Rigorous justification for non-relativistic QM perturbation theory assumptions?The expansion of a function in...
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Rigorous justification for non-relativistic QM perturbation theory assumptions?
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$begingroup$
In perturbation theory for non-relativistic quantum mechanics, you begin with a Hamiltonian of the form $$H=H_0+lambda H'$$
and assume that the perturbed eigenstates and eigenvalues can be written as power series in $lambda$:
$$left|psi_nright>=left|psi_n^0right>+lambdaleft|psi_n^1right>+lambda^2left|psi_n^2right>+dots;$$
$$E_n=E_n^0+lambda E_n^1+lambda^2 E_n^2+dots.$$
Then you plug these expansions into the eigenvalue equation, $Hleft|psiright>=lambdaleft|psiright>$, and set the coefficients of like powers of $lambda$ equal to each other.
At least three concerns arise:
Why are we justified in assuming these Taylor expansions exist, i.e., have nonzero radii of convergence?
Why are we justified in Taylor expanding $left|psiright>$, as it is an abstract vector in Hilbert space and not a scalar-valued function?
Why are we justified in setting like coefficients of $lambda$ equal to each other?
For #3, I have seen some mathematically precise arguments for why, if $P(x)=Q(x)$ for all $x$, where $P$ and $Q$ are finite polynomials, then their coefficients must be equal. But we now have infinite Taylor series, as well as coefficients of $lambda$ that are not scalars but rather abstract vectors in infinite-dimensional Hilbert space—should it still be obvious that we can do this?
EDIT: To clarify question 1, by "convergence" it don't just mean convergence at all, but even stronger: convergence to the correct value. There are non-analytic functions, like $f(x) = {0$ if $x=0, e^{-1/x^2}$ otherwise$}$ which have Taylor expansions that converge (to some value) everywhere, but to the correct value only at the single point of expansion (at zero, for this example).
quantum-mechanics hilbert-space perturbation-theory
asked 1 hour ago
WillGWillG
452110
$endgroup$
add a comment |
$begingroup$
In perturbation theory for non-relativistic quantum mechanics, you begin with a Hamiltonian of the form $$H=H_0+lambda H'$$
and assume that the perturbed eigenstates and eigenvalues can be written as power series in $lambda$:
$$left|psi_nright>=left|psi_n^0right>+lambdaleft|psi_n^1right>+lambda^2left|psi_n^2right>+dots;$$
$$E_n=E_n^0+lambda E_n^1+lambda^2 E_n^2+dots.$$
Then you plug these expansions into the eigenvalue equation, $Hleft|psiright>=lambdaleft|psiright>$, and set the coefficients of like powers of $lambda$ equal to each other.
At least three concerns arise:
Why are we justified in assuming these Taylor expansions exist, i.e., have nonzero radii of convergence?
Why are we justified in Taylor expanding $left|psiright>$, as it is an abstract vector in Hilbert space and not a scalar-valued function?
Why are we justified in setting like coefficients of $lambda$ equal to each other?
For #3, I have seen some mathematically precise arguments for why, if $P(x)=Q(x)$ for all $x$, where $P$ and $Q$ are finite polynomials, then their coefficients must be equal. But we now have infinite Taylor series, as well as coefficients of $lambda$ that are not scalars but rather abstract vectors in infinite-dimensional Hilbert space—should it still be obvious that we can do this?
EDIT: To clarify question 1, by "convergence" it don't just mean convergence at all, but even stronger: convergence to the correct value. There are non-analytic functions, like $f(x) = {0$ if $x=0, e^{-1/x^2}$ otherwise$}$ which have Taylor expansions that converge (to some value) everywhere, but to the correct value only at the single point of expansion (at zero, for this example).
quantum-mechanics hilbert-space perturbation-theory
asked 1 hour ago
WillGWillG
452110
$endgroup$
add a comment |
$begingroup$
In perturbation theory for non-relativistic quantum mechanics, you begin with a Hamiltonian of the form $$H=H_0+lambda H'$$
and assume that the perturbed eigenstates and eigenvalues can be written as power series in $lambda$:
$$left|psi_nright>=left|psi_n^0right>+lambdaleft|psi_n^1right>+lambda^2left|psi_n^2right>+dots;$$
$$E_n=E_n^0+lambda E_n^1+lambda^2 E_n^2+dots.$$
Then you plug these expansions into the eigenvalue equation, $Hleft|psiright>=lambdaleft|psiright>$, and set the coefficients of like powers of $lambda$ equal to each other.
