How come there are Schrödinger Picture operators with explicit time dependence?
up vote
4
down vote
favorite
In the Schrödinger picture, observables are said to be time independent (see Cohen, for example) operators. However, when deriving the Heisenberg Equation of Motion $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]+ihbarBig(frac{partial}{partial t}A_S(t)Big)_H.$$ a term with an explicit time dependence of the operator in the Schrödinger picture appears. I looked at other related questions and some argued that in the S-picture, only operators that are related to observables are time-independent. Is this really the case? If so, is this equation a general description of dynamics of operators and reduces to $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]$$ if $A_S$ is an observable? Furthermore, is the existence of (explicit) time dependence equivalent to time evolution?
quantum-mechanics operators observables time-evolution
New contributor
add a comment |
up vote
4
down vote
favorite
In the Schrödinger picture, observables are said to be time independent (see Cohen, for example) operators. However, when deriving the Heisenberg Equation of Motion $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]+ihbarBig(frac{partial}{partial t}A_S(t)Big)_H.$$ a term with an explicit time dependence of the operator in the Schrödinger picture appears. I looked at other related questions and some argued that in the S-picture, only operators that are related to observables are time-independent. Is this really the case? If so, is this equation a general description of dynamics of operators and reduces to $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]$$ if $A_S$ is an observable? Furthermore, is the existence of (explicit) time dependence equivalent to time evolution?
quantum-mechanics operators observables time-evolution
New contributor
Which other related question?
– Qmechanic♦
2 days ago
@Qmechanic this one physics.stackexchange.com/q/351020
– João Pedro Gomide
2 days ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In the Schrödinger picture, observables are said to be time independent (see Cohen, for example) operators. However, when deriving the Heisenberg Equation of Motion $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]+ihbarBig(frac{partial}{partial t}A_S(t)Big)_H.$$ a term with an explicit time dependence of the operator in the Schrödinger picture appears. I looked at other related questions and some argued that in the S-picture, only operators that are related to observables are time-independent. Is this really the case? If so, is this equation a general description of dynamics of operators and reduces to $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]$$ if $A_S$ is an observable? Furthermore, is the existence of (explicit) time dependence equivalent to time evolution?
quantum-mechanics operators observables time-evolution
New contributor
In the Schrödinger picture, observables are said to be time independent (see Cohen, for example) operators. However, when deriving the Heisenberg Equation of Motion $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]+ihbarBig(frac{partial}{partial t}A_S(t)Big)_H.$$ a term with an explicit time dependence of the operator in the Schrödinger picture appears. I looked at other related questions and some argued that in the S-picture, only operators that are related to observables are time-independent. Is this really the case? If so, is this equation a general description of dynamics of operators and reduces to $$ihbarfrac{d}{dt}A_H(t)=[A_H(t),H_H(t)]$$ if $A_S$ is an observable? Furthermore, is the existence of (explicit) time dependence equivalent to time evolution?
quantum-mechanics operators observables time-evolution
quantum-mechanics operators observables time-evolution
New contributor
New contributor
edited 2 days ago
Qmechanic♦
99.5k121781113
99.5k121781113
New contributor
asked Nov 18 at 3:15
João Pedro Gomide
235
235
New contributor
New contributor
Which other related question?
– Qmechanic♦
2 days ago
@Qmechanic this one physics.stackexchange.com/q/351020
– João Pedro Gomide
2 days ago
add a comment |
Which other related question?
– Qmechanic♦
2 days ago
@Qmechanic this one physics.stackexchange.com/q/351020
– João Pedro Gomide
2 days ago
Which other related question?
– Qmechanic♦
2 days ago
Which other related question?
– Qmechanic♦
2 days ago
@Qmechanic this one physics.stackexchange.com/q/351020
– João Pedro Gomide
2 days ago
@Qmechanic this one physics.stackexchange.com/q/351020
– João Pedro Gomide
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
In the Schrodinger picture there is no time dependence of operators due to unitary transformations. Operators in the Schrodinger picture can still have a time dependence if something is physically changing$^*$. An example of this is if we have a particle in a time dependent electric field. The Hamiltonian will have time dependence due to the field actually changing, not because of a unitary time evolution (if we treat the field as external to the system). The eigenvalues (possible measurement outcomes) in this case can have a time dependence.
