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How do base kets satisfy Schrödinger's equation in Schrödinger picture and why don't they evolve with time?
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According to Sakurai, eigenvalue equation for an operator $A$, $A|a'rangle=a'|a'rangle$. In the Schrödinger picture, $A$ does not change, so the base kets, obtained as the solutions to this eigenvalue equation at t=0, for instance, must remain unchanged.
Since base kets do not evolve with time $|a',trangle=|a'rangle$ and is independent of t.
Schrödinger equation
$$ihbarfracpartial partial t=H|a',trangle,$$
the LHS is zero and RHS is non-zero. Why is the Schrödinger equation not satisfied?Suppose $A$ commutes with $H$ (Hamiltonian).
$A|a'rangle=a'|a'rangle$ and evolution operator is $U(t,0)=exp(-fraciHthbar)$
$$UA|a'rangle=Ua'|a'rangle$$
Since $H$ and $A$ commute, $U$ and $A$ also commute.
$$AU|a'rangle=a'U|a'rangle$$
So the eigenvalue remains same and eigenket is now $U|a'rangle$ and evolves with time, which reduces to $|a'rangle$ at t=0.
So, I can conclude that base kets evolve with time when $A$ commutes with Hamiltonian. This has an additional advantage that Schrödinger Equation is now satisfied.
As stated in the book, the base kets do not change in the Schrödinger picture. Is this statement wrong in the above case?
quantum-mechanics operators hilbert-space schroedinger-equation time-evolution
edited 1 hour ago
Qmechanic♦
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asked 2 hours ago
Asit Srivastava
467
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up vote
1
down vote
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According to Sakurai, eigenvalue equation for an operator $A$, $A|a'rangle=a'|a'rangle$. In the Schrödinger picture, $A$ does not change, so the base kets, obtained as the solutions to this eigenvalue equation at t=0, for instance, must remain unchanged.
Since base kets do not evolve with time $|a',trangle=|a'rangle$ and is independent of t.
Schrödinger equation
$$ihbarfracpartial partial t=H|a',trangle,$$
the LHS is zero and RHS is non-zero. Why is the Schrödinger equation not satisfied?Suppose $A$ commutes with $H$ (Hamiltonian).
$A|a'rangle=a'|a'rangle$ and evolution operator is $U(t,0)=exp(-fraciHthbar)$
$$UA|a'rangle=Ua'|a'rangle$$
Since $H$ and $A$ commute, $U$ and $A$ also commute.
$$AU|a'rangle=a'U|a'rangle$$
So the eigenvalue remains same and eigenket is now $U|a'rangle$ and evolves with time, which reduces to $|a'rangle$ at t=0.
So, I can conclude that base kets evolve with time when $A$ commutes with Hamiltonian. This has an additional advantage that Schrödinger Equation is now satisfied.
As stated in the book, the base kets do not change in the Schrödinger picture. Is this statement wrong in the above case?
quantum-mechanics operators hilbert-space schroedinger-equation time-evolution
edited 1 hour ago
Qmechanic♦
97.8k121641053
asked 2 hours ago
Asit Srivastava
467
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
According to Sakurai, eigenvalue equation for an operator $A$, $A|a'rangle=a'|a'rangle$. In the Schrödinger picture, $A$ does not change, so the base kets, obtained as the solutions to this eigenvalue equation at t=0, for instance, must remain unchanged.
Since base kets do not evolve with time $|a',trangle=|a'rangle$ and is independent of t.
Schrödinger equation
$$ihbarfracpartial partial t=H|a',trangle,$$
the LHS is zero and RHS is non-zero. Why is the Schrödinger equation not satisfied?Suppose $A$ commutes with $H$ (Hamiltonian).
$A|a'rangle=a'|a'rangle$ and evolution operator is $U(t,0)=exp(-fraciHthbar)$
$$UA|a'rangle=Ua'|a'rangle$$
Since $H$ and $A$ commute, $U$ and $A$ also commute.
$$AU|a'rangle=a'U|a'rangle$$
So the eigenvalue remains same and eigenket is now $U|a'rangle$ and evolves with time, which reduces to $|a'rangle$ at t=0.
