measure in the computational basis because we can restrict $x$ to $[0, frac{pi}{2}]$ and when $x=0$ the states are indistinguishable and when $x=frac{pi}{2}$ the states are orthogonal, but I'm just not sure how to prove it.
$endgroup$
– ahelwer
Jan 28 at 4:15
add a comment |
$begingroup$
Consider the following game:
I flip a fair coin, and depending on the outcome (either heads/tails), I'll give you one of the following states:
$$|0rangle text{ or } cos(x)|0rangle + sin(x)|1rangle.$$
Here, $x$ is a known constant angle. But, I don't tell you which state I give you.
How can I describe a measurement procedure (i.e. an orthonormal qubit basis) to guess which state I'm given, while maximizing the chance of being right? Is there an optimal solution?
I've been self-studying quantum computing, and I came across this exercise. I don't really know how to even start, and I would really appreciate some help.
I think that a good strategy would be to perform an orthogonal transformation with
$$begin{bmatrix}
cos(x) & -sin(theta)\
sin(x) & cos(theta)
end{bmatrix}.$$
Can't make much progress...
quantum-state measurement games
edited Jan 28 at 10:49
Blue♦
6,57541555
asked Jan 27 at 22:55
jjbidjjbid
513
$endgroup$
Consider the following game:
I flip a fair coin, and depending on the outcome (either heads/tails), I'll give you one of the following states:
$$|0rangle text{ or } cos(x)|0rangle + sin(x)|1rangle.$$
Here, $x$ is a known constant angle. But, I don't tell you which state I give you.
How can I describe a measurement procedure (i.e. an orthonormal qubit basis) to guess which state I'm given, while maximizing the chance of being right? Is there an optimal solution?
I've been self-studying quantum computing, and I came across this exercise. I don't really know how to even start, and I would really appreciate some help.
I think that a good strategy would be to perform an orthogonal transformation with
$$begin{bmatrix}
cos(x) & -sin(theta)\
sin(x) & cos(theta)
end{bmatrix}.$$
Can't make much progress...
quantum-state measurement games
quantum-state measurement games
edited Jan 28 at 10:49
Blue♦
6,57541555
asked Jan 27 at 22:55
jjbidjjbid
513
edited Jan 28 at 10:49
Blue♦
6,57541555
asked Jan 27 at 22:55
jjbidjjbid
513
edited Jan 28 at 10:49
Blue♦
6,57541555
edited Jan 28 at 10:49
Blue♦
6,57541555
edited Jan 28 at 10:49
Blue♦
6,57541555
6,57541555
asked Jan 27 at 22:55
jjbidjjbid
513
asked Jan 27 at 22:55
jjbidjjbid
513
asked Jan 27 at 22:55
jjbidjjbid
513
513
$begingroup$
Intuitively the answer is to measure in the computational basis because we can restrict $x$ to $[0, frac{pi}{2}]$ and when $x=0$ the states are indistinguishable and when $x=frac{pi}{2}$ the states are orthogonal, but I'm just not sure how to prove it.
$endgroup$
– ahelwer
Jan 28 at 4:15
add a comment |
$begingroup$
Intuitively the answer is to measure in the computational basis because we can restrict $x$ to $[0, frac{pi}{2}]$ and when $x=0$ the states are indistinguishable and when $x=frac{pi}{2}$ the states are orthogonal, but I'm just not sure how to prove it.
$endgroup$
– ahelwer
Jan 28 at 4:15
$begingroup$
Intuitively the answer is to measure in the computational basis because we can restrict $x$ to $[0, frac{pi}{2}]$ and when $x=0$ the states are indistinguishable and when $x=frac{pi}{2}$ the states are orthogonal, but I'm just not sure how to prove it.
$endgroup$
– ahelwer
Jan 28 at 4:15
$begingroup$
Intuitively the answer is to measure in the computational basis because we can restrict $x$ to $[0, frac{pi}{2}]$ and when $x=0$ the states are indistinguishable and when $x=frac{pi}{2}$ the states are orthogonal, but I'm just not sure how to prove it.
$endgroup$
– ahelwer
Jan 28 at 4:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere).