So, in the Schrodinger picture, unitary transformations are what cause the state vector to change over time. Operators do not evolve through time in this way. If an operator does have explicit time dependence, it is due to something physical the operator depends on that is physically changing. This time dependence does not have to be described as a unitary transformation.
$^*$ This seems to have gained some confusion here. I am not saying that unitary transformations do not have physical consequences. I am saying they do not represent physical changes themselves; they only represent changes in the probability of measuring the system to be in some state. State vectors and operators are not physical things, so unitary transformations that cause them to change are not direct physical changes. On the other hand, in my example fields are physical, directly measurable things. In the Schrodinger picture operators that depend on the field can have explicit time dependence, and so can the eigenvalues associated with those operators.
1
The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking?
– João Pedro Gomide
Nov 18 at 3:40
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
– Aaron Stevens
Nov 18 at 4:37
1
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
– Aaron Stevens
2 days ago
2
This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations.
– Andrew Steane
2 days ago
1
@AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you!
– João Pedro Gomide
2 days ago
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
In the Schrodinger picture there is no time dependence of operators due to unitary transformations. Operators in the Schrodinger picture can still have a time dependence if something is physically changing$^*$. An example of this is if we have a particle in a time dependent electric field. The Hamiltonian will have time dependence due to the field actually changing, not because of a unitary time evolution (if we treat the field as external to the system). The eigenvalues (possible measurement outcomes) in this case can have a time dependence.
So, in the Schrodinger picture, unitary transformations are what cause the state vector to change over time. Operators do not evolve through time in this way. If an operator does have explicit time dependence, it is due to something physical the operator depends on that is physically changing. This time dependence does not have to be described as a unitary transformation.
$^*$ This seems to have gained some confusion here. I am not saying that unitary transformations do not have physical consequences. I am saying they do not represent physical changes themselves; they only represent changes in the probability of measuring the system to be in some state. State vectors and operators are not physical things, so unitary transformations that cause them to change are not direct physical changes. On the other hand, in my example fields are physical, directly measurable things. In the Schrodinger picture operators that depend on the field can have explicit time dependence, and so can the eigenvalues associated with those operators.
1
The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking?
– João Pedro Gomide
Nov 18 at 3:40
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
– Aaron Stevens
Nov 18 at 4:37
1
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
– Aaron Stevens
2 days ago
2
This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations.
– Andrew Steane
2 days ago
1
@AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you!
– João Pedro Gomide
2 days ago
|
show 3 more comments
up vote
4
down vote
accepted
In the Schrodinger picture there is no time dependence of operators due to unitary transformations. Operators in the Schrodinger picture can still have a time dependence if something is physically changing$^*$. An example of this is if we have a particle in a time dependent electric field. The Hamiltonian will have time dependence due to the field actually changing, not because of a unitary time evolution (if we treat the field as external to the system). The eigenvalues (possible measurement outcomes) in this case can have a time dependence.
So, in the Schrodinger picture, unitary transformations are what cause the state vector to change over time. Operators do not evolve through time in this way. If an operator does have explicit time dependence, it is due to something physical the operator depends on that is physically changing. This time dependence does not have to be described as a unitary transformation.
$^*$ This seems to have gained some confusion here. I am not saying that unitary transformations do not have physical consequences. I am saying they do not represent physical changes themselves; they only represent changes in the probability of measuring the system to be in some state. State vectors and operators are not physical things, so unitary transformations that cause them to change are not direct physical changes. On the other hand, in my example fields are physical, directly measurable things. In the Schrodinger picture operators that depend on the field can have explicit time dependence, and so can the eigenvalues associated with those operators.
1
The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking?
– João Pedro Gomide
Nov 18 at 3:40
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
– Aaron Stevens
Nov 18 at 4:37
1
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
– Aaron Stevens
2 days ago
2
This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations.
– Andrew Steane
2 days ago
1
@AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you!