So, I can conclude that base kets evolve with time when $A$ commutes with Hamiltonian. This has an additional advantage that Schrödinger Equation is now satisfied.
As stated in the book, the base kets do not change in the Schrödinger picture. Is this statement wrong in the above case?
quantum-mechanics operators hilbert-space schroedinger-equation time-evolution
edited 1 hour ago
Qmechanic♦
97.8k121641053
asked 2 hours ago
Asit Srivastava
467
According to Sakurai, eigenvalue equation for an operator $A$, $A|a'rangle=a'|a'rangle$. In the Schrödinger picture, $A$ does not change, so the base kets, obtained as the solutions to this eigenvalue equation at t=0, for instance, must remain unchanged.
Since base kets do not evolve with time $|a',trangle=|a'rangle$ and is independent of t.
Schrödinger equation
$$ihbarfracpartial partial t=H|a',trangle,$$
the LHS is zero and RHS is non-zero. Why is the Schrödinger equation not satisfied?Suppose $A$ commutes with $H$ (Hamiltonian).
$A|a'rangle=a'|a'rangle$ and evolution operator is $U(t,0)=exp(-fraciHthbar)$
$$UA|a'rangle=Ua'|a'rangle$$
Since $H$ and $A$ commute, $U$ and $A$ also commute.
$$AU|a'rangle=a'U|a'rangle$$
So the eigenvalue remains same and eigenket is now $U|a'rangle$ and evolves with time, which reduces to $|a'rangle$ at t=0.
So, I can conclude that base kets evolve with time when $A$ commutes with Hamiltonian. This has an additional advantage that Schrödinger Equation is now satisfied.
As stated in the book, the base kets do not change in the Schrödinger picture. Is this statement wrong in the above case?
quantum-mechanics operators hilbert-space schroedinger-equation time-evolution
quantum-mechanics operators hilbert-space schroedinger-equation time-evolution
edited 1 hour ago
Qmechanic♦
97.8k121641053
asked 2 hours ago
Asit Srivastava
467
edited 1 hour ago
Qmechanic♦
97.8k121641053
asked 2 hours ago
Asit Srivastava
467
edited 1 hour ago
Qmechanic♦
97.8k121641053
edited 1 hour ago
Qmechanic♦
97.8k121641053
edited 1 hour ago
Qmechanic♦
97.8k121641053
97.8k121641053
asked 2 hours ago
Asit Srivastava
467
asked 2 hours ago
Asit Srivastava
467
asked 2 hours ago
Asit Srivastava
467
467
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3 Answers
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up vote
2
down vote
Since base kets do not evolve with time $|a',trangle=|a'rangle$
This is a misreading of that statement. When we say that the energy-eigenbasis kets of a time-independent hamiltonian to not evolve with time, what we mean is that if $H|Erangle = E|Erangle$ then
$$
|Psi_E(t)rangle = e^-iEt/hbar|Erangle
$$
is a solution of the Schrödinger equation, which starts off at $|Psi_E(0)rangle = |Erangle$ and which maintains a unit inner product with its initial condition,
$$
|langle Psi(0)|Psi(t)rangle| = |langle E |Psi(t)rangle| = 1.
$$
The phase factor $e^-iEt/hbar$ is absolutely crucial for the Schrödinger equation to be satisfied. On the other hand, at any given time $t$ it acts as a global phase factor, which is irrelevant for any physical observable, which is why it is justified to say that the base ket does not 'really' evolve with time.
For your second question, your initial manipulations are correct, and they are best understood in the form
$$
A U|arangle = U A|arangle = U a|arangle = a U|arangle
$$
(where I've dropped the primes, to $A|arangle = a|arangle$, because yours didn't make any sense). In other words, if you start off in an eigenvector $|arangle$ of $A$ with eigenvalue $a$ and $A$ commutes with $H$, then the time-evolved state $U(t)|arangle$ will always be an eigenstate of $A$.
Now, if you know that $A$ is non-degenerate on that eigenspace, then that combination allows you to conclude that $|arangle$ is also an eigenstate of $H$ and the time evolution will keep $U(t)|arangle$ as a multiple of $|arangle$.
