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What's the role of the tensor product in quantum mechanics?
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There are several ways to define the tensor product. So there are multiple ways to look at it. I've seen it being used as an object to calculate metric, and I also seen in ring module theories.
I'm studying a bit of quantum comptuing and I've seen qubits being represented as tensor products. I don't understand why.
It seems that tensor products make a good framework for quantum mechanics, but to represent what?
linear-algebra tensor-calculus quantum-mechanics
asked Jul 16 at 0:11
Guerlando OCs
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migrated from math.stackexchange.com Jul 16 at 22:24
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up vote
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There are several ways to define the tensor product. So there are multiple ways to look at it. I've seen it being used as an object to calculate metric, and I also seen in ring module theories.
I'm studying a bit of quantum comptuing and I've seen qubits being represented as tensor products. I don't understand why.
It seems that tensor products make a good framework for quantum mechanics, but to represent what?
linear-algebra tensor-calculus quantum-mechanics
asked Jul 16 at 0:11
Guerlando OCs
1443
migrated from math.stackexchange.com Jul 16 at 22:24
This question came from our site for people studying math at any level and professionals in related fields.
2
In a nutshell: just as a Hilbert space is used for the "state space" of a system, so is the tensor product of Hilbert spaces used for the "state space" of two systems combined
– Omnomnomnom
Jul 16 at 0:58
1
My guess is that since a Hilbert space is in particular a vector space, (multi)linear maps just naturally appear, since they preserve the linear structure. And tensor products are (multi)linear maps.
– Jackozee Hakkiuz
Jul 16 at 6:34
"I've seen qubits being represented as tensor products." What makes a notation useful? I suggest you narrow your Question to the field with which you are at least a "bit" familiar, rather than throwing it open to all of quantum mechanics.
– hardmath
Jul 16 at 22:23
add a comment |Â
up vote
4
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up vote
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down vote
favorite
There are several ways to define the tensor product. So there are multiple ways to look at it. I've seen it being used as an object to calculate metric, and I also seen in ring module theories.
I'm studying a bit of quantum comptuing and I've seen qubits being represented as tensor products. I don't understand why.
It seems that tensor products make a good framework for quantum mechanics, but to represent what?
linear-algebra tensor-calculus quantum-mechanics
asked Jul 16 at 0:11
Guerlando OCs
1443
There are several ways to define the tensor product. So there are multiple ways to look at it. I've seen it being used as an object to calculate metric, and I also seen in ring module theories.
I'm studying a bit of quantum comptuing and I've seen qubits being represented as tensor products. I don't understand why.
It seems that tensor products make a good framework for quantum mechanics, but to represent what?
linear-algebra tensor-calculus quantum-mechanics
asked Jul 16 at 0:11
Guerlando OCs
1443
asked Jul 16 at 0:11
Guerlando OCs
1443
asked Jul 16 at 0:11
Guerlando OCs
1443
asked Jul 16 at 0:11
Guerlando OCs
1443
1443
migrated from math.stackexchange.com Jul 16 at 22:24
This question came from our site for people studying math at any level and professionals in related fields.
migrated from math.stackexchange.com Jul 16 at 22:24
This question came from our site for people studying math at any level and professionals in related fields.
2
In a nutshell: just as a Hilbert space is used for the "state space" of a system, so is the tensor product of Hilbert spaces used for the "state space" of two systems combined
– Omnomnomnom
Jul 16 at 0:58
1
My guess is that since a Hilbert space is in particular a vector space, (multi)linear maps just naturally appear, since they preserve the linear structure. And tensor products are (multi)linear maps.
– Jackozee Hakkiuz
Jul 16 at 6:34
"I've seen qubits being represented as tensor products." What makes a notation useful? I suggest you narrow your Question to the field with which you are at least a "bit" familiar, rather than throwing it open to all of quantum mechanics.
– hardmath
Jul 16 at 22:23
add a comment |Â
2
In a nutshell: just as a Hilbert space is used for the "state space" of a system, so is the tensor product of Hilbert spaces used for the "state space" of two systems combined
– Omnomnomnom
Jul 16 at 0:58
1
My guess is that since a Hilbert space is in particular a vector space, (multi)linear maps just naturally appear, since they preserve the linear structure. And tensor products are (multi)linear maps.
– Jackozee Hakkiuz
Jul 16 at 6:34
"I've seen qubits being represented as tensor products." What makes a notation useful? I suggest you narrow your Question to the field with which you are at least a "bit" familiar, rather than throwing it open to all of quantum mechanics.