First, something that we won't really need, the precise description of the state. The full state of the system that depends both on superpositions a well as a classical fair coin may be encoded in the density matrix
$$ rho = frac 12 pmatrix{1&0\0&0} + frac 12 pmatrix{cos^2x &sin x cos x\ sin xcos x & sin^2 x} $$
where the left column and upper row corresponds to the basis state "zero" and the remaining ones to "one". It's helpful to rewrite the density matrix in terms of the 4-element basis of the $2times 2$ matrices,
$$ rho = frac 12+ frac{sin x cos x}{2} sigma_x + left(frac{cos^2 x - sin^2 x}{4}+frac 14right) sigma_z $$
That may be written in terms of the angle $2x$:
$$rho = frac 12 + frac {sin 2x}{4} sigma_x + frac{cos 2x +1}{4} sigma_z $$
Now, regardless of the mixed state, this is still a two-level system and all measurements on the two-dimensional Hilbert space are either trivial (measurements of a $c$-number) or equivalent to the measurement of the spin along an axis, i.e. measurements of
$$ V = vec n cdot vec sigma $$
which is a unit 3D vector multiplied by the vector of Pauli matrices. OK, what happens if we measure $V$? The eigenvalues of $V$ are plus one or minus one. The probability of each may be obtained from the expectation value of $V$ which is
$$langle V rangle = {rm Tr} (V rho) $$
The traces of products only contribute if $1$ meets $1$ (but we assume there was no term in $V$) or $sigma_x$ meets $sigma_x$ etc., in which cases the trace of the matrix gives an extra factor of 2. So we have
$$langle V rangle = frac{sin 2x}{2}n_x + frac{cos 2x +1}{2} n_z $$
We get the eigenvalue $pm 1$ with the probabilities $(1pmlangle V rangle) / 2$, respectively. Exactly when $cos x = 0$, the two initial "head and tail" states are orthogonal to one another (basically $|0rangle$ and $|1rangle$) and we may fully discriminate them. To make the probabilities $0,1$, we must simply choose $vec n=(0,0,pm 1)$; note that the overall sign of $vec n$ doesn't matter for the procedure.
Now, for $cos x neq 0$, the states are non-orthogonal i.e. "not mutually exclusive" in the quantum sense and we can't measure directly whether the coin was tails or heads because those possibilities were mixed in the density matrix. In fact, the density matrix contains all probabilities of all measurements, so if we could get the same density matrix by a different mixture of possible states from coin tosses, the states of the qubit would be strictly indistinguishable.
Our probability of success will be below 100% if $cos xneq 0$. But the only meaningful way to use the classical bit $V=pm 1$ from the measurement is to directly translate it to our guess about the initial state. Without a loss of generality, our translation may be chosen to be
$$ (V = +1) to |irangle = |0rangle \ $$
and
$$ (V = -1) to |irangle = cos x |0rangle + sin x |1 rangle.$$
If we wanted the opposite, cross-identification of the heads-tails and the signs of $V$, we could simply achieve it by flipping the overall sign of $vec n to -vec n$.
Let's call the first simple initial state "heads" (the zero) and the second harder one "tails" (the cosine-sine superposition). The probability of success is, given our translation from $+1$ to heads and $-1$ to tails,
$$ P_{rm success} = P(H) P(+1|H) + P(T) P(-1|T). $$
Because it's a fair coin, the two factors included above are $P(H)=P(T)=1/2$. The most difficult calculation among the four probabilities is $P(-1|T)$. But we have already made a harder calculation above, it was the $(1-langle Vrangle) / 2$. Here we just omit the constant term proportional to $n_z$ and multiply by two:
$$ P(-1|T ) = frac 12 - sin 2x frac{ n_x}2 - cos 2x frac{ n_z}2 $$
The result for "heads" is simply obtained by setting $x=0$ because the "heads" state equals "tails" states with $x=0$ substituted. So
$$ P(-1|H) = frac{1-n_z}{2} $$
and the complementary $1-P$ probability is
$$ P(+1|H) = frac{1+n_z}{2} $$
Substitute those results to our "success probability" to get
$$P_{rm success} = frac{1+n_z +1 - (sin 2x)n_x - (cos 2x)n_z}{4} $$
or
$$P_{rm success} = frac 12 - frac{n_x}4 sin 2x + frac{n_z}{4} (1-cos 2x ) $$
If we define $(n_x,n_z)=(-cos alpha,-sinalpha)$, we may also write it as
$$ P_{rm success} = frac 12 +frac{sin(2x+alpha)-sin alpha}{4} =frac 12+frac{sin x cos(x+alpha)}{2} $$
We want to maximize that over $alpha$. Clearly, the maximum is for $cos(x+alpha)=pm 1$ where the sign agrees with that of $sin x$ i.e. $alpha=-x$ or $alpha=pi -x $ and the value at this maximum is
$$ P_{rm success} = frac{1+|sin x|}{2}$$
which sits in the interval 50% and 100%.
That's a nice measurement which is really quantum mechanical. We use a different measurement than that of $sigma_z$, i.e. the classical measurement of the bit. Instead, we measure the spin along the axis in the $xz$-plane that is defined by the same nonzero angle as the angle $x$ at the beginning, with some correct signs and shifts by multiples of $pi/2$. Note that if you measured simply $sigma_z$, the classical bit, the success rate would be just $(3-cos 2x)/4$, also between 50% and 100%, but smaller than our result. In particular, for a small $x=0+epsilon$, our optimal result would be Taylor-expanded as $1/2+|x|/2$ while the non-optimum result using the classical measurement would increase above $1/2$ more slowly, as $1/2+x^2/2$.
