Mixing
Are the squared absolute values of the eigenvalues of a unitary matrix always 1?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I'm going through the phase estimation algorithm, and wanted to sanity-check my calculations by making sure the state I'd calculated was still normalized. It is, assuming the square of the absolute value of the eigenvalue of the arbitrary unitary operator I'm analyzing equals 1. So, does it? Assuming that the eigenvector of the eigenvalue is normalized.
phase-estimation
asked 9 hours ago
ahelwer
6539
add a comment |Â
up vote
1
down vote
favorite
I'm going through the phase estimation algorithm, and wanted to sanity-check my calculations by making sure the state I'd calculated was still normalized. It is, assuming the square of the absolute value of the eigenvalue of the arbitrary unitary operator I'm analyzing equals 1. So, does it? Assuming that the eigenvector of the eigenvalue is normalized.
phase-estimation
asked 9 hours ago
ahelwer
6539
1
Note that your question doesn't depend on whether the eigenvector is normalised. If you have a longer or shorter eigenvector, then that longer or shorter eigenvector has its norm changed by the same scalar factor as if it were normalised.
– Niel de Beaudrap
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm going through the phase estimation algorithm, and wanted to sanity-check my calculations by making sure the state I'd calculated was still normalized. It is, assuming the square of the absolute value of the eigenvalue of the arbitrary unitary operator I'm analyzing equals 1. So, does it? Assuming that the eigenvector of the eigenvalue is normalized.
phase-estimation
asked 9 hours ago
ahelwer
6539
I'm going through the phase estimation algorithm, and wanted to sanity-check my calculations by making sure the state I'd calculated was still normalized. It is, assuming the square of the absolute value of the eigenvalue of the arbitrary unitary operator I'm analyzing equals 1. So, does it? Assuming that the eigenvector of the eigenvalue is normalized.
phase-estimation
phase-estimation
asked 9 hours ago
ahelwer
6539
asked 9 hours ago
ahelwer
6539
edited 9 hours ago
asked 9 hours ago
ahelwer
6539
asked 9 hours ago
ahelwer
6539
asked 9 hours ago
ahelwer
6539
6539
1
Note that your question doesn't depend on whether the eigenvector is normalised. If you have a longer or shorter eigenvector, then that longer or shorter eigenvector has its norm changed by the same scalar factor as if it were normalised.
– Niel de Beaudrap
1 hour ago
add a comment |Â
1
Note that your question doesn't depend on whether the eigenvector is normalised. If you have a longer or shorter eigenvector, then that longer or shorter eigenvector has its norm changed by the same scalar factor as if it were normalised.
– Niel de Beaudrap
1 hour ago
1
1
Note that your question doesn't depend on whether the eigenvector is normalised. If you have a longer or shorter eigenvector, then that longer or shorter eigenvector has its norm changed by the same scalar factor as if it were normalised.
– Niel de Beaudrap
1 hour ago
Note that your question doesn't depend on whether the eigenvector is normalised. If you have a longer or shorter eigenvector, then that longer or shorter eigenvector has its norm changed by the same scalar factor as if it were normalised.
– Niel de Beaudrap
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Good question. The answer turns out to be Yes.
Start with the definition of eigenvalues and eigenvectors:
$$
beginalign
U|psirangle &= lambda |psirangle\
langlepsi|U^dagger &= langle psi| barlambda.
endalign
$$
Left multiply each side of line 1 by the corresponding side of line 2.
$$
beginalign
langle psi|U^daggercdot U|psi rangle &= langle psi | barlambda lambda |psirangle \
langle psi |psi rangle &= |lambda |^2 langle psi | |psi rangle \
1 &= |lambda|^2
endalign
$$
edited 5 hours ago
AHusain
715127
answered 8 hours ago
user1271772
4,552230
1
I get every step except for how $langle psi | U cdot U | psi rangle = langle psi | psi rangle$; doesn't that only work if the matrix is hermitian as well as unitary?
– ahelwer
7 hours ago
1
I'm also not sure how we get from $langle psi | lambda lambda | psi rangle$ to $| lambda | ^2 langle psi || psi rangle$, isn't $| lambda | ^2$ the same as $lambda cdot lambda ^*$? Where did the conjugate come from?
– ahelwer
7 hours ago
Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it!
– user1271772
5 hours ago
Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain!
– user1271772
5 hours ago
add a comment |Â
up vote
1
down vote
@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation,
$$
ifracddt=H|psirangle.
$$
For a time-invariant Hamiltonian, the solution is
$$
|psi(t)rangle=e^-iHt|psi(0)rangle,
$$
where $e^-iHt$ is unitary because $e^-iHte^iHt=mathbbI$. Just stating this solution actually skips over the thing I really want to focus on.
