Mixing
Prove that if the expectation value of an operator in any state is 1, the operator is Identity
$begingroup$
I want to prove that if $ langle psi | A | psi rangle = 1$ for all $ psi ,$ then $A=I .$
Let's write $A$ and $psi$ in the same basis.
$$
begin{alignat}{7}
leftlangle psi middle| A middle| psi rightrangle & ~=~ && left( sum_w alpha_w^* langle w | right) left(sum_{pq} gamma_{pq} |prangle langle q| right) left(sum_v alpha_v | v rangle right) \[5px]
& ~=~ && sum_{wvpq} alpha_w^* gamma_{pq} alpha_v langle w | p rangle langle q | v rangle \[5px]
& ~=~ && sum_{wv} alpha_w^* gamma_{wv} alpha_v
end{alignat}
$$
which is equal to $1 .$
We know that the $alpha$'s can be anything, and we need to prove that the $gamma$'s where $w = v$ are $1$ and when $w neq v$ is $0 .$
How do I proceed?
quantum-mechanics homework-and-exercises operators hilbert-space
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
$endgroup$
add a comment |
$begingroup$
I want to prove that if $ langle psi | A | psi rangle = 1$ for all $ psi ,$ then $A=I .$
Let's write $A$ and $psi$ in the same basis.
$$
begin{alignat}{7}
leftlangle psi middle| A middle| psi rightrangle & ~=~ && left( sum_w alpha_w^* langle w | right) left(sum_{pq} gamma_{pq} |prangle langle q| right) left(sum_v alpha_v | v rangle right) \[5px]
& ~=~ && sum_{wvpq} alpha_w^* gamma_{pq} alpha_v langle w | p rangle langle q | v rangle \[5px]
& ~=~ && sum_{wv} alpha_w^* gamma_{wv} alpha_v
end{alignat}
$$
which is equal to $1 .$
We know that the $alpha$'s can be anything, and we need to prove that the $gamma$'s where $w = v$ are $1$ and when $w neq v$ is $0 .$
How do I proceed?
quantum-mechanics homework-and-exercises operators hilbert-space
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
$endgroup$
1
$begingroup$
Can we prove it in the opposite direction (i.e. contra-positive)? if $A neq I$, then $langle psi | A | psi rangle neq 1$ for some $psi$
$endgroup$
– K_inverse
Jan 21 at 8:37
1
$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
Jan 21 at 10:17
$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
Jan 21 at 10:18
$begingroup$
@MahathiVempati if $alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=sum alpha^*_w delta_{wv}alpha_v$.
$endgroup$
– jacob1729
Jan 21 at 11:31
add a comment |
$begingroup$
I want to prove that if $ langle psi | A | psi rangle = 1$ for all $ psi ,$ then $A=I .$
Let's write $A$ and $psi$ in the same basis.
$$
begin{alignat}{7}
leftlangle psi middle| A middle| psi rightrangle & ~=~ && left( sum_w alpha_w^* langle w | right) left(sum_{pq} gamma_{pq} |prangle langle q| right) left(sum_v alpha_v | v rangle right) \[5px]
& ~=~ && sum_{wvpq} alpha_w^* gamma_{pq} alpha_v langle w | p rangle langle q | v rangle \[5px]
& ~=~ && sum_{wv} alpha_w^* gamma_{wv} alpha_v
end{alignat}
$$
which is equal to $1 .$
We know that the $alpha$'s can be anything, and we need to prove that the $gamma$'s where $w = v$ are $1$ and when $w neq v$ is $0 .$
How do I proceed?
quantum-mechanics homework-and-exercises operators hilbert-space
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
$endgroup$
I want to prove that if $ langle psi | A | psi rangle = 1$ for all $ psi ,$ then $A=I .$
Let's write $A$ and $psi$ in the same basis.
$$
begin{alignat}{7}
leftlangle psi middle| A middle| psi rightrangle & ~=~ && left( sum_w alpha_w^* langle w | right) left(sum_{pq} gamma_{pq} |prangle langle q| right) left(sum_v alpha_v | v rangle right) \[5px]
& ~=~ && sum_{wvpq} alpha_w^* gamma_{pq} alpha_v langle w | p rangle langle q | v rangle \[5px]
& ~=~ && sum_{wv} alpha_w^* gamma_{wv} alpha_v
end{alignat}
$$
which is equal to $1 .$
We know that the $alpha$'s can be anything, and we need to prove that the $gamma$'s where $w = v$ are $1$ and when $w neq v$ is $0 .$
How do I proceed?
