Mixing
Schrödinger's cat bra-ket interpretation
Clash Royale CLAN TAG#URR8PPP
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Let $verttext#rangle$ be the vector state of the cat, $vert1rangle$ the "alive" state, and $vert0rangle$ the "dead" state. Using the normalization condition $langle text#verttext#rangle=1$:
beginequation
vert text#rangle=avert1rangle+bvert0rangle
endequation
becomes
beginequation
vert avert^2+a^*blangle 1vert0rangle+b^*alangle 0vert1rangle+vert b vert^2=1
endequation
where $vert avert^2$ is the probability of the cat being at the state $vert1rangle$ (alive).
The equation leads to $a=b=frac1sqrt2$.
However, why is this? And how should $langle 1vert0rangle$ and $langle 0vert1rangle$ be interpreted?
quantum-mechanics hilbert-space schroedinger-equation superposition schroedingers-cat
edited 2 hours ago
Qmechanic♦
98.1k121701066
asked 3 hours ago
IchVerloren
286
add a comment |Â
up vote
1
down vote
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Let $verttext#rangle$ be the vector state of the cat, $vert1rangle$ the "alive" state, and $vert0rangle$ the "dead" state. Using the normalization condition $langle text#verttext#rangle=1$:
beginequation
vert text#rangle=avert1rangle+bvert0rangle
endequation
becomes
beginequation
vert avert^2+a^*blangle 1vert0rangle+b^*alangle 0vert1rangle+vert b vert^2=1
endequation
where $vert avert^2$ is the probability of the cat being at the state $vert1rangle$ (alive).
The equation leads to $a=b=frac1sqrt2$.
However, why is this? And how should $langle 1vert0rangle$ and $langle 0vert1rangle$ be interpreted?
quantum-mechanics hilbert-space schroedinger-equation superposition schroedingers-cat
edited 2 hours ago
Qmechanic♦
98.1k121701066
asked 3 hours ago
IchVerloren
286
In my interpretation with your notation, $langle 1 vert$ is the representation of the (probability) measurement captured the signal of alive.
– ChoMedit
23 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $verttext#rangle$ be the vector state of the cat, $vert1rangle$ the "alive" state, and $vert0rangle$ the "dead" state. Using the normalization condition $langle text#verttext#rangle=1$:
beginequation
vert text#rangle=avert1rangle+bvert0rangle
endequation
becomes
beginequation
vert avert^2+a^*blangle 1vert0rangle+b^*alangle 0vert1rangle+vert b vert^2=1
endequation
where $vert avert^2$ is the probability of the cat being at the state $vert1rangle$ (alive).
The equation leads to $a=b=frac1sqrt2$.
However, why is this? And how should $langle 1vert0rangle$ and $langle 0vert1rangle$ be interpreted?
quantum-mechanics hilbert-space schroedinger-equation superposition schroedingers-cat
edited 2 hours ago
Qmechanic♦
98.1k121701066
asked 3 hours ago
IchVerloren
286
Let $verttext#rangle$ be the vector state of the cat, $vert1rangle$ the "alive" state, and $vert0rangle$ the "dead" state. Using the normalization condition $langle text#verttext#rangle=1$:
beginequation
vert text#rangle=avert1rangle+bvert0rangle
endequation
becomes
beginequation
vert avert^2+a^*blangle 1vert0rangle+b^*alangle 0vert1rangle+vert b vert^2=1
endequation
where $vert avert^2$ is the probability of the cat being at the state $vert1rangle$ (alive).
The equation leads to $a=b=frac1sqrt2$.
However, why is this? And how should $langle 1vert0rangle$ and $langle 0vert1rangle$ be interpreted?
quantum-mechanics hilbert-space schroedinger-equation superposition schroedingers-cat
quantum-mechanics hilbert-space schroedinger-equation superposition schroedingers-cat
edited 2 hours ago
Qmechanic♦
98.1k121701066
asked 3 hours ago
IchVerloren
286
edited 2 hours ago
Qmechanic♦
98.1k121701066
asked 3 hours ago
IchVerloren
286
edited 2 hours ago
Qmechanic♦
98.1k121701066
edited 2 hours ago
Qmechanic♦
98.1k121701066
edited 2 hours ago
Qmechanic♦
98.1k121701066
98.1k121701066
asked 3 hours ago
IchVerloren
286
asked 3 hours ago
IchVerloren
286
asked 3 hours ago
IchVerloren
286
286
In my interpretation with your notation, $langle 1 vert$ is the representation of the (probability) measurement captured the signal of alive.
– ChoMedit
23 mins ago
add a comment |Â
In my interpretation with your notation, $langle 1 vert$ is the representation of the (probability) measurement captured the signal of alive.