For many hours, a wrong answer (a mistake in the final portions) was posted here, despite the fact that I had previously fixed many wrong factors of two. I posted a slightly edited version of this answer on my weblog where some discussion may take place:
The Reference Frame: A fun simple problem in quantum computing
On that page, I also write the eigenstates of the measured operator in the appendix. The arguments in the angles may be surprising for some folks who think that this problem is obvious in terms of the wave functions or that the wave functions after the measurement have to be simple.
edited Jan 29 at 13:58
Blue♦
6,57541555
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
$endgroup$
4
$begingroup$
Hi Luboš. Welcome to Quantum Computing! It's nice to see you here. :)
$endgroup$
– Blue♦
Jan 28 at 9:42
$begingroup$
I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?
$endgroup$
– glS
Jan 30 at 19:18
$begingroup$
Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.
$endgroup$
– Luboš Motl
Jan 31 at 12:28
add a comment |
$begingroup$
The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.
edited Jan 28 at 7:45
Blue♦
6,57541555
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere).
First, something that we won't really need, the precise description of the state. The full state of the system that depends both on superpositions a well as a classical fair coin may be encoded in the density matrix
$$ rho = frac 12 pmatrix{1&0\0&0} + frac 12 pmatrix{cos^2x &sin x cos x\ sin xcos x & sin^2 x} $$
where the left column and upper row corresponds to the basis state "zero" and the remaining ones to "one". It's helpful to rewrite the density matrix in terms of the 4-element basis of the $2times 2$ matrices,
$$ rho = frac 12+ frac{sin x cos x}{2} sigma_x + left(frac{cos^2 x - sin^2 x}{4}+frac 14right) sigma_z $$
That may be written in terms of the angle $2x$:
$$rho = frac 12 + frac {sin 2x}{4} sigma_x + frac{cos 2x +1}{4} sigma_z $$
Now, regardless of the mixed state, this is still a two-level system and all measurements on the two-dimensional Hilbert space are either trivial (measurements of a $c$-number) or equivalent to the measurement of the spin along an axis, i.e. measurements of
$$ V = vec n cdot vec sigma $$
which is a unit 3D vector multiplied by the vector of Pauli matrices. OK, what happens if we measure $V$? The eigenvalues of $V$ are plus one or minus one. The probability of each may be obtained from the expectation value of $V$ which is
$$langle V rangle = {rm Tr} (V rho) $$
The traces of products only contribute if $1$ meets $1$ (but we assume there was no term in $V$) or $sigma_x$ meets $sigma_x$ etc., in which cases the trace of the matrix gives an extra factor of 2. So we have
$$langle V rangle = frac{sin 2x}{2}n_x + frac{cos 2x +1}{2} n_z $$
We get the eigenvalue $pm 1$ with the probabilities $(1pmlangle V rangle) / 2$, respectively. Exactly when $cos x = 0$, the two initial "head and tail" states are orthogonal to one another (basically $|0rangle$ and $|1rangle$) and we may fully discriminate them. To make the probabilities $0,1$, we must simply choose $vec n=(0,0,pm 1)$; note that the overall sign of $vec n$ doesn't matter for the procedure.
Now, for $cos x neq 0$, the states are non-orthogonal i.e. "not mutually exclusive" in the quantum sense and we can't measure directly whether the coin was tails or heads because those possibilities were mixed in the density matrix. In fact, the density matrix contains all probabilities of all measurements, so if we could get the same density matrix by a different mixture of possible states from coin tosses, the states of the qubit would be strictly indistinguishable.
Our probability of success will be below 100% if $cos xneq 0$. But the only meaningful way to use the classical bit $V=pm 1$ from the measurement is to directly translate it to our guess about the initial state. Without a loss of generality, our translation may be chosen to be
$$ (V = +1) to |irangle = |0rangle \ $$
and
$$ (V = -1) to |irangle = cos x |0rangle + sin x |1 rangle.$$
If we wanted the opposite, cross-identification of the heads-tails and the signs of $V$, we could simply achieve it by flipping the overall sign of $vec n to -vec n$.