At least three concerns arise:
Why are we justified in assuming these Taylor expansions exist, i.e., have nonzero radii of convergence?
Why are we justified in Taylor expanding $left|psiright>$, as it is an abstract vector in Hilbert space and not a scalar-valued function?
Why are we justified in setting like coefficients of $lambda$ equal to each other?
For #3, I have seen some mathematically precise arguments for why, if $P(x)=Q(x)$ for all $x$, where $P$ and $Q$ are finite polynomials, then their coefficients must be equal. But we now have infinite Taylor series, as well as coefficients of $lambda$ that are not scalars but rather abstract vectors in infinite-dimensional Hilbert space—should it still be obvious that we can do this?
EDIT: To clarify question 1, by "convergence" it don't just mean convergence at all, but even stronger: convergence to the correct value. There are non-analytic functions, like $f(x) = {0$ if $x=0, e^{-1/x^2}$ otherwise$}$ which have Taylor expansions that converge (to some value) everywhere, but to the correct value only at the single point of expansion (at zero, for this example).
quantum-mechanics hilbert-space perturbation-theory
asked 1 hour ago
WillGWillG
452110
$endgroup$
In perturbation theory for non-relativistic quantum mechanics, you begin with a Hamiltonian of the form $$H=H_0+lambda H'$$
and assume that the perturbed eigenstates and eigenvalues can be written as power series in $lambda$:
$$left|psi_nright>=left|psi_n^0right>+lambdaleft|psi_n^1right>+lambda^2left|psi_n^2right>+dots;$$
$$E_n=E_n^0+lambda E_n^1+lambda^2 E_n^2+dots.$$
Then you plug these expansions into the eigenvalue equation, $Hleft|psiright>=lambdaleft|psiright>$, and set the coefficients of like powers of $lambda$ equal to each other.
At least three concerns arise:
Why are we justified in assuming these Taylor expansions exist, i.e., have nonzero radii of convergence?
Why are we justified in Taylor expanding $left|psiright>$, as it is an abstract vector in Hilbert space and not a scalar-valued function?
Why are we justified in setting like coefficients of $lambda$ equal to each other?
For #3, I have seen some mathematically precise arguments for why, if $P(x)=Q(x)$ for all $x$, where $P$ and $Q$ are finite polynomials, then their coefficients must be equal. But we now have infinite Taylor series, as well as coefficients of $lambda$ that are not scalars but rather abstract vectors in infinite-dimensional Hilbert space—should it still be obvious that we can do this?
EDIT: To clarify question 1, by "convergence" it don't just mean convergence at all, but even stronger: convergence to the correct value. There are non-analytic functions, like $f(x) = {0$ if $x=0, e^{-1/x^2}$ otherwise$}$ which have Taylor expansions that converge (to some value) everywhere, but to the correct value only at the single point of expansion (at zero, for this example).
quantum-mechanics hilbert-space perturbation-theory
quantum-mechanics hilbert-space perturbation-theory
asked 1 hour ago
WillGWillG
452110
asked 1 hour ago
WillGWillG
452110
edited 10 mins ago
WillG
asked 1 hour ago
WillGWillG
452110
asked 1 hour ago
WillGWillG
452110
asked 1 hour ago
WillGWillG
452110
452110
add a comment |
add a comment |
1 Answer
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$begingroup$
By the definition of a Hilbert space, any element of a Hilbert space can be represented as a sum $sum_n c_n lvert psi_nrangle$ with the sequence $(c_n)_{ninmathbb{N}}$ being square summable and the $lvert psi_nrangle$ being normalized. So if $lambda$ is small enough that the sequence $(lambda^n)_{ninmathbb{N}}$ is square summable, then these series converge. This series is a geometric series and converges if and only if $lambda < 1$.
Taylor's theorem holds in arbitrary Banach spaces, cf. e.g. this random search result for "Taylor series in Banach spaces". A Hilbert space is a Banach space.
A power series is zero on an open subset if and only if all its coefficients are zero, cf. e.g. this math.SE post, this is also the underpinning of the holomorphic identity theorem. So if two power series agree on an open interval of $lambda$ (not just a point), their coefficients are equal.