On the other hand, it is perfectly possible for $A$ to have a degenerate eigenspace there, in which case $U(t)|arangle$ can have a nontrivial time dependence. If you want an explicit example, try
$$
A = beginpmatrix 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 endpmatrix
textunder
H = beginpmatrix 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 3 endpmatrix
$$
with $|arangle = (1,0,0)$.
answered 1 hour ago
Emilio Pisanty
78.5k21187385
add a comment |Â
up vote
1
down vote
I think there is a lot of confusion in what you wrote. The Schrodinger equation is satisfied,of course not by the eigenkets themselves, but by the expansion coefficients of any state expanded on the basis of eigenkets. The eigenkets are defined as being time independent exactly because, for a general state, the time dependence of each expansion coefficient on the eigenket basis is really simple. To make this clear, for any state $|psirangle$, take a complete set of eigenkets $(|a_1rangle,|a_2rangle,|a_3rangle,...)$, then the state $|psirangle$ can be expressed as:
$$|psirangle=sum_n c_n(t)*|a_nrangle$$
So $|psirangle$ is actually $|psi(t)rangle$, and it does satisfy Schrodingers equation. So if the $|a_nrangle$ are the eigenkets of the Hamiltonian, then the time dependence of the expansion coefficients is just the usual $e^-ifrac E_nhbart$, and the expansion of the state is just:
$$|psi(t)rangle=sum_n c_n(0)*e^-ifrac E_nhbart*|a_nrangle$$
answered 1 hour ago
Hugo V
1436
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
Only kets that represent physical systems ("state vectors") satisfy the Schrodinger equation. Basis kets don't represent physical systems, but just a system of coordinates, so they don't.
Your question is analogous to asking why the coordinates of a random point in space don't satisfy Hamilton's or the Euler-Lagrange equations. There's just nothing there to time-evolve.
answered 1 hour ago
tparker
21.4k143114
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since base kets do not evolve with time $|a',trangle=|a'rangle$
This is a misreading of that statement. When we say that the energy-eigenbasis kets of a time-independent hamiltonian to not evolve with time, what we mean is that if $H|Erangle = E|Erangle$ then
$$
|Psi_E(t)rangle = e^-iEt/hbar|Erangle
$$
is a solution of the Schrödinger equation, which starts off at $|Psi_E(0)rangle = |Erangle$ and which maintains a unit inner product with its initial condition,
$$
|langle Psi(0)|Psi(t)rangle| = |langle E |Psi(t)rangle| = 1.
$$
The phase factor $e^-iEt/hbar$ is absolutely crucial for the Schrödinger equation to be satisfied. On the other hand, at any given time $t$ it acts as a global phase factor, which is irrelevant for any physical observable, which is why it is justified to say that the base ket does not 'really' evolve with time.
For your second question, your initial manipulations are correct, and they are best understood in the form
$$
A U|arangle = U A|arangle = U a|arangle = a U|arangle
$$
(where I've dropped the primes, to $A|arangle = a|arangle$, because yours didn't make any sense). In other words, if you start off in an eigenvector $|arangle$ of $A$ with eigenvalue $a$ and $A$ commutes with $H$, then the time-evolved state $U(t)|arangle$ will always be an eigenstate of $A$.
Now, if you know that $A$ is non-degenerate on that eigenspace, then that combination allows you to conclude that $|arangle$ is also an eigenstate of $H$ and the time evolution will keep $U(t)|arangle$ as a multiple of $|arangle$.
