Mixing
Transformation of a bell state
Clash Royale CLAN TAG#URR8PPP
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I am relatively new and interested in quantum computing.
Specifically, I am interested in transforming an equation that I found on Wikipedia. But I did not quite understand the transformation.
$ frac1sqrt2(left|0right>_xleft|0right>_y-left|1right>_xleft|1right>_y) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
My idea is so far to use Hadamard transform for the two qubits:
$ frac1sqrt2(H(left|0right>_xleft|0right>_y)-H(left|1right>_xleft|1right>_y)) $
I have used the Hadamard transformation and now come to this:
$ = frac1sqrt2(frac12[(left|00right>+left|10right>+left|01right>+left|11right>) -(left|00right>-left|10right>-left|01right>+left|11right>)]) $
If I simplify that a bit now then I have that as a result:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) $
But the result looks different now than the equation I wrote down at the beginning:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
I do not know if the forming is allowed that way. If somebody knows how the transformation of the equation works, so that I get what I wrote in the beginning, I would be very happy if somebody could explain it!
I hope that my question is understandable :)
quantum-gate qubit qubit-state
asked 38 mins ago
P_Gate
113
New contributor
P_Gate 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
2
down vote
favorite
I am relatively new and interested in quantum computing.
Specifically, I am interested in transforming an equation that I found on Wikipedia. But I did not quite understand the transformation.
$ frac1sqrt2(left|0right>_xleft|0right>_y-left|1right>_xleft|1right>_y) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
My idea is so far to use Hadamard transform for the two qubits:
$ frac1sqrt2(H(left|0right>_xleft|0right>_y)-H(left|1right>_xleft|1right>_y)) $
I have used the Hadamard transformation and now come to this:
$ = frac1sqrt2(frac12[(left|00right>+left|10right>+left|01right>+left|11right>) -(left|00right>-left|10right>-left|01right>+left|11right>)]) $
If I simplify that a bit now then I have that as a result:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) $
But the result looks different now than the equation I wrote down at the beginning:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
I do not know if the forming is allowed that way. If somebody knows how the transformation of the equation works, so that I get what I wrote in the beginning, I would be very happy if somebody could explain it!
I hope that my question is understandable :)
quantum-gate qubit qubit-state
asked 38 mins ago
P_Gate
113
New contributor
P_Gate 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
2
down vote
favorite
up vote
2
down vote
favorite
I am relatively new and interested in quantum computing.
Specifically, I am interested in transforming an equation that I found on Wikipedia. But I did not quite understand the transformation.
$ frac1sqrt2(left|0right>_xleft|0right>_y-left|1right>_xleft|1right>_y) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
My idea is so far to use Hadamard transform for the two qubits:
$ frac1sqrt2(H(left|0right>_xleft|0right>_y)-H(left|1right>_xleft|1right>_y)) $
I have used the Hadamard transformation and now come to this:
$ = frac1sqrt2(frac12[(left|00right>+left|10right>+left|01right>+left|11right>) -(left|00right>-left|10right>-left|01right>+left|11right>)]) $
If I simplify that a bit now then I have that as a result:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) $
But the result looks different now than the equation I wrote down at the beginning:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
I do not know if the forming is allowed that way. If somebody knows how the transformation of the equation works, so that I get what I wrote in the beginning, I would be very happy if somebody could explain it!
I hope that my question is understandable :)
quantum-gate qubit qubit-state
asked 38 mins ago
P_Gate
113
New contributor
P_Gate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am relatively new and interested in quantum computing.
Specifically, I am interested in transforming an equation that I found on Wikipedia. But I did not quite understand the transformation.
$ frac1sqrt2(left|0right>_xleft|0right>_y-left|1right>_xleft|1right>_y) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
My idea is so far to use Hadamard transform for the two qubits:
$ frac1sqrt2(H(left|0right>_xleft|0right>_y)-H(left|1right>_xleft|1right>_y)) $
I have used the Hadamard transformation and now come to this:
$ = frac1sqrt2(frac12[(left|00right>+left|10right>+left|01right>+left|11right>) -(left|00right>-left|10right>-left|01right>+left|11right>)]) $
If I simplify that a bit now then I have that as a result:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) $
But the result looks different now than the equation I wrote down at the beginning:
$ = frac1sqrt2(left|1_x0_yright>+left|0_x1_yright>) = frac1sqrt2(|+rangle_x|-rangle_y+|-rangle_x|+rangle_y) $
I do not know if the forming is allowed that way. If somebody knows how the transformation of the equation works, so that I get what I wrote in the beginning, I would be very happy if somebody could explain it!