quantum-mechanics homework-and-exercises operators hilbert-space
quantum-mechanics homework-and-exercises operators hilbert-space
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
edited Jan 21 at 9:38
Qmechanic♦
106k121921211
106k121921211
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
asked Jan 21 at 7:55
Mahathi VempatiMahathi Vempati
2961521
2961521
1
$begingroup$
Can we prove it in the opposite direction (i.e. contra-positive)? if $A neq I$, then $langle psi | A | psi rangle neq 1$ for some $psi$
$endgroup$
– K_inverse
Jan 21 at 8:37
1
$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
Jan 21 at 10:17
$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
Jan 21 at 10:18
$begingroup$
@MahathiVempati if $alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=sum alpha^*_w delta_{wv}alpha_v$.
$endgroup$
– jacob1729
Jan 21 at 11:31
add a comment |
1
$begingroup$
Can we prove it in the opposite direction (i.e. contra-positive)? if $A neq I$, then $langle psi | A | psi rangle neq 1$ for some $psi$
$endgroup$
– K_inverse
Jan 21 at 8:37
1
$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
Jan 21 at 10:17
$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
Jan 21 at 10:18
$begingroup$
@MahathiVempati if $alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=sum alpha^*_w delta_{wv}alpha_v$.
$endgroup$
– jacob1729
Jan 21 at 11:31
1
1
$begingroup$
Can we prove it in the opposite direction (i.e. contra-positive)? if $A neq I$, then $langle psi | A | psi rangle neq 1$ for some $psi$
$endgroup$
– K_inverse
Jan 21 at 8:37
$begingroup$
Can we prove it in the opposite direction (i.e. contra-positive)? if $A neq I$, then $langle psi | A | psi rangle neq 1$ for some $psi$
$endgroup$
– K_inverse
Jan 21 at 8:37
1
1
$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
Jan 21 at 10:17
$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
Jan 21 at 10:17
$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
Jan 21 at 10:18
$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
Jan 21 at 10:18
$begingroup$
@MahathiVempati if $alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=sum alpha^*_w delta_{wv}alpha_v$.
$endgroup$
– jacob1729
Jan 21 at 11:31
$begingroup$
@MahathiVempati if $alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=sum alpha^*_w delta_{wv}alpha_v$.
$endgroup$
– jacob1729
Jan 21 at 11:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have the freedom to expand $A$ in its eigenbasis, then
begin{equation}
A = sum_{alpha} vertalpharangle langle alphavert Avertalpharangle langle alphavert = sum_{alpha} vertalpharangle langle alphavert = 1,
end{equation}
where in the second equality we used that by assumption $langle alphavert Avertalpharangle=1$.
answered Jan 21 at 8:40
loeweloewe
420213
$endgroup$
$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
Jan 21 at 9:12
3
$begingroup$
This is slick but assumes that $A$ can be diagonalized..
$endgroup$
– lcv
Jan 21 at 19:19
$begingroup$
@MahathiVempati I sort of addressed your comment.
$endgroup$
– lcv
Jan 22 at 9:03
add a comment |
$begingroup$
Loewe's proof is very slick but assumes that $A$ can be diagonalized.
Here is a proof with no such restriction.
First, setting $B=A-I$ the statements becomes
$$
langle psi, B psi rangle=0 , forall parallel psi parallel=1 Rightarrow B = 0.
$$
By going to a basis, reasoning as the OP, one obtains
$$
sum_{i,j} psi_i^ast psi_j B_{i,j} =0
$$
from which one obtains that the hermitian part of $B$ must be zero. I.e., the statements holds if $B$ (and hence $A$) is hermitian. Loewe's proves it for the more general case for which $A$ can be diagonalized.
For full generality the trick is to realize that we also have information on the off-diagonal elements of $B$ via the polarization identity.
Define
$$
q(psi) = langle psi, Bpsi rangle
$$
then
$$
langle phi, B psi rangle = frac{1}{4} sum_{n=0}^3 i^n q(phi + i^n psi)
$$
but by assumption $q(xi)=0$ for all $xi$, hence $langle phi, B psi rangle =0$ for all $phi,psi$, hence $B=0$.
answered Jan 21 at 19:16
lcvlcv
57326
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have the freedom to expand $A$ in its eigenbasis, then
begin{equation}
A = sum_{alpha} vertalpharangle langle alphavert Avertalpharangle langle alphavert = sum_{alpha} vertalpharangle langle alphavert = 1,
end{equation}
where in the second equality we used that by assumption $langle alphavert Avertalpharangle=1$.
answered Jan 21 at 8:40
loeweloewe
420213
$endgroup$
$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
Jan 21 at 9:12
3
$begingroup$
This is slick but assumes that $A$ can be diagonalized..