– ChoMedit
23 mins ago
In my interpretation with your notation, $langle 1 vert$ is the representation of the (probability) measurement captured the signal of alive.
– ChoMedit
23 mins ago
In my interpretation with your notation, $langle 1 vert$ is the representation of the (probability) measurement captured the signal of alive.
– ChoMedit
23 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
When you write:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
you are assuming there exists an alive operator and that this operator has eigenstates $|1rangle$ and $|0rangle$ that you can use as a basis for writing the wavefunction of the cat. Neither of these assumptions seem reasonable so as it stands the question makes no sense.
However you could replace the cat by a spin that can be in a superposition of up and down states. In that case we are writing the $z$ component of the spin using the eigenstates of $hatL_z$ as a basis, and we'd get the same equation:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
where now our $|1rangle$ and $|0rangle$ states are well defined because they are the eigenstates of $hatL_z$. And now it's clear that $langle 0 | 1 rangle$ and $langle 1 | 0 rangle$ are zero because eigenstates are orthogonal.
answered 3 hours ago
John Rennie
265k41517765
I was thinking about $vert 1 rangle$ and $vert 0 rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states.
– IchVerloren
2 hours ago
@IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense.
– John Rennie
1 hour ago
add a comment |Â
up vote
2
down vote
Following John Rennie's analogy, let the cat be a spin with up (alive) or down (dead) state. Note that you've expanded $|#rangle$ in terms of the orthonormal basis vectors:
$$|#rangle = a|1rangle + b|0rangle$$
By definition of orthogonality (and the idea that "alive" and "dead" are orthogonal), $langle 0|1rangle = langle 1|0rangle = 0$.
Now you're computing the probability that the cat is alive (hopefully it always is) or dead:
$$ |#rangle |^2 = |a|^2, quad |langle 0 |#rangle|^2 = |b|^2 $$
To determine the range of values $a$ and $b$ can take, compute:
$$ |langle#|#rangle|^2 = |a|^2 + |b|^2 = 1$$
This is $1$ because $|#rangle$ is assumed to be normalised. So the range of values of $a$ and $b$ are not strictly $1/sqrt2$ for both, but $(a,b) in S^1$, where $S^1$ is the unit circle.
answered 3 hours ago
Arjit Seth
506210
add a comment |Â
up vote
0
down vote
First of all, I do not think $a^2 + b^2 =1$ tells you that a = $frac1sqrt 2$. Basic trignometry.
And to tell the truth, no one knows why $a^2$ is the probability(Or more precisely, the square of the coefficients). It's Born's Postulate.
<0|1> can be viewed as what is the probability density to find state |1> in |0>. And those two happens to be orthogonal if $|0>$ and |1> are a set of basis.
By the way, I think your question can be found in any quantum physics textbooks. Hope this helps you!
answered 3 hours ago
Simon241
1
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
When you write:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
you are assuming there exists an alive operator and that this operator has eigenstates $|1rangle$ and $|0rangle$ that you can use as a basis for writing the wavefunction of the cat. Neither of these assumptions seem reasonable so as it stands the question makes no sense.
However you could replace the cat by a spin that can be in a superposition of up and down states. In that case we are writing the $z$ component of the spin using the eigenstates of $hatL_z$ as a basis, and we'd get the same equation:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
where now our $|1rangle$ and $|0rangle$ states are well defined because they are the eigenstates of $hatL_z$. And now it's clear that $langle 0 | 1 rangle$ and $langle 1 | 0 rangle$ are zero because eigenstates are orthogonal.
answered 3 hours ago
John Rennie
265k41517765
I was thinking about $vert 1 rangle$ and $vert 0 rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states.
– IchVerloren
2 hours ago
@IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense.
– John Rennie
1 hour ago
add a comment |Â
up vote
3
down vote
When you write:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
you are assuming there exists an alive operator and that this operator has eigenstates $|1rangle$ and $|0rangle$ that you can use as a basis for writing the wavefunction of the cat. Neither of these assumptions seem reasonable so as it stands the question makes no sense.
However you could replace the cat by a spin that can be in a superposition of up and down states. In that case we are writing the $z$ component of the spin using the eigenstates of $hatL_z$ as a basis, and we'd get the same equation:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
where now our $|1rangle$ and $|0rangle$ states are well defined because they are the eigenstates of $hatL_z$. And now it's clear that $langle 0 | 1 rangle$ and $langle 1 | 0 rangle$ are zero because eigenstates are orthogonal.
answered 3 hours ago
John Rennie
265k41517765
I was thinking about $vert 1 rangle$ and $vert 0 rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states.
– IchVerloren
2 hours ago
@IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense.