– hardmath
Jul 16 at 22:23
2
2
In a nutshell: just as a Hilbert space is used for the "state space" of a system, so is the tensor product of Hilbert spaces used for the "state space" of two systems combined
– Omnomnomnom
Jul 16 at 0:58
In a nutshell: just as a Hilbert space is used for the "state space" of a system, so is the tensor product of Hilbert spaces used for the "state space" of two systems combined
– Omnomnomnom
Jul 16 at 0:58
1
1
My guess is that since a Hilbert space is in particular a vector space, (multi)linear maps just naturally appear, since they preserve the linear structure. And tensor products are (multi)linear maps.
– Jackozee Hakkiuz
Jul 16 at 6:34
My guess is that since a Hilbert space is in particular a vector space, (multi)linear maps just naturally appear, since they preserve the linear structure. And tensor products are (multi)linear maps.
– Jackozee Hakkiuz
Jul 16 at 6:34
"I've seen qubits being represented as tensor products." What makes a notation useful? I suggest you narrow your Question to the field with which you are at least a "bit" familiar, rather than throwing it open to all of quantum mechanics.
– hardmath
Jul 16 at 22:23
"I've seen qubits being represented as tensor products." What makes a notation useful? I suggest you narrow your Question to the field with which you are at least a "bit" familiar, rather than throwing it open to all of quantum mechanics.
– hardmath
Jul 16 at 22:23
add a comment |Â
1 Answer
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I like Nielsen and Chuang's approach to this question (from "Quantum Computation and Quantum Information"). Ultimately, the connection between quantum mechanics and linear algebra requires some set of "postulates". The postulates chosen by Nielson and Chuang which address your concerns are as follows:
Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system's state space.
I presume that this is a fact you're already more or less comfortable with. Intuitively, I think about it as follows: the number of dimensions in a state space corresponds to the number of "mutually exclusive" configurations. For instance, a photon propogating at a fixed frequency along a fixed axis has a two-dimensional state space of polarizations. Light which is vertically polarized has a $0$ percent chance of being horizontally polarized and vice versa. Each possible polarization can be characterized as a complex linear combination of these two states.
Regarding composite systems, we have the following.
Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $|psi_irangle$, then the joint state of the total system is $|psi_1rangle ⊗ |psi_2rangle ⊗ · · · ⊗ |psi_nrangle$.
I'll paraphrase their heuristic justification, focusing on the case of two systems. Suppose that system $mathcal A$ has a state space spanned by basis $psi_1,dots,psi_n$ and that system $mathcal B$ has a state space spanned by $phi_1,dots,phi_m$. We can take this to mean that system $mathcal A$ has $n$ mutually exclusive configurations and system $mathcal B$ has $m$ such configurations.
If $mathcal A$ is prepared in state $psi_i$ and $mathcal B$ is prepared in state $phi_j$, then we should expect that there is a $0$ probability that the combined system will be measured in state $phi_p,psi_q$ whenever $i neq p$ or $j neq q$. With this intuition, we can deduce that our composite system (if it can be described with some state space) has a state space with orthonormal basis
$$
psi_i phi_j rangle : 1 leq i leq n, 1 leq j leq m
$$
where the state $|psi_iphi_j rangle$ characterizes the composite system $mathcalAB$ in which system $mathcal A$ is prepared in state $psi_i$ and system $mathcal B$ is prepared in state $phi_j$.
Loosely speaking, this construction (allowing for "superpositions") leads to the characterization of the combined system with the new Hilbert space $H_mathcal A otimes H_mathcal B$.
answered Jul 16 at 16:56
Omnomnomnom
1506
add a comment |Â
1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I like Nielsen and Chuang's approach to this question (from "Quantum Computation and Quantum Information"). Ultimately, the connection between quantum mechanics and linear algebra requires some set of "postulates". The postulates chosen by Nielson and Chuang which address your concerns are as follows:
Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system's state space.
I presume that this is a fact you're already more or less comfortable with. Intuitively, I think about it as follows: the number of dimensions in a state space corresponds to the number of "mutually exclusive" configurations. For instance, a photon propogating at a fixed frequency along a fixed axis has a two-dimensional state space of polarizations. Light which is vertically polarized has a $0$ percent chance of being horizontally polarized and vice versa. Each possible polarization can be characterized as a complex linear combination of these two states.
Regarding composite systems, we have the following.
Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $|psi_irangle$, then the joint state of the total system is $|psi_1rangle ⊗ |psi_2rangle ⊗ · · · ⊗ |psi_nrangle$.
I'll paraphrase their heuristic justification, focusing on the case of two systems. Suppose that system $mathcal A$ has a state space spanned by basis $psi_1,dots,psi_n$ and that system $mathcal B$ has a state space spanned by $phi_1,dots,phi_m$. We can take this to mean that system $mathcal A$ has $n$ mutually exclusive configurations and system $mathcal B$ has $m$ such configurations.
If $mathcal A$ is prepared in state $psi_i$ and $mathcal B$ is prepared in state $phi_j$, then we should expect that there is a $0$ probability that the combined system will be measured in state $phi_p,psi_q$ whenever $i neq p$ or $j neq q$. With this intuition, we can deduce that our composite system (if it can be described with some state space) has a state space with orthonormal basis
$$
psi_i phi_j rangle : 1 leq i leq n, 1 leq j leq m
$$
where the state $|psi_iphi_j rangle$ characterizes the composite system $mathcalAB$ in which system $mathcal A$ is prepared in state $psi_i$ and system $mathcal B$ is prepared in state $phi_j$.
Loosely speaking, this construction (allowing for "superpositions") leads to the characterization of the combined system with the new Hilbert space $H_mathcal A otimes H_mathcal B$.
answered Jul 16 at 16:56
Omnomnomnom
1506
add a comment |Â
up vote
2
down vote
I like Nielsen and Chuang's approach to this question (from "Quantum Computation and Quantum Information"). Ultimately, the connection between quantum mechanics and linear algebra requires some set of "postulates". The postulates chosen by Nielson and Chuang which address your concerns are as follows:
Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system's state space.
I presume that this is a fact you're already more or less comfortable with. Intuitively, I think about it as follows: the number of dimensions in a state space corresponds to the number of "mutually exclusive" configurations. For instance, a photon propogating at a fixed frequency along a fixed axis has a two-dimensional state space of polarizations. Light which is vertically polarized has a $0$ percent chance of being horizontally polarized and vice versa. Each possible polarization can be characterized as a complex linear combination of these two states.
Regarding composite systems, we have the following.
Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $|psi_irangle$, then the joint state of the total system is $|psi_1rangle ⊗ |psi_2rangle ⊗ · · · ⊗ |psi_nrangle$.
I'll paraphrase their heuristic justification, focusing on the case of two systems. Suppose that system $mathcal A$ has a state space spanned by basis $psi_1,dots,psi_n$ and that system $mathcal B$ has a state space spanned by $phi_1,dots,phi_m$. We can take this to mean that system $mathcal A$ has $n$ mutually exclusive configurations and system $mathcal B$ has $m$ such configurations.
If $mathcal A$ is prepared in state $psi_i$ and $mathcal B$ is prepared in state $phi_j$, then we should expect that there is a $0$ probability that the combined system will be measured in state $phi_p,psi_q$ whenever $i neq p$ or $j neq q$. With this intuition, we can deduce that our composite system (if it can be described with some state space) has a state space with orthonormal basis
$$
psi_i phi_j rangle : 1 leq i leq n, 1 leq j leq m
$$
where the state $|psi_iphi_j rangle$ characterizes the composite system $mathcalAB$ in which system $mathcal A$ is prepared in state $psi_i$ and system $mathcal B$ is prepared in state $phi_j$.
Loosely speaking, this construction (allowing for "superpositions") leads to the characterization of the combined system with the new Hilbert space $H_mathcal A otimes H_mathcal B$.
answered Jul 16 at 16:56
Omnomnomnom
1506
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I like Nielsen and Chuang's approach to this question (from "Quantum Computation and Quantum Information"). Ultimately, the connection between quantum mechanics and linear algebra requires some set of "postulates". The postulates chosen by Nielson and Chuang which address your concerns are as follows:
Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system's state space.
I presume that this is a fact you're already more or less comfortable with. Intuitively, I think about it as follows: the number of dimensions in a state space corresponds to the number of "mutually exclusive" configurations. For instance, a photon propogating at a fixed frequency along a fixed axis has a two-dimensional state space of polarizations. Light which is vertically polarized has a $0$ percent chance of being horizontally polarized and vice versa. Each possible polarization can be characterized as a complex linear combination of these two states.
Regarding composite systems, we have the following.
Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $|psi_irangle$, then the joint state of the total system is $|psi_1rangle ⊗ |psi_2rangle ⊗ · · · ⊗ |psi_nrangle$.
I'll paraphrase their heuristic justification, focusing on the case of two systems. Suppose that system $mathcal A$ has a state space spanned by basis $psi_1,dots,psi_n$ and that system $mathcal B$ has a state space spanned by $phi_1,dots,phi_m$. We can take this to mean that system $mathcal A$ has $n$ mutually exclusive configurations and system $mathcal B$ has $m$ such configurations.
If $mathcal A$ is prepared in state $psi_i$ and $mathcal B$ is prepared in state $phi_j$, then we should expect that there is a $0$ probability that the combined system will be measured in state $phi_p,psi_q$ whenever $i neq p$ or $j neq q$. With this intuition, we can deduce that our composite system (if it can be described with some state space) has a state space with orthonormal basis
$$
psi_i phi_j rangle : 1 leq i leq n, 1 leq j leq m
$$
where the state $|psi_iphi_j rangle$ characterizes the composite system $mathcalAB$ in which system $mathcal A$ is prepared in state $psi_i$ and system $mathcal B$ is prepared in state $phi_j$.
Loosely speaking, this construction (allowing for "superpositions") leads to the characterization of the combined system with the new Hilbert space $H_mathcal A otimes H_mathcal B$.
answered Jul 16 at 16:56
Omnomnomnom
1506
I like Nielsen and Chuang's approach to this question (from "Quantum Computation and Quantum Information"). Ultimately, the connection between quantum mechanics and linear algebra requires some set of "postulates". The postulates chosen by Nielson and Chuang which address your concerns are as follows:
Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system's state space.
I presume that this is a fact you're already more or less comfortable with. Intuitively, I think about it as follows: the number of dimensions in a state space corresponds to the number of "mutually exclusive" configurations. For instance, a photon propogating at a fixed frequency along a fixed axis has a two-dimensional state space of polarizations. Light which is vertically polarized has a $0$ percent chance of being horizontally polarized and vice versa. Each possible polarization can be characterized as a complex linear combination of these two states.
Regarding composite systems, we have the following.
Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $|psi_irangle$, then the joint state of the total system is $|psi_1rangle ⊗ |psi_2rangle ⊗ · · · ⊗ |psi_nrangle$.
I'll paraphrase their heuristic justification, focusing on the case of two systems. Suppose that system $mathcal A$ has a state space spanned by basis $psi_1,dots,psi_n$ and that system $mathcal B$ has a state space spanned by $phi_1,dots,phi_m$. We can take this to mean that system $mathcal A$ has $n$ mutually exclusive configurations and system $mathcal B$ has $m$ such configurations.
If $mathcal A$ is prepared in state $psi_i$ and $mathcal B$ is prepared in state $phi_j$, then we should expect that there is a $0$ probability that the combined system will be measured in state $phi_p,psi_q$ whenever $i neq p$ or $j neq q$. With this intuition, we can deduce that our composite system (if it can be described with some state space) has a state space with orthonormal basis
$$
psi_i phi_j rangle : 1 leq i leq n, 1 leq j leq m
$$
where the state $|psi_iphi_j rangle$ characterizes the composite system $mathcalAB$ in which system $mathcal A$ is prepared in state $psi_i$ and system $mathcal B$ is prepared in state $phi_j$.
Loosely speaking, this construction (allowing for "superpositions") leads to the characterization of the combined system with the new Hilbert space $H_mathcal A otimes H_mathcal B$.
answered Jul 16 at 16:56
Omnomnomnom
1506
answered Jul 16 at 16:56
Omnomnomnom
1506
answered Jul 16 at 16:56
Omnomnomnom
1506
answered Jul 16 at 16:56
Omnomnomnom
1506
1506
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2
In a nutshell: just as a Hilbert space is used for the "state space" of a system, so is the tensor product of Hilbert spaces used for the "state space" of two systems combined
– Omnomnomnom
Jul 16 at 0:58
1
My guess is that since a Hilbert space is in particular a vector space, (multi)linear maps just naturally appear, since they preserve the linear structure. And tensor products are (multi)linear maps.
– Jackozee Hakkiuz
Jul 16 at 6:34
"I've seen qubits being represented as tensor products." What makes a notation useful? I suggest you narrow your Question to the field with which you are at least a "bit" familiar, rather than throwing it open to all of quantum mechanics.
– hardmath
Jul 16 at 22:23