I hope that my question is understandable :)
quantum-gate qubit qubit-state
quantum-gate qubit qubit-state
asked 38 mins ago
P_Gate
113
New contributor
P_Gate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 38 mins ago
P_Gate
113
New contributor
P_Gate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 38 mins ago
P_Gate
113
New contributor
P_Gate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 38 mins ago
P_Gate
113
asked 38 mins ago
P_Gate
113
113
New contributor
P_Gate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P_Gate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P_Gate 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
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2 Answers
2
active
oldest
votes
up vote
2
down vote
You've started wanting to talk about
$$
(|00rangle-|11rangle)/sqrt2,
$$
but you've then gone ahead and calculated
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2.
$$
You should not expect these to be equal.
On the other hand, the calculation that you've done is sensible if you understand what you're doing. What you actually want to see, in order to verify your original statement is that
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2=(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2.
$$
You've already calculated the left-hand side. The calculation on the right-hand side is, I suspect, the whole reason why you've chosen to apply the Hadamard transform - you know that the Hadamard converts $|+rangle$ to $|0rangle$, and $|-rangle$ to $|1rangle$. So, you can immediately read that
$$
(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2=(|01rangle+|10rangle)/sqrt2,
$$
so you can see that the left-hand side is equal to the right-hand side, as desired.
answered 31 mins ago
DaftWullie
9,5871332
Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show?
– P_Gate
26 mins ago
@P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides.
– DaftWullie
18 mins ago
Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep)
– P_Gate
16 mins ago
add a comment |Â
up vote
1
down vote
You are not proving the equality in a correct way. By multiplying by the Hadamard matrix, you are changing the state you are trying to calculate, not demonstrating the equality you want to prove.$defket#1lvert#1rangle$
In order to prove what you state at the beginning of the question, I would use the facts that $$
ket+ =frac1sqrt2bigl(ket0 + ket1bigr), qquadket-=frac1sqrt2bigl(ket0 - ket1bigr),$$
and then develop the equality in the inverse order. Consequently:
$$
beginalignat2
frac1sqrt2&bigl(ket+_x ket-_y - ket-_x ket+_ybigr)mspace-128mu
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_x+ket1_xbigr)bigl(ket0_y-ket1_ybigr)+bigl(ket0_x-ket1_xbigr)bigl(ket0_y+ket1_ybigr)Bigr)
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_xket0_y+ket1_xket-_y-ket0_xket1_y-ket1_xket1_y
\[-1ex]&&&+ket0_xket0_y+ket0_xket1_y-ket1_xket0_y-ket1_xket1_ybigr)Bigr)
\=;&
frac1sqrt2bigl(ket0_xket0_y - ket1_xket1_ybigr).mspace-128mu
endalignat
$$
And so you prove the equality you were trying to solve at the beginning of the question.
edited 6 mins ago
Niel de Beaudrap
5,027830
answered 19 mins ago
Josu Etxezarreta Martinez
926116
1
Thanks, your answer is also helpful for me! I would upvote but I have not enough reputation :(
– P_Gate
15 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You've started wanting to talk about
$$
(|00rangle-|11rangle)/sqrt2,
$$
but you've then gone ahead and calculated
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2.
$$
You should not expect these to be equal.
On the other hand, the calculation that you've done is sensible if you understand what you're doing. What you actually want to see, in order to verify your original statement is that
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2=(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2.
$$
You've already calculated the left-hand side. The calculation on the right-hand side is, I suspect, the whole reason why you've chosen to apply the Hadamard transform - you know that the Hadamard converts $|+rangle$ to $|0rangle$, and $|-rangle$ to $|1rangle$. So, you can immediately read that
$$
(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2=(|01rangle+|10rangle)/sqrt2,
$$
so you can see that the left-hand side is equal to the right-hand side, as desired.
answered 31 mins ago
DaftWullie
9,5871332
Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show?
– P_Gate
26 mins ago
@P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides.
– DaftWullie
18 mins ago
Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep)
– P_Gate
16 mins ago
add a comment |Â
up vote
2
down vote
You've started wanting to talk about
$$
(|00rangle-|11rangle)/sqrt2,
$$
but you've then gone ahead and calculated
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2.
$$
You should not expect these to be equal.
On the other hand, the calculation that you've done is sensible if you understand what you're doing. What you actually want to see, in order to verify your original statement is that
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2=(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2.
$$
You've already calculated the left-hand side. The calculation on the right-hand side is, I suspect, the whole reason why you've chosen to apply the Hadamard transform - you know that the Hadamard converts $|+rangle$ to $|0rangle$, and $|-rangle$ to $|1rangle$. So, you can immediately read that
$$
(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2=(|01rangle+|10rangle)/sqrt2,
$$
so you can see that the left-hand side is equal to the right-hand side, as desired.
answered 31 mins ago
DaftWullie
9,5871332
Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show?
– P_Gate
26 mins ago
@P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides.
– DaftWullie
18 mins ago
Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep)
– P_Gate
16 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You've started wanting to talk about
$$
(|00rangle-|11rangle)/sqrt2,
$$
but you've then gone ahead and calculated
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2.
$$
You should not expect these to be equal.