– John Rennie
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
When you write:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
you are assuming there exists an alive operator and that this operator has eigenstates $|1rangle$ and $|0rangle$ that you can use as a basis for writing the wavefunction of the cat. Neither of these assumptions seem reasonable so as it stands the question makes no sense.
However you could replace the cat by a spin that can be in a superposition of up and down states. In that case we are writing the $z$ component of the spin using the eigenstates of $hatL_z$ as a basis, and we'd get the same equation:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
where now our $|1rangle$ and $|0rangle$ states are well defined because they are the eigenstates of $hatL_z$. And now it's clear that $langle 0 | 1 rangle$ and $langle 1 | 0 rangle$ are zero because eigenstates are orthogonal.
answered 3 hours ago
John Rennie
265k41517765
When you write:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
you are assuming there exists an alive operator and that this operator has eigenstates $|1rangle$ and $|0rangle$ that you can use as a basis for writing the wavefunction of the cat. Neither of these assumptions seem reasonable so as it stands the question makes no sense.
However you could replace the cat by a spin that can be in a superposition of up and down states. In that case we are writing the $z$ component of the spin using the eigenstates of $hatL_z$ as a basis, and we'd get the same equation:
$$ vert text#rangle=avert1rangle+bvert0rangle $$
where now our $|1rangle$ and $|0rangle$ states are well defined because they are the eigenstates of $hatL_z$. And now it's clear that $langle 0 | 1 rangle$ and $langle 1 | 0 rangle$ are zero because eigenstates are orthogonal.
answered 3 hours ago
John Rennie
265k41517765
answered 3 hours ago
John Rennie
265k41517765
answered 3 hours ago
John Rennie
265k41517765
answered 3 hours ago
John Rennie
265k41517765
265k41517765
I was thinking about $vert 1 rangle$ and $vert 0 rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states.
– IchVerloren
2 hours ago
@IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense.
– John Rennie
1 hour ago
add a comment |Â
I was thinking about $vert 1 rangle$ and $vert 0 rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states.
– IchVerloren
2 hours ago
@IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense.
– John Rennie
1 hour ago
I was thinking about $vert 1 rangle$ and $vert 0 rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states.
– IchVerloren
2 hours ago
I was thinking about $vert 1 rangle$ and $vert 0 rangle$ as "alive" and "dead" states. Does that make sense? The cat sould be in a superposition of dead and alive states.
– IchVerloren
2 hours ago
@IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense.
– John Rennie
1 hour ago
@IchVerloren a cat is a massively complex system, and being alive is also a complicated business. There isn't a precise distinction between alive and dead. So trying to define alive and dead states and use them as a basis doesn't make any sense.
– John Rennie
1 hour ago
add a comment |Â
up vote
2
down vote
Following John Rennie's analogy, let the cat be a spin with up (alive) or down (dead) state. Note that you've expanded $|#rangle$ in terms of the orthonormal basis vectors:
$$|#rangle = a|1rangle + b|0rangle$$
By definition of orthogonality (and the idea that "alive" and "dead" are orthogonal), $langle 0|1rangle = langle 1|0rangle = 0$.
Now you're computing the probability that the cat is alive (hopefully it always is) or dead:
$$ |#rangle |^2 = |a|^2, quad |langle 0 |#rangle|^2 = |b|^2 $$
To determine the range of values $a$ and $b$ can take, compute:
$$ |langle#|#rangle|^2 = |a|^2 + |b|^2 = 1$$
This is $1$ because $|#rangle$ is assumed to be normalised. So the range of values of $a$ and $b$ are not strictly $1/sqrt2$ for both, but $(a,b) in S^1$, where $S^1$ is the unit circle.
answered 3 hours ago
Arjit Seth
506210
add a comment |Â
up vote
2
down vote
Following John Rennie's analogy, let the cat be a spin with up (alive) or down (dead) state. Note that you've expanded $|#rangle$ in terms of the orthonormal basis vectors:
$$|#rangle = a|1rangle + b|0rangle$$
By definition of orthogonality (and the idea that "alive" and "dead" are orthogonal), $langle 0|1rangle = langle 1|0rangle = 0$.
Now you're computing the probability that the cat is alive (hopefully it always is) or dead:
$$ |#rangle |^2 = |a|^2, quad |langle 0 |#rangle|^2 = |b|^2 $$
To determine the range of values $a$ and $b$ can take, compute:
$$ |langle#|#rangle|^2 = |a|^2 + |b|^2 = 1$$
This is $1$ because $|#rangle$ is assumed to be normalised. So the range of values of $a$ and $b$ are not strictly $1/sqrt2$ for both, but $(a,b) in S^1$, where $S^1$ is the unit circle.
answered 3 hours ago
Arjit Seth
506210
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Following John Rennie's analogy, let the cat be a spin with up (alive) or down (dead) state. Note that you've expanded $|#rangle$ in terms of the orthonormal basis vectors:
$$|#rangle = a|1rangle + b|0rangle$$
By definition of orthogonality (and the idea that "alive" and "dead" are orthogonal), $langle 0|1rangle = langle 1|0rangle = 0$.