Let's call the first simple initial state "heads" (the zero) and the second harder one "tails" (the cosine-sine superposition). The probability of success is, given our translation from $+1$ to heads and $-1$ to tails,
$$ P_{rm success} = P(H) P(+1|H) + P(T) P(-1|T). $$
Because it's a fair coin, the two factors included above are $P(H)=P(T)=1/2$. The most difficult calculation among the four probabilities is $P(-1|T)$. But we have already made a harder calculation above, it was the $(1-langle Vrangle) / 2$. Here we just omit the constant term proportional to $n_z$ and multiply by two:
$$ P(-1|T ) = frac 12 - sin 2x frac{ n_x}2 - cos 2x frac{ n_z}2 $$
The result for "heads" is simply obtained by setting $x=0$ because the "heads" state equals "tails" states with $x=0$ substituted. So
$$ P(-1|H) = frac{1-n_z}{2} $$
and the complementary $1-P$ probability is
$$ P(+1|H) = frac{1+n_z}{2} $$
Substitute those results to our "success probability" to get
$$P_{rm success} = frac{1+n_z +1 - (sin 2x)n_x - (cos 2x)n_z}{4} $$
or
$$P_{rm success} = frac 12 - frac{n_x}4 sin 2x + frac{n_z}{4} (1-cos 2x ) $$
If we define $(n_x,n_z)=(-cos alpha,-sinalpha)$, we may also write it as
$$ P_{rm success} = frac 12 +frac{sin(2x+alpha)-sin alpha}{4} =frac 12+frac{sin x cos(x+alpha)}{2} $$
We want to maximize that over $alpha$. Clearly, the maximum is for $cos(x+alpha)=pm 1$ where the sign agrees with that of $sin x$ i.e. $alpha=-x$ or $alpha=pi -x $ and the value at this maximum is
$$ P_{rm success} = frac{1+|sin x|}{2}$$
which sits in the interval 50% and 100%.
That's a nice measurement which is really quantum mechanical. We use a different measurement than that of $sigma_z$, i.e. the classical measurement of the bit. Instead, we measure the spin along the axis in the $xz$-plane that is defined by the same nonzero angle as the angle $x$ at the beginning, with some correct signs and shifts by multiples of $pi/2$. Note that if you measured simply $sigma_z$, the classical bit, the success rate would be just $(3-cos 2x)/4$, also between 50% and 100%, but smaller than our result. In particular, for a small $x=0+epsilon$, our optimal result would be Taylor-expanded as $1/2+|x|/2$ while the non-optimum result using the classical measurement would increase above $1/2$ more slowly, as $1/2+x^2/2$.
For many hours, a wrong answer (a mistake in the final portions) was posted here, despite the fact that I had previously fixed many wrong factors of two. I posted a slightly edited version of this answer on my weblog where some discussion may take place:
The Reference Frame: A fun simple problem in quantum computing
On that page, I also write the eigenstates of the measured operator in the appendix. The arguments in the angles may be surprising for some folks who think that this problem is obvious in terms of the wave functions or that the wave functions after the measurement have to be simple.
edited Jan 29 at 13:58
Blue♦
6,57541555
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
$endgroup$
4
$begingroup$
Hi Luboš. Welcome to Quantum Computing! It's nice to see you here. :)
$endgroup$
– Blue♦
Jan 28 at 9:42
$begingroup$
I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?
$endgroup$
– glS
Jan 30 at 19:18
$begingroup$
Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.
$endgroup$
– Luboš Motl
Jan 31 at 12:28
add a comment |
$begingroup$
We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere).
First, something that we won't really need, the precise description of the state. The full state of the system that depends both on superpositions a well as a classical fair coin may be encoded in the density matrix
$$ rho = frac 12 pmatrix{1&0\0&0} + frac 12 pmatrix{cos^2x &sin x cos x\ sin xcos x & sin^2 x} $$
where the left column and upper row corresponds to the basis state "zero" and the remaining ones to "one". It's helpful to rewrite the density matrix in terms of the 4-element basis of the $2times 2$ matrices,
$$ rho = frac 12+ frac{sin x cos x}{2} sigma_x + left(frac{cos^2 x - sin^2 x}{4}+frac 14right) sigma_z $$
That may be written in terms of the angle $2x$:
$$rho = frac 12 + frac {sin 2x}{4} sigma_x + frac{cos 2x +1}{4} sigma_z $$
Now, regardless of the mixed state, this is still a two-level system and all measurements on the two-dimensional Hilbert space are either trivial (measurements of a $c$-number) or equivalent to the measurement of the spin along an axis, i.e. measurements of
$$ V = vec n cdot vec sigma $$
which is a unit 3D vector multiplied by the vector of Pauli matrices. OK, what happens if we measure $V$? The eigenvalues of $V$ are plus one or minus one. The probability of each may be obtained from the expectation value of $V$ which is
$$langle V rangle = {rm Tr} (V rho) $$
The traces of products only contribute if $1$ meets $1$ (but we assume there was no term in $V$) or $sigma_x$ meets $sigma_x$ etc., in which cases the trace of the matrix gives an extra factor of 2. So we have
$$langle V rangle = frac{sin 2x}{2}n_x + frac{cos 2x +1}{2} n_z $$
We get the eigenvalue $pm 1$ with the probabilities $(1pmlangle V rangle) / 2$, respectively. Exactly when $cos x = 0$, the two initial "head and tail" states are orthogonal to one another (basically $|0rangle$ and $|1rangle$) and we may fully discriminate them. To make the probabilities $0,1$, we must simply choose $vec n=(0,0,pm 1)$; note that the overall sign of $vec n$ doesn't matter for the procedure.