We could expand a generic $|psirangle=sum_ia_i|irangle$, so the Schrödinger equation becomes a series of simultaneous differential equations for the $a_i$:
$$
ifracda_idt=sum_jH_ija_j.
$$
Now consider what happens to an eigenvector $|lambdarangle=sum_ib_i|irangle$ of $H$:
$$
sum_jH_ijb_i^star=lambda b_i^star.
$$
We can take linear combinations of the $a_i$:
$$
ifracdsum_ib_i^star a_idt=sum_ijb_i^star H_ija_j=lambdasum_jb_j^star a_j.
$$
Hence, we see that the component $x=sum_jb_j^star a_j$ simply satisfies
$$
ifracdxdt=lambda x,
$$
so $x(t)=e^-ilambda tx(0)$. In other words, a state initially created as an eigenvector $|lambdarangle$ stays in that state and just acquires a phase over time $e^-ilambda t|lambdarangle$. Hence, eigenvectors of $H$ are also eigenvectors of the unitary $U$, with eigenvalues $e^-ilambda t$, and these have modulus 1.
answered 2 hours ago
DaftWullie
8,7171331
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Good question. The answer turns out to be Yes.
Start with the definition of eigenvalues and eigenvectors:
$$
beginalign
U|psirangle &= lambda |psirangle\
langlepsi|U^dagger &= langle psi| barlambda.
endalign
$$
Left multiply each side of line 1 by the corresponding side of line 2.
$$
beginalign
langle psi|U^daggercdot U|psi rangle &= langle psi | barlambda lambda |psirangle \
langle psi |psi rangle &= |lambda |^2 langle psi | |psi rangle \
1 &= |lambda|^2
endalign
$$
edited 5 hours ago
AHusain
715127
answered 8 hours ago
user1271772
4,552230
1
I get every step except for how $langle psi | U cdot U | psi rangle = langle psi | psi rangle$; doesn't that only work if the matrix is hermitian as well as unitary?
– ahelwer
7 hours ago
1
I'm also not sure how we get from $langle psi | lambda lambda | psi rangle$ to $| lambda | ^2 langle psi || psi rangle$, isn't $| lambda | ^2$ the same as $lambda cdot lambda ^*$? Where did the conjugate come from?
– ahelwer
7 hours ago
Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it!
– user1271772
5 hours ago
Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain!
– user1271772
5 hours ago
add a comment |Â
up vote
4
down vote
accepted
Good question. The answer turns out to be Yes.
Start with the definition of eigenvalues and eigenvectors:
$$
beginalign
U|psirangle &= lambda |psirangle\
langlepsi|U^dagger &= langle psi| barlambda.
endalign
$$
Left multiply each side of line 1 by the corresponding side of line 2.
$$
beginalign
langle psi|U^daggercdot U|psi rangle &= langle psi | barlambda lambda |psirangle \
langle psi |psi rangle &= |lambda |^2 langle psi | |psi rangle \
1 &= |lambda|^2
endalign
$$
edited 5 hours ago
AHusain
715127
answered 8 hours ago
user1271772
4,552230
1
I get every step except for how $langle psi | U cdot U | psi rangle = langle psi | psi rangle$; doesn't that only work if the matrix is hermitian as well as unitary?
– ahelwer
7 hours ago
1
I'm also not sure how we get from $langle psi | lambda lambda | psi rangle$ to $| lambda | ^2 langle psi || psi rangle$, isn't $| lambda | ^2$ the same as $lambda cdot lambda ^*$? Where did the conjugate come from?
– ahelwer
7 hours ago
Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it!
– user1271772
5 hours ago
Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain!
– user1271772
5 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Good question. The answer turns out to be Yes.
Start with the definition of eigenvalues and eigenvectors:
$$
beginalign
U|psirangle &= lambda |psirangle\
langlepsi|U^dagger &= langle psi| barlambda.
endalign
$$
Left multiply each side of line 1 by the corresponding side of line 2.
$$
beginalign
langle psi|U^daggercdot U|psi rangle &= langle psi | barlambda lambda |psirangle \
langle psi |psi rangle &= |lambda |^2 langle psi | |psi rangle \
1 &= |lambda|^2
endalign
$$
edited 5 hours ago
AHusain
715127
answered 8 hours ago
user1271772
4,552230
Good question. The answer turns out to be Yes.