On the other hand, it is perfectly possible for $A$ to have a degenerate eigenspace there, in which case $U(t)|arangle$ can have a nontrivial time dependence. If you want an explicit example, try
$$
A = beginpmatrix 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 endpmatrix
textunder
H = beginpmatrix 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 3 endpmatrix
$$
with $|arangle = (1,0,0)$.
answered 1 hour ago
Emilio Pisanty
78.5k21187385
add a comment |Â
up vote
2
down vote
Since base kets do not evolve with time $|a',trangle=|a'rangle$
This is a misreading of that statement. When we say that the energy-eigenbasis kets of a time-independent hamiltonian to not evolve with time, what we mean is that if $H|Erangle = E|Erangle$ then
$$
|Psi_E(t)rangle = e^-iEt/hbar|Erangle
$$
is a solution of the Schrödinger equation, which starts off at $|Psi_E(0)rangle = |Erangle$ and which maintains a unit inner product with its initial condition,
$$
|langle Psi(0)|Psi(t)rangle| = |langle E |Psi(t)rangle| = 1.
$$
The phase factor $e^-iEt/hbar$ is absolutely crucial for the Schrödinger equation to be satisfied. On the other hand, at any given time $t$ it acts as a global phase factor, which is irrelevant for any physical observable, which is why it is justified to say that the base ket does not 'really' evolve with time.
For your second question, your initial manipulations are correct, and they are best understood in the form
$$
A U|arangle = U A|arangle = U a|arangle = a U|arangle
$$
(where I've dropped the primes, to $A|arangle = a|arangle$, because yours didn't make any sense). In other words, if you start off in an eigenvector $|arangle$ of $A$ with eigenvalue $a$ and $A$ commutes with $H$, then the time-evolved state $U(t)|arangle$ will always be an eigenstate of $A$.
Now, if you know that $A$ is non-degenerate on that eigenspace, then that combination allows you to conclude that $|arangle$ is also an eigenstate of $H$ and the time evolution will keep $U(t)|arangle$ as a multiple of $|arangle$.
On the other hand, it is perfectly possible for $A$ to have a degenerate eigenspace there, in which case $U(t)|arangle$ can have a nontrivial time dependence. If you want an explicit example, try
$$
A = beginpmatrix 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 endpmatrix
textunder
H = beginpmatrix 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 3 endpmatrix
$$
with $|arangle = (1,0,0)$.
answered 1 hour ago
Emilio Pisanty
78.5k21187385
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since base kets do not evolve with time $|a',trangle=|a'rangle$
This is a misreading of that statement. When we say that the energy-eigenbasis kets of a time-independent hamiltonian to not evolve with time, what we mean is that if $H|Erangle = E|Erangle$ then
$$
|Psi_E(t)rangle = e^-iEt/hbar|Erangle
$$
is a solution of the Schrödinger equation, which starts off at $|Psi_E(0)rangle = |Erangle$ and which maintains a unit inner product with its initial condition,
$$
|langle Psi(0)|Psi(t)rangle| = |langle E |Psi(t)rangle| = 1.
$$
The phase factor $e^-iEt/hbar$ is absolutely crucial for the Schrödinger equation to be satisfied. On the other hand, at any given time $t$ it acts as a global phase factor, which is irrelevant for any physical observable, which is why it is justified to say that the base ket does not 'really' evolve with time.
For your second question, your initial manipulations are correct, and they are best understood in the form
$$
A U|arangle = U A|arangle = U a|arangle = a U|arangle
$$
(where I've dropped the primes, to $A|arangle = a|arangle$, because yours didn't make any sense). In other words, if you start off in an eigenvector $|arangle$ of $A$ with eigenvalue $a$ and $A$ commutes with $H$, then the time-evolved state $U(t)|arangle$ will always be an eigenstate of $A$.
Now, if you know that $A$ is non-degenerate on that eigenspace, then that combination allows you to conclude that $|arangle$ is also an eigenstate of $H$ and the time evolution will keep $U(t)|arangle$ as a multiple of $|arangle$.
