Prove that if the expectation value of an operator in any state is 1, the operator is Identity












3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    3 hours ago








  • 1




    $begingroup$
    You say the $alpha$'s can be anything but surely they need to be normalised?
    $endgroup$
    – jacob1729
    1 hour ago










  • $begingroup$
    @jacob1729, true, they need to be normalised
    $endgroup$
    – Mahathi Vempati
    1 hour ago










  • $begingroup$
    @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
    41 mins ago


















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    3 hours ago








  • 1




    $begingroup$
    You say the $alpha$'s can be anything but surely they need to be normalised?
    $endgroup$
    – jacob1729
    1 hour ago










  • $begingroup$
    @jacob1729, true, they need to be normalised
    $endgroup$
    – Mahathi Vempati
    1 hour ago










  • $begingroup$
    @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
    41 mins ago
















3












3








3





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Qmechanic♦

103k121851174




103k121851174










asked 4 hours ago









Mahathi VempatiMahathi Vempati

2961520




2961520








  • 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
    3 hours ago








  • 1




    $begingroup$
    You say the $alpha$'s can be anything but surely they need to be normalised?
    $endgroup$
    – jacob1729
    1 hour ago










  • $begingroup$
    @jacob1729, true, they need to be normalised
    $endgroup$
    – Mahathi Vempati
    1 hour ago










  • $begingroup$
    @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
    41 mins ago
















  • 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
    3 hours ago








  • 1




    $begingroup$
    You say the $alpha$'s can be anything but surely they need to be normalised?
    $endgroup$
    – jacob1729
    1 hour ago










  • $begingroup$
    @jacob1729, true, they need to be normalised
    $endgroup$
    – Mahathi Vempati
    1 hour ago










  • $begingroup$
    @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
    41 mins ago










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
3 hours ago






$begingroup$
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
3 hours ago






1




1




$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
1 hour ago




$begingroup$
You say the $alpha$'s can be anything but surely they need to be normalised?
$endgroup$
– jacob1729
1 hour ago












$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
1 hour ago




$begingroup$
@jacob1729, true, they need to be normalised
$endgroup$
– Mahathi Vempati
1 hour ago












$begingroup$
@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
41 mins ago






$begingroup$
@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
41 mins ago












1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $langle alphavert Avertalpharangle=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Is it possible to continue my line of proof, though?
    $endgroup$
    – Mahathi Vempati
    3 hours ago











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









4












$begingroup$

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 $langle alphavert Avertalpharangle=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Is it possible to continue my line of proof, though?
    $endgroup$
    – Mahathi Vempati
    3 hours ago
















4












$begingroup$

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 $langle alphavert Avertalpharangle=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Is it possible to continue my line of proof, though?
    $endgroup$
    – Mahathi Vempati
    3 hours ago














4












4








4





$begingroup$

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 $langle alphavert Avertalpharangle=1$.






share|cite|improve this answer









$endgroup$



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 $langle alphavert Avertalpharangle=1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









loeweloewe

350113




350113












  • $begingroup$
    Thank you! Is it possible to continue my line of proof, though?
    $endgroup$
    – Mahathi Vempati
    3 hours ago


















  • $begingroup$
    Thank you! Is it possible to continue my line of proof, though?
    $endgroup$
    – Mahathi Vempati
    3 hours ago
















$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
3 hours ago




$begingroup$
Thank you! Is it possible to continue my line of proof, though?
$endgroup$
– Mahathi Vempati
3 hours ago


















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