Start with the definition of eigenvalues and eigenvectors:
$$
beginalign
U|psirangle &= lambda |psirangle\
langlepsi|U^dagger &= langle psi| barlambda.
endalign
$$
Left multiply each side of line 1 by the corresponding side of line 2.
$$
beginalign
langle psi|U^daggercdot U|psi rangle &= langle psi | barlambda lambda |psirangle \
langle psi |psi rangle &= |lambda |^2 langle psi | |psi rangle \
1 &= |lambda|^2
endalign
$$
edited 5 hours ago
AHusain
715127
answered 8 hours ago
user1271772
4,552230
edited 5 hours ago
AHusain
715127
edited 5 hours ago
AHusain
715127
edited 5 hours ago
AHusain
715127
715127
answered 8 hours ago
user1271772
4,552230
answered 8 hours ago
user1271772
4,552230
answered 8 hours ago
user1271772
4,552230
4,552230
1
I get every step except for how $langle psi | U cdot U | psi rangle = langle psi | psi rangle$; doesn't that only work if the matrix is hermitian as well as unitary?
– ahelwer
7 hours ago
1
I'm also not sure how we get from $langle psi | lambda lambda | psi rangle$ to $| lambda | ^2 langle psi || psi rangle$, isn't $| lambda | ^2$ the same as $lambda cdot lambda ^*$? Where did the conjugate come from?
– ahelwer
7 hours ago
Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it!
– user1271772
5 hours ago
Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain!
– user1271772
5 hours ago
add a comment |Â
1
I get every step except for how $langle psi | U cdot U | psi rangle = langle psi | psi rangle$; doesn't that only work if the matrix is hermitian as well as unitary?
– ahelwer
7 hours ago
1
I'm also not sure how we get from $langle psi | lambda lambda | psi rangle$ to $| lambda | ^2 langle psi || psi rangle$, isn't $| lambda | ^2$ the same as $lambda cdot lambda ^*$? Where did the conjugate come from?
– ahelwer
7 hours ago
Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it!
– user1271772
5 hours ago
Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain!
– user1271772
5 hours ago
1
1
I get every step except for how $langle psi | U cdot U | psi rangle = langle psi | psi rangle$; doesn't that only work if the matrix is hermitian as well as unitary?
– ahelwer
7 hours ago
I get every step except for how $langle psi | U cdot U | psi rangle = langle psi | psi rangle$; doesn't that only work if the matrix is hermitian as well as unitary?
– ahelwer
7 hours ago
1
1
I'm also not sure how we get from $langle psi | lambda lambda | psi rangle$ to $| lambda | ^2 langle psi || psi rangle$, isn't $| lambda | ^2$ the same as $lambda cdot lambda ^*$? Where did the conjugate come from?
– ahelwer
7 hours ago
I'm also not sure how we get from $langle psi | lambda lambda | psi rangle$ to $| lambda | ^2 langle psi || psi rangle$, isn't $| lambda | ^2$ the same as $lambda cdot lambda ^*$? Where did the conjugate come from?
– ahelwer
7 hours ago
Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it!
– user1271772
5 hours ago
Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it!
– user1271772
5 hours ago
Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain!
– user1271772
5 hours ago
Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain!
– user1271772
5 hours ago
add a comment |Â
up vote
1
down vote
@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation,
$$
ifracddt=H|psirangle.
$$
For a time-invariant Hamiltonian, the solution is
$$
|psi(t)rangle=e^-iHt|psi(0)rangle,
$$
where $e^-iHt$ is unitary because $e^-iHte^iHt=mathbbI$. Just stating this solution actually skips over the thing I really want to focus on.
We could expand a generic $|psirangle=sum_ia_i|irangle$, so the Schrödinger equation becomes a series of simultaneous differential equations for the $a_i$:
$$
ifracda_idt=sum_jH_ija_j.
$$
Now consider what happens to an eigenvector $|lambdarangle=sum_ib_i|irangle$ of $H$:
$$
sum_jH_ijb_i^star=lambda b_i^star.
$$
We can take linear combinations of the $a_i$:
$$
ifracdsum_ib_i^star a_idt=sum_ijb_i^star H_ija_j=lambdasum_jb_j^star a_j.
$$
Hence, we see that the component $x=sum_jb_j^star a_j$ simply satisfies
$$
ifracdxdt=lambda x,
$$
so $x(t)=e^-ilambda tx(0)$. In other words, a state initially created as an eigenvector $|lambdarangle$ stays in that state and just acquires a phase over time $e^-ilambda t|lambdarangle$. Hence, eigenvectors of $H$ are also eigenvectors of the unitary $U$, with eigenvalues $e^-ilambda t$, and these have modulus 1.
answered 2 hours ago
DaftWullie
8,7171331
add a comment |Â
up vote
1
down vote
@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation,
$$
ifracddt=H|psirangle.
$$
For a time-invariant Hamiltonian, the solution is
$$
|psi(t)rangle=e^-iHt|psi(0)rangle,
$$
where $e^-iHt$ is unitary because $e^-iHte^iHt=mathbbI$. Just stating this solution actually skips over the thing I really want to focus on.