Now, for $cos x neq 0$, the states are non-orthogonal i.e. "not mutually exclusive" in the quantum sense and we can't measure directly whether the coin was tails or heads because those possibilities were mixed in the density matrix. In fact, the density matrix contains all probabilities of all measurements, so if we could get the same density matrix by a different mixture of possible states from coin tosses, the states of the qubit would be strictly indistinguishable.
Our probability of success will be below 100% if $cos xneq 0$. But the only meaningful way to use the classical bit $V=pm 1$ from the measurement is to directly translate it to our guess about the initial state. Without a loss of generality, our translation may be chosen to be
$$ (V = +1) to |irangle = |0rangle \ $$
and
$$ (V = -1) to |irangle = cos x |0rangle + sin x |1 rangle.$$
If we wanted the opposite, cross-identification of the heads-tails and the signs of $V$, we could simply achieve it by flipping the overall sign of $vec n to -vec n$.
Let's call the first simple initial state "heads" (the zero) and the second harder one "tails" (the cosine-sine superposition). The probability of success is, given our translation from $+1$ to heads and $-1$ to tails,
$$ P_{rm success} = P(H) P(+1|H) + P(T) P(-1|T). $$
Because it's a fair coin, the two factors included above are $P(H)=P(T)=1/2$. The most difficult calculation among the four probabilities is $P(-1|T)$. But we have already made a harder calculation above, it was the $(1-langle Vrangle) / 2$. Here we just omit the constant term proportional to $n_z$ and multiply by two:
$$ P(-1|T ) = frac 12 - sin 2x frac{ n_x}2 - cos 2x frac{ n_z}2 $$
The result for "heads" is simply obtained by setting $x=0$ because the "heads" state equals "tails" states with $x=0$ substituted. So
$$ P(-1|H) = frac{1-n_z}{2} $$
and the complementary $1-P$ probability is
$$ P(+1|H) = frac{1+n_z}{2} $$
Substitute those results to our "success probability" to get
$$P_{rm success} = frac{1+n_z +1 - (sin 2x)n_x - (cos 2x)n_z}{4} $$
or
$$P_{rm success} = frac 12 - frac{n_x}4 sin 2x + frac{n_z}{4} (1-cos 2x ) $$
If we define $(n_x,n_z)=(-cos alpha,-sinalpha)$, we may also write it as
$$ P_{rm success} = frac 12 +frac{sin(2x+alpha)-sin alpha}{4} =frac 12+frac{sin x cos(x+alpha)}{2} $$
We want to maximize that over $alpha$. Clearly, the maximum is for $cos(x+alpha)=pm 1$ where the sign agrees with that of $sin x$ i.e. $alpha=-x$ or $alpha=pi -x $ and the value at this maximum is
$$ P_{rm success} = frac{1+|sin x|}{2}$$
which sits in the interval 50% and 100%.
That's a nice measurement which is really quantum mechanical. We use a different measurement than that of $sigma_z$, i.e. the classical measurement of the bit. Instead, we measure the spin along the axis in the $xz$-plane that is defined by the same nonzero angle as the angle $x$ at the beginning, with some correct signs and shifts by multiples of $pi/2$. Note that if you measured simply $sigma_z$, the classical bit, the success rate would be just $(3-cos 2x)/4$, also between 50% and 100%, but smaller than our result. In particular, for a small $x=0+epsilon$, our optimal result would be Taylor-expanded as $1/2+|x|/2$ while the non-optimum result using the classical measurement would increase above $1/2$ more slowly, as $1/2+x^2/2$.
For many hours, a wrong answer (a mistake in the final portions) was posted here, despite the fact that I had previously fixed many wrong factors of two. I posted a slightly edited version of this answer on my weblog where some discussion may take place:
The Reference Frame: A fun simple problem in quantum computing
On that page, I also write the eigenstates of the measured operator in the appendix. The arguments in the angles may be surprising for some folks who think that this problem is obvious in terms of the wave functions or that the wave functions after the measurement have to be simple.
edited Jan 29 at 13:58
Blue♦
6,57541555
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
$endgroup$
4
$begingroup$
Hi Luboš. Welcome to Quantum Computing! It's nice to see you here. :)
$endgroup$
– Blue♦
Jan 28 at 9:42
$begingroup$
I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?
$endgroup$
– glS
Jan 30 at 19:18
$begingroup$
Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.
$endgroup$
– Luboš Motl
Jan 31 at 12:28
add a comment |
$begingroup$
We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere).