On the other hand, it is perfectly possible for $A$ to have a degenerate eigenspace there, in which case $U(t)|arangle$ can have a nontrivial time dependence. If you want an explicit example, try
$$
A = beginpmatrix 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 endpmatrix
textunder
H = beginpmatrix 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 3 endpmatrix
$$
with $|arangle = (1,0,0)$.
answered 1 hour ago
Emilio Pisanty
78.5k21187385
Since base kets do not evolve with time $|a',trangle=|a'rangle$
This is a misreading of that statement. When we say that the energy-eigenbasis kets of a time-independent hamiltonian to not evolve with time, what we mean is that if $H|Erangle = E|Erangle$ then
$$
|Psi_E(t)rangle = e^-iEt/hbar|Erangle
$$
is a solution of the Schrödinger equation, which starts off at $|Psi_E(0)rangle = |Erangle$ and which maintains a unit inner product with its initial condition,
$$
|langle Psi(0)|Psi(t)rangle| = |langle E |Psi(t)rangle| = 1.
$$
The phase factor $e^-iEt/hbar$ is absolutely crucial for the Schrödinger equation to be satisfied. On the other hand, at any given time $t$ it acts as a global phase factor, which is irrelevant for any physical observable, which is why it is justified to say that the base ket does not 'really' evolve with time.
For your second question, your initial manipulations are correct, and they are best understood in the form
$$
A U|arangle = U A|arangle = U a|arangle = a U|arangle
$$
(where I've dropped the primes, to $A|arangle = a|arangle$, because yours didn't make any sense). In other words, if you start off in an eigenvector $|arangle$ of $A$ with eigenvalue $a$ and $A$ commutes with $H$, then the time-evolved state $U(t)|arangle$ will always be an eigenstate of $A$.
Now, if you know that $A$ is non-degenerate on that eigenspace, then that combination allows you to conclude that $|arangle$ is also an eigenstate of $H$ and the time evolution will keep $U(t)|arangle$ as a multiple of $|arangle$.
On the other hand, it is perfectly possible for $A$ to have a degenerate eigenspace there, in which case $U(t)|arangle$ can have a nontrivial time dependence. If you want an explicit example, try
$$
A = beginpmatrix 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 endpmatrix
textunder
H = beginpmatrix 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 3 endpmatrix
$$
with $|arangle = (1,0,0)$.
answered 1 hour ago
Emilio Pisanty
78.5k21187385
answered 1 hour ago
Emilio Pisanty
78.5k21187385
answered 1 hour ago
Emilio Pisanty
78.5k21187385
answered 1 hour ago
Emilio Pisanty
78.5k21187385
78.5k21187385
add a comment |Â
add a comment |Â
up vote
1
down vote
I think there is a lot of confusion in what you wrote. The Schrodinger equation is satisfied,of course not by the eigenkets themselves, but by the expansion coefficients of any state expanded on the basis of eigenkets. The eigenkets are defined as being time independent exactly because, for a general state, the time dependence of each expansion coefficient on the eigenket basis is really simple. To make this clear, for any state $|psirangle$, take a complete set of eigenkets $(|a_1rangle,|a_2rangle,|a_3rangle,...)$, then the state $|psirangle$ can be expressed as:
$$|psirangle=sum_n c_n(t)*|a_nrangle$$
So $|psirangle$ is actually $|psi(t)rangle$, and it does satisfy Schrodingers equation. So if the $|a_nrangle$ are the eigenkets of the Hamiltonian, then the time dependence of the expansion coefficients is just the usual $e^-ifrac E_nhbart$, and the expansion of the state is just:
$$|psi(t)rangle=sum_n c_n(0)*e^-ifrac E_nhbart*|a_nrangle$$
answered 1 hour ago
Hugo V
1436
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
I think there is a lot of confusion in what you wrote. The Schrodinger equation is satisfied,of course not by the eigenkets themselves, but by the expansion coefficients of any state expanded on the basis of eigenkets. The eigenkets are defined as being time independent exactly because, for a general state, the time dependence of each expansion coefficient on the eigenket basis is really simple. To make this clear, for any state $|psirangle$, take a complete set of eigenkets $(|a_1rangle,|a_2rangle,|a_3rangle,...)$, then the state $|psirangle$ can be expressed as:
$$|psirangle=sum_n c_n(t)*|a_nrangle$$
So $|psirangle$ is actually $|psi(t)rangle$, and it does satisfy Schrodingers equation. So if the $|a_nrangle$ are the eigenkets of the Hamiltonian, then the time dependence of the expansion coefficients is just the usual $e^-ifrac E_nhbart$, and the expansion of the state is just:
$$|psi(t)rangle=sum_n c_n(0)*e^-ifrac E_nhbart*|a_nrangle$$
answered 1 hour ago
Hugo V
1436
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think there is a lot of confusion in what you wrote. The Schrodinger equation is satisfied,of course not by the eigenkets themselves, but by the expansion coefficients of any state expanded on the basis of eigenkets. The eigenkets are defined as being time independent exactly because, for a general state, the time dependence of each expansion coefficient on the eigenket basis is really simple. To make this clear, for any state $|psirangle$, take a complete set of eigenkets $(|a_1rangle,|a_2rangle,|a_3rangle,...)$, then the state $|psirangle$ can be expressed as:
$$|psirangle=sum_n c_n(t)*|a_nrangle$$
So $|psirangle$ is actually $|psi(t)rangle$, and it does satisfy Schrodingers equation. So if the $|a_nrangle$ are the eigenkets of the Hamiltonian, then the time dependence of the expansion coefficients is just the usual $e^-ifrac E_nhbart$, and the expansion of the state is just:
$$|psi(t)rangle=sum_n c_n(0)*e^-ifrac E_nhbart*|a_nrangle$$
answered 1 hour ago
Hugo V
1436
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think there is a lot of confusion in what you wrote. The Schrodinger equation is satisfied,of course not by the eigenkets themselves, but by the expansion coefficients of any state expanded on the basis of eigenkets. The eigenkets are defined as being time independent exactly because, for a general state, the time dependence of each expansion coefficient on the eigenket basis is really simple. To make this clear, for any state $|psirangle$, take a complete set of eigenkets $(|a_1rangle,|a_2rangle,|a_3rangle,...)$, then the state $|psirangle$ can be expressed as:
$$|psirangle=sum_n c_n(t)*|a_nrangle$$
So $|psirangle$ is actually $|psi(t)rangle$, and it does satisfy Schrodingers equation. So if the $|a_nrangle$ are the eigenkets of the Hamiltonian, then the time dependence of the expansion coefficients is just the usual $e^-ifrac E_nhbart$, and the expansion of the state is just:
$$|psi(t)rangle=sum_n c_n(0)*e^-ifrac E_nhbart*|a_nrangle$$
answered 1 hour ago
Hugo V
1436
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Hugo V
1436
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Hugo V
1436
answered 1 hour ago
Hugo V
1436
1436
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
1
down vote
Only kets that represent physical systems ("state vectors") satisfy the Schrodinger equation. Basis kets don't represent physical systems, but just a system of coordinates, so they don't.
Your question is analogous to asking why the coordinates of a random point in space don't satisfy Hamilton's or the Euler-Lagrange equations. There's just nothing there to time-evolve.
answered 1 hour ago
tparker
21.4k143114
add a comment |Â
up vote
1
down vote
Only kets that represent physical systems ("state vectors") satisfy the Schrodinger equation. Basis kets don't represent physical systems, but just a system of coordinates, so they don't.
Your question is analogous to asking why the coordinates of a random point in space don't satisfy Hamilton's or the Euler-Lagrange equations. There's just nothing there to time-evolve.
answered 1 hour ago
tparker
21.4k143114
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Only kets that represent physical systems ("state vectors") satisfy the Schrodinger equation. Basis kets don't represent physical systems, but just a system of coordinates, so they don't.
Your question is analogous to asking why the coordinates of a random point in space don't satisfy Hamilton's or the Euler-Lagrange equations. There's just nothing there to time-evolve.
answered 1 hour ago
tparker
21.4k143114
Only kets that represent physical systems ("state vectors") satisfy the Schrodinger equation. Basis kets don't represent physical systems, but just a system of coordinates, so they don't.
Your question is analogous to asking why the coordinates of a random point in space don't satisfy Hamilton's or the Euler-Lagrange equations. There's just nothing there to time-evolve.
answered 1 hour ago
tparker
21.4k143114
answered 1 hour ago
tparker
21.4k143114
answered 1 hour ago
tparker
21.4k143114
answered 1 hour ago
tparker
21.4k143114
21.4k143114
add a comment |Â
add a comment |Â
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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
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