We could expand a generic $|psirangle=sum_ia_i|irangle$, so the Schrödinger equation becomes a series of simultaneous differential equations for the $a_i$:
$$
ifracda_idt=sum_jH_ija_j.
$$
Now consider what happens to an eigenvector $|lambdarangle=sum_ib_i|irangle$ of $H$:
$$
sum_jH_ijb_i^star=lambda b_i^star.
$$
We can take linear combinations of the $a_i$:
$$
ifracdsum_ib_i^star a_idt=sum_ijb_i^star H_ija_j=lambdasum_jb_j^star a_j.
$$
Hence, we see that the component $x=sum_jb_j^star a_j$ simply satisfies
$$
ifracdxdt=lambda x,
$$
so $x(t)=e^-ilambda tx(0)$. In other words, a state initially created as an eigenvector $|lambdarangle$ stays in that state and just acquires a phase over time $e^-ilambda t|lambdarangle$. Hence, eigenvectors of $H$ are also eigenvectors of the unitary $U$, with eigenvalues $e^-ilambda t$, and these have modulus 1.
answered 2 hours ago
DaftWullie
8,7171331
add a comment |Â
up vote
1
down vote
up vote
1
down vote
@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation,
$$
ifracddt=H|psirangle.
$$
For a time-invariant Hamiltonian, the solution is
$$
|psi(t)rangle=e^-iHt|psi(0)rangle,
$$
where $e^-iHt$ is unitary because $e^-iHte^iHt=mathbbI$. Just stating this solution actually skips over the thing I really want to focus on.
We could expand a generic $|psirangle=sum_ia_i|irangle$, so the Schrödinger equation becomes a series of simultaneous differential equations for the $a_i$:
$$
ifracda_idt=sum_jH_ija_j.
$$
Now consider what happens to an eigenvector $|lambdarangle=sum_ib_i|irangle$ of $H$:
$$
sum_jH_ijb_i^star=lambda b_i^star.
$$
We can take linear combinations of the $a_i$:
$$
ifracdsum_ib_i^star a_idt=sum_ijb_i^star H_ija_j=lambdasum_jb_j^star a_j.
$$
Hence, we see that the component $x=sum_jb_j^star a_j$ simply satisfies
$$
ifracdxdt=lambda x,
$$
so $x(t)=e^-ilambda tx(0)$. In other words, a state initially created as an eigenvector $|lambdarangle$ stays in that state and just acquires a phase over time $e^-ilambda t|lambdarangle$. Hence, eigenvectors of $H$ are also eigenvectors of the unitary $U$, with eigenvalues $e^-ilambda t$, and these have modulus 1.
answered 2 hours ago
DaftWullie
8,7171331
@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation,
$$
ifracddt=H|psirangle.
$$
For a time-invariant Hamiltonian, the solution is
$$
|psi(t)rangle=e^-iHt|psi(0)rangle,
$$
where $e^-iHt$ is unitary because $e^-iHte^iHt=mathbbI$. Just stating this solution actually skips over the thing I really want to focus on.
We could expand a generic $|psirangle=sum_ia_i|irangle$, so the Schrödinger equation becomes a series of simultaneous differential equations for the $a_i$:
$$
ifracda_idt=sum_jH_ija_j.
$$
Now consider what happens to an eigenvector $|lambdarangle=sum_ib_i|irangle$ of $H$:
$$
sum_jH_ijb_i^star=lambda b_i^star.
$$
We can take linear combinations of the $a_i$:
$$
ifracdsum_ib_i^star a_idt=sum_ijb_i^star H_ija_j=lambdasum_jb_j^star a_j.
$$
Hence, we see that the component $x=sum_jb_j^star a_j$ simply satisfies
$$
ifracdxdt=lambda x,
$$
so $x(t)=e^-ilambda tx(0)$. In other words, a state initially created as an eigenvector $|lambdarangle$ stays in that state and just acquires a phase over time $e^-ilambda t|lambdarangle$. Hence, eigenvectors of $H$ are also eigenvectors of the unitary $U$, with eigenvalues $e^-ilambda t$, and these have modulus 1.
answered 2 hours ago
DaftWullie
8,7171331
answered 2 hours ago
DaftWullie
8,7171331
answered 2 hours ago
DaftWullie
8,7171331
answered 2 hours ago
DaftWullie
8,7171331
8,7171331
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f4380%2fare-the-squared-absolute-values-of-the-eigenvalues-of-a-unitary-matrix-always-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Note that your question doesn't depend on whether the eigenvector is normalised. If you have a longer or shorter eigenvector, then that longer or shorter eigenvector has its norm changed by the same scalar factor as if it were normalised.
– Niel de Beaudrap
1 hour ago