First, something that we won't really need, the precise description of the state. The full state of the system that depends both on superpositions a well as a classical fair coin may be encoded in the density matrix
$$ rho = frac 12 pmatrix{1&0\0&0} + frac 12 pmatrix{cos^2x &sin x cos x\ sin xcos x & sin^2 x} $$
where the left column and upper row corresponds to the basis state "zero" and the remaining ones to "one". It's helpful to rewrite the density matrix in terms of the 4-element basis of the $2times 2$ matrices,
$$ rho = frac 12+ frac{sin x cos x}{2} sigma_x + left(frac{cos^2 x - sin^2 x}{4}+frac 14right) sigma_z $$
That may be written in terms of the angle $2x$:
$$rho = frac 12 + frac {sin 2x}{4} sigma_x + frac{cos 2x +1}{4} sigma_z $$
Now, regardless of the mixed state, this is still a two-level system and all measurements on the two-dimensional Hilbert space are either trivial (measurements of a $c$-number) or equivalent to the measurement of the spin along an axis, i.e. measurements of
$$ V = vec n cdot vec sigma $$
which is a unit 3D vector multiplied by the vector of Pauli matrices. OK, what happens if we measure $V$? The eigenvalues of $V$ are plus one or minus one. The probability of each may be obtained from the expectation value of $V$ which is
$$langle V rangle = {rm Tr} (V rho) $$
The traces of products only contribute if $1$ meets $1$ (but we assume there was no term in $V$) or $sigma_x$ meets $sigma_x$ etc., in which cases the trace of the matrix gives an extra factor of 2. So we have
$$langle V rangle = frac{sin 2x}{2}n_x + frac{cos 2x +1}{2} n_z $$
We get the eigenvalue $pm 1$ with the probabilities $(1pmlangle V rangle) / 2$, respectively. Exactly when $cos x = 0$, the two initial "head and tail" states are orthogonal to one another (basically $|0rangle$ and $|1rangle$) and we may fully discriminate them. To make the probabilities $0,1$, we must simply choose $vec n=(0,0,pm 1)$; note that the overall sign of $vec n$ doesn't matter for the procedure.
Now, for $cos x neq 0$, the states are non-orthogonal i.e. "not mutually exclusive" in the quantum sense and we can't measure directly whether the coin was tails or heads because those possibilities were mixed in the density matrix. In fact, the density matrix contains all probabilities of all measurements, so if we could get the same density matrix by a different mixture of possible states from coin tosses, the states of the qubit would be strictly indistinguishable.
Our probability of success will be below 100% if $cos xneq 0$. But the only meaningful way to use the classical bit $V=pm 1$ from the measurement is to directly translate it to our guess about the initial state. Without a loss of generality, our translation may be chosen to be
$$ (V = +1) to |irangle = |0rangle \ $$
and
$$ (V = -1) to |irangle = cos x |0rangle + sin x |1 rangle.$$
If we wanted the opposite, cross-identification of the heads-tails and the signs of $V$, we could simply achieve it by flipping the overall sign of $vec n to -vec n$.
Let's call the first simple initial state "heads" (the zero) and the second harder one "tails" (the cosine-sine superposition). The probability of success is, given our translation from $+1$ to heads and $-1$ to tails,
$$ P_{rm success} = P(H) P(+1|H) + P(T) P(-1|T). $$
Because it's a fair coin, the two factors included above are $P(H)=P(T)=1/2$. The most difficult calculation among the four probabilities is $P(-1|T)$. But we have already made a harder calculation above, it was the $(1-langle Vrangle) / 2$. Here we just omit the constant term proportional to $n_z$ and multiply by two:
$$ P(-1|T ) = frac 12 - sin 2x frac{ n_x}2 - cos 2x frac{ n_z}2 $$
The result for "heads" is simply obtained by setting $x=0$ because the "heads" state equals "tails" states with $x=0$ substituted. So
$$ P(-1|H) = frac{1-n_z}{2} $$
and the complementary $1-P$ probability is
$$ P(+1|H) = frac{1+n_z}{2} $$
Substitute those results to our "success probability" to get
$$P_{rm success} = frac{1+n_z +1 - (sin 2x)n_x - (cos 2x)n_z}{4} $$
or
$$P_{rm success} = frac 12 - frac{n_x}4 sin 2x + frac{n_z}{4} (1-cos 2x ) $$
If we define $(n_x,n_z)=(-cos alpha,-sinalpha)$, we may also write it as
$$ P_{rm success} = frac 12 +frac{sin(2x+alpha)-sin alpha}{4} =frac 12+frac{sin x cos(x+alpha)}{2} $$
We want to maximize that over $alpha$. Clearly, the maximum is for $cos(x+alpha)=pm 1$ where the sign agrees with that of $sin x$ i.e. $alpha=-x$ or $alpha=pi -x $ and the value at this maximum is
$$ P_{rm success} = frac{1+|sin x|}{2}$$
which sits in the interval 50% and 100%.
That's a nice measurement which is really quantum mechanical. We use a different measurement than that of $sigma_z$, i.e. the classical measurement of the bit. Instead, we measure the spin along the axis in the $xz$-plane that is defined by the same nonzero angle as the angle $x$ at the beginning, with some correct signs and shifts by multiples of $pi/2$. Note that if you measured simply $sigma_z$, the classical bit, the success rate would be just $(3-cos 2x)/4$, also between 50% and 100%, but smaller than our result. In particular, for a small $x=0+epsilon$, our optimal result would be Taylor-expanded as $1/2+|x|/2$ while the non-optimum result using the classical measurement would increase above $1/2$ more slowly, as $1/2+x^2/2$.
