How come there are Schrödinger Picture operators with explicit time dependence?












5












$begingroup$


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







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$endgroup$












  • $begingroup$
    Which other related question?
    $endgroup$
    – Qmechanic♦
    Nov 18 '18 at 12:26










  • $begingroup$
    @Qmechanic this one physics.stackexchange.com/q/351020
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:27


















5












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which other related question?
    $endgroup$
    – Qmechanic♦
    Nov 18 '18 at 12:26










  • $begingroup$
    @Qmechanic this one physics.stackexchange.com/q/351020
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:27
















5












5








5





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 18 '18 at 5:52









Qmechanic♦

104k121871192




104k121871192










asked Nov 18 '18 at 3:15









João Pedro GomideJoão Pedro Gomide

285




285












  • $begingroup$
    Which other related question?
    $endgroup$
    – Qmechanic♦
    Nov 18 '18 at 12:26










  • $begingroup$
    @Qmechanic this one physics.stackexchange.com/q/351020
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:27




















  • $begingroup$
    Which other related question?
    $endgroup$
    – Qmechanic♦
    Nov 18 '18 at 12:26










  • $begingroup$
    @Qmechanic this one physics.stackexchange.com/q/351020
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:27


















$begingroup$
Which other related question?
$endgroup$
– Qmechanic♦
Nov 18 '18 at 12:26




$begingroup$
Which other related question?
$endgroup$
– Qmechanic♦
Nov 18 '18 at 12:26












$begingroup$
@Qmechanic this one physics.stackexchange.com/q/351020
$endgroup$
– João Pedro Gomide
Nov 18 '18 at 13:27






$begingroup$
@Qmechanic this one physics.stackexchange.com/q/351020
$endgroup$
– João Pedro Gomide
Nov 18 '18 at 13:27












1 Answer
1






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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?
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 3:40












  • $begingroup$
    @JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:37






  • 1




    $begingroup$
    @JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:51






  • 2




    $begingroup$
    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.
    $endgroup$
    – Andrew Steane
    Nov 18 '18 at 12:28






  • 1




    $begingroup$
    @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!
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:31











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$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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?
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 3:40












  • $begingroup$
    @JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:37






  • 1




    $begingroup$
    @JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:51






  • 2




    $begingroup$
    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.
    $endgroup$
    – Andrew Steane
    Nov 18 '18 at 12:28






  • 1




    $begingroup$
    @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!
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:31
















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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?
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 3:40












  • $begingroup$
    @JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:37






  • 1




    $begingroup$
    @JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:51






  • 2




    $begingroup$
    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.
    $endgroup$
    – Andrew Steane
    Nov 18 '18 at 12:28






  • 1




    $begingroup$
    @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!
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:31














5












5








5





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 18 '18 at 12:46

























answered Nov 18 '18 at 3:30









Aaron StevensAaron Stevens

10.2k31742




10.2k31742








  • 1




    $begingroup$
    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?
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 3:40












  • $begingroup$
    @JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:37






  • 1




    $begingroup$
    @JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:51






  • 2




    $begingroup$
    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.
    $endgroup$
    – Andrew Steane
    Nov 18 '18 at 12:28






  • 1




    $begingroup$
    @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!
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:31














  • 1




    $begingroup$
    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?
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 3:40












  • $begingroup$
    @JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:37






  • 1




    $begingroup$
    @JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
    $endgroup$
    – Aaron Stevens
    Nov 18 '18 at 4:51






  • 2




    $begingroup$
    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.
    $endgroup$
    – Andrew Steane
    Nov 18 '18 at 12:28






  • 1




    $begingroup$
    @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!
    $endgroup$
    – João Pedro Gomide
    Nov 18 '18 at 13:31








1




1




$begingroup$
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?
$endgroup$
– João Pedro Gomide
Nov 18 '18 at 3:40






$begingroup$
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?
$endgroup$
– João Pedro Gomide
Nov 18 '18 at 3:40














$begingroup$
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
$endgroup$
– Aaron Stevens
Nov 18 '18 at 4:37




$begingroup$
@JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are?
$endgroup$
– Aaron Stevens
Nov 18 '18 at 4:37




1




1




$begingroup$
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
$endgroup$
– Aaron Stevens
Nov 18 '18 at 4:51




$begingroup$
@JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution?
$endgroup$
– Aaron Stevens
Nov 18 '18 at 4:51




2




2




$begingroup$
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.
$endgroup$
– Andrew Steane
Nov 18 '18 at 12:28




$begingroup$
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.
$endgroup$
– Andrew Steane
Nov 18 '18 at 12:28




1




1




$begingroup$
@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!
$endgroup$
– João Pedro Gomide
Nov 18 '18 at 13:31




$begingroup$
@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!
$endgroup$
– João Pedro Gomide
Nov 18 '18 at 13:31


















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