Now you're computing the probability that the cat is alive (hopefully it always is) or dead:
$$ |#rangle |^2 = |a|^2, quad |langle 0 |#rangle|^2 = |b|^2 $$
To determine the range of values $a$ and $b$ can take, compute:
$$ |langle#|#rangle|^2 = |a|^2 + |b|^2 = 1$$
This is $1$ because $|#rangle$ is assumed to be normalised. So the range of values of $a$ and $b$ are not strictly $1/sqrt2$ for both, but $(a,b) in S^1$, where $S^1$ is the unit circle.
answered 3 hours ago
Arjit Seth
506210
Following John Rennie's analogy, let the cat be a spin with up (alive) or down (dead) state. Note that you've expanded $|#rangle$ in terms of the orthonormal basis vectors:
$$|#rangle = a|1rangle + b|0rangle$$
By definition of orthogonality (and the idea that "alive" and "dead" are orthogonal), $langle 0|1rangle = langle 1|0rangle = 0$.
Now you're computing the probability that the cat is alive (hopefully it always is) or dead:
$$ |#rangle |^2 = |a|^2, quad |langle 0 |#rangle|^2 = |b|^2 $$
To determine the range of values $a$ and $b$ can take, compute:
$$ |langle#|#rangle|^2 = |a|^2 + |b|^2 = 1$$
This is $1$ because $|#rangle$ is assumed to be normalised. So the range of values of $a$ and $b$ are not strictly $1/sqrt2$ for both, but $(a,b) in S^1$, where $S^1$ is the unit circle.
answered 3 hours ago
Arjit Seth
506210
answered 3 hours ago
Arjit Seth
506210
answered 3 hours ago
Arjit Seth
506210
answered 3 hours ago
Arjit Seth
506210
506210
add a comment |Â
add a comment |Â
up vote
0
down vote
First of all, I do not think $a^2 + b^2 =1$ tells you that a = $frac1sqrt 2$. Basic trignometry.
And to tell the truth, no one knows why $a^2$ is the probability(Or more precisely, the square of the coefficients). It's Born's Postulate.
<0|1> can be viewed as what is the probability density to find state |1> in |0>. And those two happens to be orthogonal if $|0>$ and |1> are a set of basis.
By the way, I think your question can be found in any quantum physics textbooks. Hope this helps you!
answered 3 hours ago
Simon241
1
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
First of all, I do not think $a^2 + b^2 =1$ tells you that a = $frac1sqrt 2$. Basic trignometry.
And to tell the truth, no one knows why $a^2$ is the probability(Or more precisely, the square of the coefficients). It's Born's Postulate.
<0|1> can be viewed as what is the probability density to find state |1> in |0>. And those two happens to be orthogonal if $|0>$ and |1> are a set of basis.
By the way, I think your question can be found in any quantum physics textbooks. Hope this helps you!
answered 3 hours ago
Simon241
1
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First of all, I do not think $a^2 + b^2 =1$ tells you that a = $frac1sqrt 2$. Basic trignometry.
And to tell the truth, no one knows why $a^2$ is the probability(Or more precisely, the square of the coefficients). It's Born's Postulate.
<0|1> can be viewed as what is the probability density to find state |1> in |0>. And those two happens to be orthogonal if $|0>$ and |1> are a set of basis.
By the way, I think your question can be found in any quantum physics textbooks. Hope this helps you!
answered 3 hours ago
Simon241
1
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First of all, I do not think $a^2 + b^2 =1$ tells you that a = $frac1sqrt 2$. Basic trignometry.
And to tell the truth, no one knows why $a^2$ is the probability(Or more precisely, the square of the coefficients). It's Born's Postulate.
<0|1> can be viewed as what is the probability density to find state |1> in |0>. And those two happens to be orthogonal if $|0>$ and |1> are a set of basis.
By the way, I think your question can be found in any quantum physics textbooks. Hope this helps you!
answered 3 hours ago
Simon241
1
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Simon241
1
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Simon241
1
answered 3 hours ago
Simon241
1
1
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Simon241 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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In my interpretation with your notation, $langle 1 vert$ is the representation of the (probability) measurement captured the signal of alive.
– ChoMedit
23 mins ago