For many hours, a wrong answer (a mistake in the final portions) was posted here, despite the fact that I had previously fixed many wrong factors of two. I posted a slightly edited version of this answer on my weblog where some discussion may take place:
The Reference Frame: A fun simple problem in quantum computing
On that page, I also write the eigenstates of the measured operator in the appendix. The arguments in the angles may be surprising for some folks who think that this problem is obvious in terms of the wave functions or that the wave functions after the measurement have to be simple.
edited Jan 29 at 13:58
Blue♦
6,57541555
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
$endgroup$
We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere).
First, something that we won't really need, the precise description of the state. The full state of the system that depends both on superpositions a well as a classical fair coin may be encoded in the density matrix
$$ rho = frac 12 pmatrix{1&0\0&0} + frac 12 pmatrix{cos^2x &sin x cos x\ sin xcos x & sin^2 x} $$
where the left column and upper row corresponds to the basis state "zero" and the remaining ones to "one". It's helpful to rewrite the density matrix in terms of the 4-element basis of the $2times 2$ matrices,
$$ rho = frac 12+ frac{sin x cos x}{2} sigma_x + left(frac{cos^2 x - sin^2 x}{4}+frac 14right) sigma_z $$
That may be written in terms of the angle $2x$:
$$rho = frac 12 + frac {sin 2x}{4} sigma_x + frac{cos 2x +1}{4} sigma_z $$
Now, regardless of the mixed state, this is still a two-level system and all measurements on the two-dimensional Hilbert space are either trivial (measurements of a $c$-number) or equivalent to the measurement of the spin along an axis, i.e. measurements of
$$ V = vec n cdot vec sigma $$
which is a unit 3D vector multiplied by the vector of Pauli matrices. OK, what happens if we measure $V$? The eigenvalues of $V$ are plus one or minus one. The probability of each may be obtained from the expectation value of $V$ which is
$$langle V rangle = {rm Tr} (V rho) $$
The traces of products only contribute if $1$ meets $1$ (but we assume there was no term in $V$) or $sigma_x$ meets $sigma_x$ etc., in which cases the trace of the matrix gives an extra factor of 2. So we have
$$langle V rangle = frac{sin 2x}{2}n_x + frac{cos 2x +1}{2} n_z $$
We get the eigenvalue $pm 1$ with the probabilities $(1pmlangle V rangle) / 2$, respectively. Exactly when $cos x = 0$, the two initial "head and tail" states are orthogonal to one another (basically $|0rangle$ and $|1rangle$) and we may fully discriminate them. To make the probabilities $0,1$, we must simply choose $vec n=(0,0,pm 1)$; note that the overall sign of $vec n$ doesn't matter for the procedure.
Now, for $cos x neq 0$, the states are non-orthogonal i.e. "not mutually exclusive" in the quantum sense and we can't measure directly whether the coin was tails or heads because those possibilities were mixed in the density matrix. In fact, the density matrix contains all probabilities of all measurements, so if we could get the same density matrix by a different mixture of possible states from coin tosses, the states of the qubit would be strictly indistinguishable.
Our probability of success will be below 100% if $cos xneq 0$. But the only meaningful way to use the classical bit $V=pm 1$ from the measurement is to directly translate it to our guess about the initial state. Without a loss of generality, our translation may be chosen to be
$$ (V = +1) to |irangle = |0rangle \ $$
and
$$ (V = -1) to |irangle = cos x |0rangle + sin x |1 rangle.$$
If we wanted the opposite, cross-identification of the heads-tails and the signs of $V$, we could simply achieve it by flipping the overall sign of $vec n to -vec n$.
Let's call the first simple initial state "heads" (the zero) and the second harder one "tails" (the cosine-sine superposition). The probability of success is, given our translation from $+1$ to heads and $-1$ to tails,
$$ P_{rm success} = P(H) P(+1|H) + P(T) P(-1|T). $$
Because it's a fair coin, the two factors included above are $P(H)=P(T)=1/2$. The most difficult calculation among the four probabilities is $P(-1|T)$. But we have already made a harder calculation above, it was the $(1-langle Vrangle) / 2$. Here we just omit the constant term proportional to $n_z$ and multiply by two:
$$ P(-1|T ) = frac 12 - sin 2x frac{ n_x}2 - cos 2x frac{ n_z}2 $$
The result for "heads" is simply obtained by setting $x=0$ because the "heads" state equals "tails" states with $x=0$ substituted. So
$$ P(-1|H) = frac{1-n_z}{2} $$
and the complementary $1-P$ probability is
$$ P(+1|H) = frac{1+n_z}{2} $$
Substitute those results to our "success probability" to get
$$P_{rm success} = frac{1+n_z +1 - (sin 2x)n_x - (cos 2x)n_z}{4} $$
or
$$P_{rm success} = frac 12 - frac{n_x}4 sin 2x + frac{n_z}{4} (1-cos 2x ) $$
If we define $(n_x,n_z)=(-cos alpha,-sinalpha)$, we may also write it as
$$ P_{rm success} = frac 12 +frac{sin(2x+alpha)-sin alpha}{4} =frac 12+frac{sin x cos(x+alpha)}{2} $$
We want to maximize that over $alpha$. Clearly, the maximum is for $cos(x+alpha)=pm 1$ where the sign agrees with that of $sin x$ i.e. $alpha=-x$ or $alpha=pi -x $ and the value at this maximum is
$$ P_{rm success} = frac{1+|sin x|}{2}$$
which sits in the interval 50% and 100%.