On the other hand, the calculation that you've done is sensible if you understand what you're doing. What you actually want to see, in order to verify your original statement is that
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2=(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2.
$$
You've already calculated the left-hand side. The calculation on the right-hand side is, I suspect, the whole reason why you've chosen to apply the Hadamard transform - you know that the Hadamard converts $|+rangle$ to $|0rangle$, and $|-rangle$ to $|1rangle$. So, you can immediately read that
$$
(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2=(|01rangle+|10rangle)/sqrt2,
$$
so you can see that the left-hand side is equal to the right-hand side, as desired.
answered 31 mins ago
DaftWullie
9,5871332
You've started wanting to talk about
$$
(|00rangle-|11rangle)/sqrt2,
$$
but you've then gone ahead and calculated
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2.
$$
You should not expect these to be equal.
On the other hand, the calculation that you've done is sensible if you understand what you're doing. What you actually want to see, in order to verify your original statement is that
$$
(Hotimes H)cdot(|00rangle-|11rangle)/sqrt2=(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2.
$$
You've already calculated the left-hand side. The calculation on the right-hand side is, I suspect, the whole reason why you've chosen to apply the Hadamard transform - you know that the Hadamard converts $|+rangle$ to $|0rangle$, and $|-rangle$ to $|1rangle$. So, you can immediately read that
$$
(Hotimes H)cdot(|+-rangle+|-+rangle)/sqrt2=(|01rangle+|10rangle)/sqrt2,
$$
so you can see that the left-hand side is equal to the right-hand side, as desired.
answered 31 mins ago
DaftWullie
9,5871332
answered 31 mins ago
DaftWullie
9,5871332
answered 31 mins ago
DaftWullie
9,5871332
answered 31 mins ago
DaftWullie
9,5871332
9,5871332
Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show?
– P_Gate
26 mins ago
@P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides.
– DaftWullie
18 mins ago
Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep)
– P_Gate
16 mins ago
add a comment |Â
Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show?
– P_Gate
26 mins ago
@P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides.
– DaftWullie
18 mins ago
Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep)
– P_Gate
16 mins ago
Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show?
– P_Gate
26 mins ago
Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show?
– P_Gate
26 mins ago
@P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides.
– DaftWullie
18 mins ago
@P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides.
– DaftWullie
18 mins ago
Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep)
– P_Gate
16 mins ago
Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep)
– P_Gate
16 mins ago
add a comment |Â
up vote
1
down vote
You are not proving the equality in a correct way. By multiplying by the Hadamard matrix, you are changing the state you are trying to calculate, not demonstrating the equality you want to prove.$defket#1lvert#1rangle$
In order to prove what you state at the beginning of the question, I would use the facts that $$
ket+ =frac1sqrt2bigl(ket0 + ket1bigr), qquadket-=frac1sqrt2bigl(ket0 - ket1bigr),$$
and then develop the equality in the inverse order. Consequently:
$$
beginalignat2
frac1sqrt2&bigl(ket+_x ket-_y - ket-_x ket+_ybigr)mspace-128mu
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_x+ket1_xbigr)bigl(ket0_y-ket1_ybigr)+bigl(ket0_x-ket1_xbigr)bigl(ket0_y+ket1_ybigr)Bigr)
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_xket0_y+ket1_xket-_y-ket0_xket1_y-ket1_xket1_y
\[-1ex]&&&+ket0_xket0_y+ket0_xket1_y-ket1_xket0_y-ket1_xket1_ybigr)Bigr)
\=;&
frac1sqrt2bigl(ket0_xket0_y - ket1_xket1_ybigr).mspace-128mu
endalignat
$$
And so you prove the equality you were trying to solve at the beginning of the question.
edited 6 mins ago
Niel de Beaudrap
5,027830
answered 19 mins ago
Josu Etxezarreta Martinez
926116
1
Thanks, your answer is also helpful for me! I would upvote but I have not enough reputation :(
– P_Gate
15 mins ago
add a comment |Â
up vote
1
down vote
You are not proving the equality in a correct way. By multiplying by the Hadamard matrix, you are changing the state you are trying to calculate, not demonstrating the equality you want to prove.$defket#1lvert#1rangle$
In order to prove what you state at the beginning of the question, I would use the facts that $$
ket+ =frac1sqrt2bigl(ket0 + ket1bigr), qquadket-=frac1sqrt2bigl(ket0 - ket1bigr),$$
and then develop the equality in the inverse order. Consequently:
$$
beginalignat2
frac1sqrt2&bigl(ket+_x ket-_y - ket-_x ket+_ybigr)mspace-128mu
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_x+ket1_xbigr)bigl(ket0_y-ket1_ybigr)+bigl(ket0_x-ket1_xbigr)bigl(ket0_y+ket1_ybigr)Bigr)
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_xket0_y+ket1_xket-_y-ket0_xket1_y-ket1_xket1_y
\[-1ex]&&&+ket0_xket0_y+ket0_xket1_y-ket1_xket0_y-ket1_xket1_ybigr)Bigr)
\=;&
frac1sqrt2bigl(ket0_xket0_y - ket1_xket1_ybigr).mspace-128mu
endalignat
$$
And so you prove the equality you were trying to solve at the beginning of the question.
