Picking the different solutions to the time independent Schrodinger eqautionSolving the time independent...
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Picking the different solutions to the time independent Schrodinger eqaution
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Picking the different solutions to the time independent Schrodinger eqaution
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The time independent Schrodinger equation
$$-frac{hbar^2}{2m} frac{d^2psi}{dx^2}+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.
For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.
I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?
The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frac{hbar}{i}frac{d}{dx})psi. $ How do we know for sure that $int a(x)^*Q(x,frac{hbar}{i}frac{d}{dx})a(x) $ and $int b(x)^*Q(x,frac{hbar}{i}frac{d}{dx})b(x)$ gives the same expectation values?
quantum-mechanics schroedinger-equation
$endgroup$
add a comment |
$begingroup$
The time independent Schrodinger equation
$$-frac{hbar^2}{2m} frac{d^2psi}{dx^2}+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.
For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.
I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?
The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frac{hbar}{i}frac{d}{dx})psi. $ How do we know for sure that $int a(x)^*Q(x,frac{hbar}{i}frac{d}{dx})a(x) $ and $int b(x)^*Q(x,frac{hbar}{i}frac{d}{dx})b(x)$ gives the same expectation values?
quantum-mechanics schroedinger-equation
$endgroup$
$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
1 hour ago
add a comment |
$begingroup$
The time independent Schrodinger equation
$$-frac{hbar^2}{2m} frac{d^2psi}{dx^2}+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.
For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.
I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?
The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frac{hbar}{i}frac{d}{dx})psi. $ How do we know for sure that $int a(x)^*Q(x,frac{hbar}{i}frac{d}{dx})a(x) $ and $int b(x)^*Q(x,frac{hbar}{i}frac{d}{dx})b(x)$ gives the same expectation values?
quantum-mechanics schroedinger-equation
$endgroup$
The time independent Schrodinger equation
$$-frac{hbar^2}{2m} frac{d^2psi}{dx^2}+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.
For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.
I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?
The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frac{hbar}{i}frac{d}{dx})psi. $ How do we know for sure that $int a(x)^*Q(x,frac{hbar}{i}frac{d}{dx})a(x) $ and $int b(x)^*Q(x,frac{hbar}{i}frac{d}{dx})b(x)$ gives the same expectation values?
quantum-mechanics schroedinger-equation
quantum-mechanics schroedinger-equation
asked 2 hours ago
TaeNyFanTaeNyFan
62414
62414
$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
1 hour ago
add a comment |
$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
1 hour ago
$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
1 hour ago
$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.
For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_{ell,m}$ with $ell=1$ and $m=-1,0,1$.
$endgroup$
add a comment |
$begingroup$
... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.
That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.
A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.
(I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)
$endgroup$
add a comment |
$begingroup$
In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbb{C}$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrm{e}^{mathrm{i}k}$ for some $kin[0,2pi)$.
If $psi(x) = a(x) + mathrm{i}b(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.
There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.
Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrm{e}^{mathrm{i}px}$ and $psi^ast(x) = mathrm{e}^{-mathrm{i}px}$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.
$endgroup$
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.
For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_{ell,m}$ with $ell=1$ and $m=-1,0,1$.
$endgroup$
add a comment |
$begingroup$
Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.
For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_{ell,m}$ with $ell=1$ and $m=-1,0,1$.
$endgroup$
add a comment |
$begingroup$
Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.
For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_{ell,m}$ with $ell=1$ and $m=-1,0,1$.
$endgroup$
Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.
For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_{ell,m}$ with $ell=1$ and $m=-1,0,1$.
answered 2 hours ago
PieterPieter
9,09231536
9,09231536
add a comment |
add a comment |
$begingroup$
... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.
That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.
A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.
(I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)
$endgroup$
add a comment |
$begingroup$
... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.
That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.
A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.
(I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)
$endgroup$
add a comment |
$begingroup$
... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.
That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.
A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.
(I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)
$endgroup$
... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.
That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.
A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.
(I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)
answered 2 hours ago
Chiral AnomalyChiral Anomaly
12.1k21540
12.1k21540
add a comment |
add a comment |
$begingroup$
In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbb{C}$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrm{e}^{mathrm{i}k}$ for some $kin[0,2pi)$.
If $psi(x) = a(x) + mathrm{i}b(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.
There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.
Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrm{e}^{mathrm{i}px}$ and $psi^ast(x) = mathrm{e}^{-mathrm{i}px}$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.
$endgroup$
add a comment |
$begingroup$
In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbb{C}$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrm{e}^{mathrm{i}k}$ for some $kin[0,2pi)$.
If $psi(x) = a(x) + mathrm{i}b(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.
There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.
Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrm{e}^{mathrm{i}px}$ and $psi^ast(x) = mathrm{e}^{-mathrm{i}px}$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.
$endgroup$
add a comment |
$begingroup$
In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbb{C}$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrm{e}^{mathrm{i}k}$ for some $kin[0,2pi)$.
If $psi(x) = a(x) + mathrm{i}b(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.
There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.
Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrm{e}^{mathrm{i}px}$ and $psi^ast(x) = mathrm{e}^{-mathrm{i}px}$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.
$endgroup$
In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbb{C}$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrm{e}^{mathrm{i}k}$ for some $kin[0,2pi)$.
If $psi(x) = a(x) + mathrm{i}b(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.
There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.
Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrm{e}^{mathrm{i}px}$ and $psi^ast(x) = mathrm{e}^{-mathrm{i}px}$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.9k18130322
72.9k18130322
add a comment |
add a comment |
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$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
1 hour ago