When can one find a canonically conjugate operator?
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Suppose one is given a self-adjoint operator $A$ acting on an infinite dimensional separable Hilbert space $mathcalH$. Under what conditions can one find an operator $B$ such that $[A,B] = i$? And if this is possible, how does one construct $B$?
For simplicity, assume that $H simeq L_2(mathbbR)$ and that $A = A(X,P)$ is a function of the position $X$ and momentum operator $P$.
quantum-mechanics operators hilbert-space canonical-conjugation
asked 1 hour ago
e4alex
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up vote
3
down vote
favorite
Suppose one is given a self-adjoint operator $A$ acting on an infinite dimensional separable Hilbert space $mathcalH$. Under what conditions can one find an operator $B$ such that $[A,B] = i$? And if this is possible, how does one construct $B$?
For simplicity, assume that $H simeq L_2(mathbbR)$ and that $A = A(X,P)$ is a function of the position $X$ and momentum operator $P$.
quantum-mechanics operators hilbert-space canonical-conjugation
asked 1 hour ago
e4alex
283
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose one is given a self-adjoint operator $A$ acting on an infinite dimensional separable Hilbert space $mathcalH$. Under what conditions can one find an operator $B$ such that $[A,B] = i$? And if this is possible, how does one construct $B$?
For simplicity, assume that $H simeq L_2(mathbbR)$ and that $A = A(X,P)$ is a function of the position $X$ and momentum operator $P$.
quantum-mechanics operators hilbert-space canonical-conjugation
asked 1 hour ago
e4alex
283
Suppose one is given a self-adjoint operator $A$ acting on an infinite dimensional separable Hilbert space $mathcalH$. Under what conditions can one find an operator $B$ such that $[A,B] = i$? And if this is possible, how does one construct $B$?
For simplicity, assume that $H simeq L_2(mathbbR)$ and that $A = A(X,P)$ is a function of the position $X$ and momentum operator $P$.
quantum-mechanics operators hilbert-space canonical-conjugation
quantum-mechanics operators hilbert-space canonical-conjugation
asked 1 hour ago
e4alex
283
asked 1 hour ago
e4alex
283
asked 1 hour ago
e4alex
283
asked 1 hour ago
e4alex
283
asked 1 hour ago
e4alex
283
283
add a comment |Â
add a comment |Â
1 Answer
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There is a conjugate operator if and only if there is a unitary operator $U$ such that $UAU^dagger = X$. This is a consequence of the Stone-von Neumann theorem - all representations of the canonical commutation relations are unitarily equivalent.
answered 1 hour ago
ACuriousMind♦
68.9k17118296
Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$?
– e4alex
1 hour ago
@e4alex I've appended the answer.
– ACuriousMind♦
1 hour ago
Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $mathbbR$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $mathbbR$.
– e4alex
37 mins ago
However, consider A = P^2, the spectrum of which coincides with the positive real line $mathbbR_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -frac14 left(frac1P X + X frac1P right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed?
– e4alex
12 mins ago
@e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators...
– ACuriousMind♦
5 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
There is a conjugate operator if and only if there is a unitary operator $U$ such that $UAU^dagger = X$. This is a consequence of the Stone-von Neumann theorem - all representations of the canonical commutation relations are unitarily equivalent.
answered 1 hour ago
ACuriousMind♦
68.9k17118296
Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$?
– e4alex
1 hour ago
@e4alex I've appended the answer.
– ACuriousMind♦
1 hour ago
Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $mathbbR$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $mathbbR$.
– e4alex
37 mins ago
However, consider A = P^2, the spectrum of which coincides with the positive real line $mathbbR_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -frac14 left(frac1P X + X frac1P right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed?
– e4alex
12 mins ago
@e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators...
– ACuriousMind♦
5 mins ago
add a comment |Â
up vote
5
down vote
There is a conjugate operator if and only if there is a unitary operator $U$ such that $UAU^dagger = X$. This is a consequence of the Stone-von Neumann theorem - all representations of the canonical commutation relations are unitarily equivalent.
answered 1 hour ago
ACuriousMind♦
68.9k17118296
Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$?
– e4alex
1 hour ago
@e4alex I've appended the answer.
– ACuriousMind♦
1 hour ago
Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $mathbbR$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $mathbbR$.
– e4alex
37 mins ago
However, consider A = P^2, the spectrum of which coincides with the positive real line $mathbbR_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -frac14 left(frac1P X + X frac1P right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed?
– e4alex
12 mins ago
@e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators...
– ACuriousMind♦
5 mins ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
There is a conjugate operator if and only if there is a unitary operator $U$ such that $UAU^dagger = X$. This is a consequence of the Stone-von Neumann theorem - all representations of the canonical commutation relations are unitarily equivalent.
answered 1 hour ago
ACuriousMind♦
68.9k17118296
There is a conjugate operator if and only if there is a unitary operator $U$ such that $UAU^dagger = X$. This is a consequence of the Stone-von Neumann theorem - all representations of the canonical commutation relations are unitarily equivalent.
answered 1 hour ago
ACuriousMind♦
68.9k17118296
edited 5 mins ago
answered 1 hour ago
ACuriousMind♦
68.9k17118296
answered 1 hour ago
ACuriousMind♦
68.9k17118296
answered 1 hour ago
ACuriousMind♦
68.9k17118296
68.9k17118296
Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$?
– e4alex
1 hour ago
@e4alex I've appended the answer.
– ACuriousMind♦
1 hour ago
Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $mathbbR$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $mathbbR$.
– e4alex
37 mins ago
However, consider A = P^2, the spectrum of which coincides with the positive real line $mathbbR_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -frac14 left(frac1P X + X frac1P right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed?
– e4alex
12 mins ago
@e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators...
– ACuriousMind♦
5 mins ago
add a comment |Â
Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$?
– e4alex
1 hour ago
@e4alex I've appended the answer.
– ACuriousMind♦
1 hour ago
Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $mathbbR$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $mathbbR$.
– e4alex
37 mins ago
However, consider A = P^2, the spectrum of which coincides with the positive real line $mathbbR_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -frac14 left(frac1P X + X frac1P right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed?
– e4alex
12 mins ago
@e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators...
– ACuriousMind♦
5 mins ago
Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$?
– e4alex
1 hour ago
Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$?
– e4alex
1 hour ago
@e4alex I've appended the answer.
– ACuriousMind♦
1 hour ago
@e4alex I've appended the answer.
– ACuriousMind♦
1 hour ago
Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $mathbbR$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $mathbbR$.
– e4alex
37 mins ago
Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $mathbbR$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $mathbbR$.
– e4alex
37 mins ago
However, consider A = P^2, the spectrum of which coincides with the positive real line $mathbbR_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -frac14 left(frac1P X + X frac1P right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed?
– e4alex
12 mins ago
However, consider A = P^2, the spectrum of which coincides with the positive real line $mathbbR_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -frac14 left(frac1P X + X frac1P right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed?
– e4alex
12 mins ago
@e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators...
– ACuriousMind♦
5 mins ago
@e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators...
– ACuriousMind♦
5 mins ago
add a comment |Â
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