edited 6 mins ago
Niel de Beaudrap
5,027830
answered 19 mins ago
Josu Etxezarreta Martinez
926116
1
Thanks, your answer is also helpful for me! I would upvote but I have not enough reputation :(
– P_Gate
15 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are not proving the equality in a correct way. By multiplying by the Hadamard matrix, you are changing the state you are trying to calculate, not demonstrating the equality you want to prove.$defket#1lvert#1rangle$
In order to prove what you state at the beginning of the question, I would use the facts that $$
ket+ =frac1sqrt2bigl(ket0 + ket1bigr), qquadket-=frac1sqrt2bigl(ket0 - ket1bigr),$$
and then develop the equality in the inverse order. Consequently:
$$
beginalignat2
frac1sqrt2&bigl(ket+_x ket-_y - ket-_x ket+_ybigr)mspace-128mu
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_x+ket1_xbigr)bigl(ket0_y-ket1_ybigr)+bigl(ket0_x-ket1_xbigr)bigl(ket0_y+ket1_ybigr)Bigr)
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_xket0_y+ket1_xket-_y-ket0_xket1_y-ket1_xket1_y
\[-1ex]&&&+ket0_xket0_y+ket0_xket1_y-ket1_xket0_y-ket1_xket1_ybigr)Bigr)
\=;&
frac1sqrt2bigl(ket0_xket0_y - ket1_xket1_ybigr).mspace-128mu
endalignat
$$
And so you prove the equality you were trying to solve at the beginning of the question.
edited 6 mins ago
Niel de Beaudrap
5,027830
answered 19 mins ago
Josu Etxezarreta Martinez
926116
You are not proving the equality in a correct way. By multiplying by the Hadamard matrix, you are changing the state you are trying to calculate, not demonstrating the equality you want to prove.$defket#1lvert#1rangle$
In order to prove what you state at the beginning of the question, I would use the facts that $$
ket+ =frac1sqrt2bigl(ket0 + ket1bigr), qquadket-=frac1sqrt2bigl(ket0 - ket1bigr),$$
and then develop the equality in the inverse order. Consequently:
$$
beginalignat2
frac1sqrt2&bigl(ket+_x ket-_y - ket-_x ket+_ybigr)mspace-128mu
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_x+ket1_xbigr)bigl(ket0_y-ket1_ybigr)+bigl(ket0_x-ket1_xbigr)bigl(ket0_y+ket1_ybigr)Bigr)
\[1ex]=;&
frac1sqrt2Bigl(frac12bigl(&&ket0_xket0_y+ket1_xket-_y-ket0_xket1_y-ket1_xket1_y
\[-1ex]&&&+ket0_xket0_y+ket0_xket1_y-ket1_xket0_y-ket1_xket1_ybigr)Bigr)
\=;&
frac1sqrt2bigl(ket0_xket0_y - ket1_xket1_ybigr).mspace-128mu
endalignat
$$
And so you prove the equality you were trying to solve at the beginning of the question.
edited 6 mins ago
Niel de Beaudrap
5,027830
answered 19 mins ago
Josu Etxezarreta Martinez
926116
edited 6 mins ago
Niel de Beaudrap
5,027830
edited 6 mins ago
Niel de Beaudrap
5,027830
edited 6 mins ago
Niel de Beaudrap
5,027830
5,027830
answered 19 mins ago
Josu Etxezarreta Martinez
926116
answered 19 mins ago
Josu Etxezarreta Martinez
926116
answered 19 mins ago
Josu Etxezarreta Martinez
926116
926116
1
Thanks, your answer is also helpful for me! I would upvote but I have not enough reputation :(
– P_Gate
15 mins ago
add a comment |Â
1
Thanks, your answer is also helpful for me! I would upvote but I have not enough reputation :(
– P_Gate
15 mins ago
1
1
Thanks, your answer is also helpful for me! I would upvote but I have not enough reputation :(
– P_Gate
15 mins ago
Thanks, your answer is also helpful for me! I would upvote but I have not enough reputation :(
– P_Gate
15 mins ago
add a comment |Â
P_Gate is a new contributor. Be nice, and check out our Code of Conduct.
P_Gate is a new contributor. Be nice, and check out our Code of Conduct.
P_Gate is a new contributor. Be nice, and check out our Code of Conduct.
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