$endgroup$
– lcv
Jan 21 at 19:19
$begingroup$
@MahathiVempati I sort of addressed your comment.
$endgroup$
– lcv
Jan 22 at 9:03
add a comment |
$begingroup$
We have the freedom to expand $A$ in its eigenbasis, then
begin{equation}
A = sum_{alpha} vertalpharangle langle alphavert Avertalpharangle langle alphavert = sum_{alpha} vertalpharangle langle alphavert = 1,
end{equation}
where in the second equality we used that by assumption $langle alphavert Avertalpharangle=1$.
answered Jan 21 at 8:40
loeweloewe
420213
$endgroup$
$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
Jan 21 at 9:12
3
$begingroup$
This is slick but assumes that $A$ can be diagonalized..
$endgroup$
– lcv
Jan 21 at 19:19
$begingroup$
@MahathiVempati I sort of addressed your comment.
$endgroup$
– lcv
Jan 22 at 9:03
add a comment |
$begingroup$
We have the freedom to expand $A$ in its eigenbasis, then
begin{equation}
A = sum_{alpha} vertalpharangle langle alphavert Avertalpharangle langle alphavert = sum_{alpha} vertalpharangle langle alphavert = 1,
end{equation}
where in the second equality we used that by assumption $langle alphavert Avertalpharangle=1$.
answered Jan 21 at 8:40
loeweloewe
420213
$endgroup$
We have the freedom to expand $A$ in its eigenbasis, then
begin{equation}
A = sum_{alpha} vertalpharangle langle alphavert Avertalpharangle langle alphavert = sum_{alpha} vertalpharangle langle alphavert = 1,
end{equation}
where in the second equality we used that by assumption $langle alphavert Avertalpharangle=1$.
answered Jan 21 at 8:40
loeweloewe
420213
answered Jan 21 at 8:40
loeweloewe
420213
answered Jan 21 at 8:40
loeweloewe
420213
answered Jan 21 at 8:40
loeweloewe
420213
420213
$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
Jan 21 at 9:12
3
$begingroup$
This is slick but assumes that $A$ can be diagonalized..
$endgroup$
– lcv
Jan 21 at 19:19
$begingroup$
@MahathiVempati I sort of addressed your comment.
$endgroup$
– lcv
Jan 22 at 9:03
add a comment |
$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
Jan 21 at 9:12
3
$begingroup$
This is slick but assumes that $A$ can be diagonalized..
$endgroup$
– lcv
Jan 21 at 19:19
$begingroup$
@MahathiVempati I sort of addressed your comment.
$endgroup$
– lcv
Jan 22 at 9:03
$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
Jan 21 at 9:12
$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
Jan 21 at 9:12
3
3
$begingroup$
This is slick but assumes that $A$ can be diagonalized..
$endgroup$
– lcv
Jan 21 at 19:19
$begingroup$
This is slick but assumes that $A$ can be diagonalized..
$endgroup$
– lcv
Jan 21 at 19:19
$begingroup$
@MahathiVempati I sort of addressed your comment.
$endgroup$
– lcv
Jan 22 at 9:03
$begingroup$
@MahathiVempati I sort of addressed your comment.
$endgroup$
– lcv
Jan 22 at 9:03
add a comment |
$begingroup$
Loewe's proof is very slick but assumes that $A$ can be diagonalized.
Here is a proof with no such restriction.
First, setting $B=A-I$ the statements becomes
$$
langle psi, B psi rangle=0 , forall parallel psi parallel=1 Rightarrow B = 0.
$$
By going to a basis, reasoning as the OP, one obtains
$$
sum_{i,j} psi_i^ast psi_j B_{i,j} =0
$$
from which one obtains that the hermitian part of $B$ must be zero. I.e., the statements holds if $B$ (and hence $A$) is hermitian. Loewe's proves it for the more general case for which $A$ can be diagonalized.
For full generality the trick is to realize that we also have information on the off-diagonal elements of $B$ via the polarization identity.