That's a nice measurement which is really quantum mechanical. We use a different measurement than that of $sigma_z$, i.e. the classical measurement of the bit. Instead, we measure the spin along the axis in the $xz$-plane that is defined by the same nonzero angle as the angle $x$ at the beginning, with some correct signs and shifts by multiples of $pi/2$. Note that if you measured simply $sigma_z$, the classical bit, the success rate would be just $(3-cos 2x)/4$, also between 50% and 100%, but smaller than our result. In particular, for a small $x=0+epsilon$, our optimal result would be Taylor-expanded as $1/2+|x|/2$ while the non-optimum result using the classical measurement would increase above $1/2$ more slowly, as $1/2+x^2/2$.
For many hours, a wrong answer (a mistake in the final portions) was posted here, despite the fact that I had previously fixed many wrong factors of two. I posted a slightly edited version of this answer on my weblog where some discussion may take place:
The Reference Frame: A fun simple problem in quantum computing
On that page, I also write the eigenstates of the measured operator in the appendix. The arguments in the angles may be surprising for some folks who think that this problem is obvious in terms of the wave functions or that the wave functions after the measurement have to be simple.
edited Jan 29 at 13:58
Blue♦
6,57541555
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
edited Jan 29 at 13:58
Blue♦
6,57541555
edited Jan 29 at 13:58
Blue♦
6,57541555
edited Jan 29 at 13:58
Blue♦
6,57541555
6,57541555
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
answered Jan 28 at 9:16
Luboš MotlLuboš Motl
1913
1913
4
$begingroup$
Hi Luboš. Welcome to Quantum Computing! It's nice to see you here. :)
$endgroup$
– Blue♦
Jan 28 at 9:42
$begingroup$
I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?
$endgroup$
– glS
Jan 30 at 19:18
$begingroup$
Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.
$endgroup$
– Luboš Motl
Jan 31 at 12:28
add a comment |
4
$begingroup$
Hi Luboš. Welcome to Quantum Computing! It's nice to see you here. :)
$endgroup$
– Blue♦
Jan 28 at 9:42
$begingroup$
I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?
$endgroup$
– glS
Jan 30 at 19:18
$begingroup$
Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.
$endgroup$
– Luboš Motl
Jan 31 at 12:28
4
4
$begingroup$
Hi Luboš. Welcome to Quantum Computing! It's nice to see you here. :)
$endgroup$
– Blue♦
Jan 28 at 9:42
$begingroup$
Hi Luboš. Welcome to Quantum Computing! It's nice to see you here. :)
$endgroup$
– Blue♦
Jan 28 at 9:42
$begingroup$
I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?
$endgroup$
– glS
Jan 30 at 19:18
$begingroup$
I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?
$endgroup$
– glS
Jan 30 at 19:18
$begingroup$
Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.
$endgroup$
– Luboš Motl
Jan 31 at 12:28
$begingroup$
Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.
$endgroup$
– Luboš Motl
Jan 31 at 12:28
add a comment |
$begingroup$
The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.
edited Jan 28 at 7:45
Blue♦
6,57541555
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
$endgroup$
add a comment |
$begingroup$
The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.
edited Jan 28 at 7:45
Blue♦
6,57541555
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
$endgroup$
add a comment |
$begingroup$
The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.
edited Jan 28 at 7:45
Blue♦
6,57541555
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
$endgroup$
The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.
edited Jan 28 at 7:45
Blue♦
6,57541555
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
edited Jan 28 at 7:45
Blue♦
6,57541555
edited Jan 28 at 7:45
Blue♦
6,57541555
edited Jan 28 at 7:45
Blue♦
6,57541555
6,57541555
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
answered Jan 28 at 7:09
DaftWullieDaftWullie
15.2k1542
15.2k1542
add a comment |
add a comment |
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Intuitively the answer is to measure in the computational basis because we can restrict $x$ to $[0, frac{pi}{2}]$ and when $x=0$ the states are indistinguishable and when $x=frac{pi}{2}$ the states are orthogonal, but I'm just not sure how to prove it.
$endgroup$
– ahelwer
Jan 28 at 4:15