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
$endgroup$
$begingroup$
Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients?
$endgroup$
– WillG
1 min ago
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
By the definition of a Hilbert space, any element of a Hilbert space can be represented as a sum $sum_n c_n lvert psi_nrangle$ with the sequence $(c_n)_{ninmathbb{N}}$ being square summable and the $lvert psi_nrangle$ being normalized. So if $lambda$ is small enough that the sequence $(lambda^n)_{ninmathbb{N}}$ is square summable, then these series converge. This series is a geometric series and converges if and only if $lambda < 1$.
Taylor's theorem holds in arbitrary Banach spaces, cf. e.g. this random search result for "Taylor series in Banach spaces". A Hilbert space is a Banach space.
A power series is zero on an open subset if and only if all its coefficients are zero, cf. e.g. this math.SE post, this is also the underpinning of the holomorphic identity theorem. So if two power series agree on an open interval of $lambda$ (not just a point), their coefficients are equal.
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
$endgroup$
$begingroup$
Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients?
$endgroup$
– WillG
1 min ago
add a comment |
$begingroup$
By the definition of a Hilbert space, any element of a Hilbert space can be represented as a sum $sum_n c_n lvert psi_nrangle$ with the sequence $(c_n)_{ninmathbb{N}}$ being square summable and the $lvert psi_nrangle$ being normalized. So if $lambda$ is small enough that the sequence $(lambda^n)_{ninmathbb{N}}$ is square summable, then these series converge. This series is a geometric series and converges if and only if $lambda < 1$.
Taylor's theorem holds in arbitrary Banach spaces, cf. e.g. this random search result for "Taylor series in Banach spaces". A Hilbert space is a Banach space.
A power series is zero on an open subset if and only if all its coefficients are zero, cf. e.g. this math.SE post, this is also the underpinning of the holomorphic identity theorem. So if two power series agree on an open interval of $lambda$ (not just a point), their coefficients are equal.
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
$endgroup$
$begingroup$
Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients?
$endgroup$
– WillG
1 min ago
add a comment |
$begingroup$
By the definition of a Hilbert space, any element of a Hilbert space can be represented as a sum $sum_n c_n lvert psi_nrangle$ with the sequence $(c_n)_{ninmathbb{N}}$ being square summable and the $lvert psi_nrangle$ being normalized. So if $lambda$ is small enough that the sequence $(lambda^n)_{ninmathbb{N}}$ is square summable, then these series converge. This series is a geometric series and converges if and only if $lambda < 1$.
Taylor's theorem holds in arbitrary Banach spaces, cf. e.g. this random search result for "Taylor series in Banach spaces". A Hilbert space is a Banach space.
A power series is zero on an open subset if and only if all its coefficients are zero, cf. e.g. this math.SE post, this is also the underpinning of the holomorphic identity theorem. So if two power series agree on an open interval of $lambda$ (not just a point), their coefficients are equal.
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
$endgroup$
By the definition of a Hilbert space, any element of a Hilbert space can be represented as a sum $sum_n c_n lvert psi_nrangle$ with the sequence $(c_n)_{ninmathbb{N}}$ being square summable and the $lvert psi_nrangle$ being normalized. So if $lambda$ is small enough that the sequence $(lambda^n)_{ninmathbb{N}}$ is square summable, then these series converge. This series is a geometric series and converges if and only if $lambda < 1$.
Taylor's theorem holds in arbitrary Banach spaces, cf. e.g. this random search result for "Taylor series in Banach spaces". A Hilbert space is a Banach space.
A power series is zero on an open subset if and only if all its coefficients are zero, cf. e.g. this math.SE post, this is also the underpinning of the holomorphic identity theorem. So if two power series agree on an open interval of $lambda$ (not just a point), their coefficients are equal.
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
answered 1 hour ago
ACuriousMind♦ACuriousMind
72.1k17126315
72.1k17126315
$begingroup$
Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients?
$endgroup$
– WillG
1 min ago
add a comment |
$begingroup$
Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients?
$endgroup$
– WillG
1 min ago
$begingroup$
Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients?
$endgroup$
– WillG
1 min ago
$begingroup$
Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients?
$endgroup$
– WillG
1 min ago
add a comment |
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