Define
$$
q(psi) = langle psi, Bpsi rangle
$$
then
$$
langle phi, B psi rangle = frac{1}{4} sum_{n=0}^3 i^n q(phi + i^n psi)
$$
but by assumption $q(xi)=0$ for all $xi$, hence $langle phi, B psi rangle =0$ for all $phi,psi$, hence $B=0$.
answered Jan 21 at 19:16
lcvlcv
57326
$endgroup$
add a comment |
$begingroup$
Loewe's proof is very slick but assumes that $A$ can be diagonalized.
Here is a proof with no such restriction.
First, setting $B=A-I$ the statements becomes
$$
langle psi, B psi rangle=0 , forall parallel psi parallel=1 Rightarrow B = 0.
$$
By going to a basis, reasoning as the OP, one obtains
$$
sum_{i,j} psi_i^ast psi_j B_{i,j} =0
$$
from which one obtains that the hermitian part of $B$ must be zero. I.e., the statements holds if $B$ (and hence $A$) is hermitian. Loewe's proves it for the more general case for which $A$ can be diagonalized.
For full generality the trick is to realize that we also have information on the off-diagonal elements of $B$ via the polarization identity.
Define
$$
q(psi) = langle psi, Bpsi rangle
$$
then
$$
langle phi, B psi rangle = frac{1}{4} sum_{n=0}^3 i^n q(phi + i^n psi)
$$
but by assumption $q(xi)=0$ for all $xi$, hence $langle phi, B psi rangle =0$ for all $phi,psi$, hence $B=0$.
answered Jan 21 at 19:16
lcvlcv
57326
$endgroup$
add a comment |
$begingroup$
Loewe's proof is very slick but assumes that $A$ can be diagonalized.
Here is a proof with no such restriction.
First, setting $B=A-I$ the statements becomes
$$
langle psi, B psi rangle=0 , forall parallel psi parallel=1 Rightarrow B = 0.
$$
By going to a basis, reasoning as the OP, one obtains
$$
sum_{i,j} psi_i^ast psi_j B_{i,j} =0
$$
from which one obtains that the hermitian part of $B$ must be zero. I.e., the statements holds if $B$ (and hence $A$) is hermitian. Loewe's proves it for the more general case for which $A$ can be diagonalized.
For full generality the trick is to realize that we also have information on the off-diagonal elements of $B$ via the polarization identity.
Define
$$
q(psi) = langle psi, Bpsi rangle
$$
then
$$
langle phi, B psi rangle = frac{1}{4} sum_{n=0}^3 i^n q(phi + i^n psi)
$$
but by assumption $q(xi)=0$ for all $xi$, hence $langle phi, B psi rangle =0$ for all $phi,psi$, hence $B=0$.
answered Jan 21 at 19:16
lcvlcv
57326
$endgroup$
Loewe's proof is very slick but assumes that $A$ can be diagonalized.
Here is a proof with no such restriction.
First, setting $B=A-I$ the statements becomes
$$
langle psi, B psi rangle=0 , forall parallel psi parallel=1 Rightarrow B = 0.
$$
By going to a basis, reasoning as the OP, one obtains
$$
sum_{i,j} psi_i^ast psi_j B_{i,j} =0
$$
from which one obtains that the hermitian part of $B$ must be zero. I.e., the statements holds if $B$ (and hence $A$) is hermitian. Loewe's proves it for the more general case for which $A$ can be diagonalized.
For full generality the trick is to realize that we also have information on the off-diagonal elements of $B$ via the polarization identity.
Define
$$
q(psi) = langle psi, Bpsi rangle
$$
then
$$
langle phi, B psi rangle = frac{1}{4} sum_{n=0}^3 i^n q(phi + i^n psi)
$$
but by assumption $q(xi)=0$ for all $xi$, hence $langle phi, B psi rangle =0$ for all $phi,psi$, hence $B=0$.
answered Jan 21 at 19:16
lcvlcv
57326
answered Jan 21 at 19:16
lcvlcv
57326
answered Jan 21 at 19:16
lcvlcv
57326
answered Jan 21 at 19:16
lcvlcv
57326
57326
add a comment |
add a comment |
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1
$begingroup$
Can we prove it in the opposite direction (i.e. contra-positive)? if $A neq I$, then $langle psi | A | psi rangle neq 1$ for some $psi$
$endgroup$
– K_inverse
Jan 21 at 8:37
1
$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
Jan 21 at 10:17
$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
Jan 21 at 10:18
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@MahathiVempati if $alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=sum alpha^*_w delta_{wv}alpha_v$.
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– jacob1729
Jan 21 at 11:31