– João Pedro Gomide
2 days ago
|
show 3 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
In the Schrodinger picture there is no time dependence of operators due to unitary transformations. Operators in the Schrodinger picture can still have a time dependence if something is physically changing$^*$. An example of this is if we have a particle in a time dependent electric field. The Hamiltonian will have time dependence due to the field actually changing, not because of a unitary time evolution (if we treat the field as external to the system). The eigenvalues (possible measurement outcomes) in this case can have a time dependence.
So, in the Schrodinger picture, unitary transformations are what cause the state vector to change over time. Operators do not evolve through time in this way. If an operator does have explicit time dependence, it is due to something physical the operator depends on that is physically changing. This time dependence does not have to be described as a unitary transformation.
$^*$ This seems to have gained some confusion here. I am not saying that unitary transformations do not have physical consequences. I am saying they do not represent physical changes themselves; they only represent changes in the probability of measuring the system to be in some state. State vectors and operators are not physical things, so unitary transformations that cause them to change are not direct physical changes. On the other hand, in my example fields are physical, directly measurable things. In the Schrodinger picture operators that depend on the field can have explicit time dependence, and so can the eigenvalues associated with those operators.
In the Schrodinger picture there is no time dependence of operators due to unitary transformations. Operators in the Schrodinger picture can still have a time dependence if something is physically changing$^*$. An example of this is if we have a particle in a time dependent electric field. The Hamiltonian will have time dependence due to the field actually changing, not because of a unitary time evolution (if we treat the field as external to the system). The eigenvalues (possible measurement outcomes) in this case can have a time dependence.
So, in the Schrodinger picture, unitary transformations are what cause the state vector to change over time. Operators do not evolve through time in this way. If an operator does have explicit time dependence, it is due to something physical the operator depends on that is physically changing. This time dependence does not have to be described as a unitary transformation.
$^*$ This seems to have gained some confusion here. I am not saying that unitary transformations do not have physical consequences. I am saying they do not represent physical changes themselves; they only represent changes in the probability of measuring the system to be in some state. State vectors and operators are not physical things, so unitary transformations that cause them to change are not direct physical changes. On the other hand, in my example fields are physical, directly measurable things. In the Schrodinger picture operators that depend on the field can have explicit time dependence, and so can the eigenvalues associated with those operators.
edited 2 days ago
answered Nov 18 at 3:30
Aaron Stevens
7,16231235
7,16231235
1
The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking?
– João Pedro Gomide
Nov 18 at 3:40
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
– Aaron Stevens
Nov 18 at 4:37
1
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
– Aaron Stevens
2 days ago
2
This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations.
– Andrew Steane
2 days ago
1
@AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you!
– João Pedro Gomide
2 days ago
|
show 3 more comments
1
The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking?
– João Pedro Gomide
Nov 18 at 3:40
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
– Aaron Stevens
Nov 18 at 4:37
1
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
– Aaron Stevens
2 days ago
2
This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations.
– Andrew Steane
2 days ago
1
@AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you!
– João Pedro Gomide
2 days ago
1
1
The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking?
– João Pedro Gomide
Nov 18 at 3:40
The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking?
– João Pedro Gomide
Nov 18 at 3:40
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
– Aaron Stevens
Nov 18 at 4:37
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
– Aaron Stevens
Nov 18 at 4:37
1
1
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
– Aaron Stevens
2 days ago
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
– Aaron Stevens
2 days ago
2
2
This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations.
– Andrew Steane
2 days ago
This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations.
– Andrew Steane
2 days ago
1
1
@AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you!
– João Pedro Gomide
2 days ago
@AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you!
– João Pedro Gomide
2 days ago
|
show 3 more comments
João Pedro Gomide is a new contributor. Be nice, and check out our Code of Conduct.
João Pedro Gomide is a new contributor. Be nice, and check out our Code of Conduct.
João Pedro Gomide is a new contributor. Be nice, and check out our Code of Conduct.
João Pedro Gomide is a new contributor. Be nice, and check out our Code of Conduct.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f441656%2fhow-come-there-are-schr%25c3%25b6dinger-picture-operators-with-explicit-time-dependence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Which other related question?
– Qmechanic♦
2 days ago
@Qmechanic this one physics.stackexchange.com/q/351020
– João Pedro Gomide
2 days ago