Mixing
What is the Hamiltonian in the “energy basis†for a simple harmonic oscillator?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
My textbook says that for a simple harmonic oscillator the Hamiltonian can be expressed in the "energy basis" in this way:
$$hat H=hbaromegabigg(hat a^daggerhat a + 1over 2bigg).$$
I know that $hat a^dagger$ and $hat a$ are the raising and lowering operators, and that they can be written in terms of $hat p_x$ and $hat x$, but how is this the "energy" basis? What does that even mean?
quantum-mechanics energy hilbert-space harmonic-oscillator hamiltonian
edited 2 hours ago
Qmechanic♦
98.8k121781090
asked 6 hours ago
matryoshka
337316
add a comment |Â
up vote
3
down vote
favorite
My textbook says that for a simple harmonic oscillator the Hamiltonian can be expressed in the "energy basis" in this way:
$$hat H=hbaromegabigg(hat a^daggerhat a + 1over 2bigg).$$
I know that $hat a^dagger$ and $hat a$ are the raising and lowering operators, and that they can be written in terms of $hat p_x$ and $hat x$, but how is this the "energy" basis? What does that even mean?
quantum-mechanics energy hilbert-space harmonic-oscillator hamiltonian
edited 2 hours ago
Qmechanic♦
98.8k121781090
asked 6 hours ago
matryoshka
337316
1
Eigenstates of $hat a^dagger hat a$ are energy eigenstates, i.e. have definite energy if $hat H$ is the Hamiltonian.
– ZeroTheHero
5 hours ago
@ZeroTheHero What do you mean by "if $hat H$ is the Hamiltonian"? Isn't $hat H$ always the Hamiltonian?
– matryoshka
4 hours ago
2
maybe I phrased it wrong. What I meant is that $hat H$ doesn't necessarily have the form you gave.
– ZeroTheHero
4 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
My textbook says that for a simple harmonic oscillator the Hamiltonian can be expressed in the "energy basis" in this way:
$$hat H=hbaromegabigg(hat a^daggerhat a + 1over 2bigg).$$
I know that $hat a^dagger$ and $hat a$ are the raising and lowering operators, and that they can be written in terms of $hat p_x$ and $hat x$, but how is this the "energy" basis? What does that even mean?
quantum-mechanics energy hilbert-space harmonic-oscillator hamiltonian
edited 2 hours ago
Qmechanic♦
98.8k121781090
asked 6 hours ago
matryoshka
337316
My textbook says that for a simple harmonic oscillator the Hamiltonian can be expressed in the "energy basis" in this way:
$$hat H=hbaromegabigg(hat a^daggerhat a + 1over 2bigg).$$
I know that $hat a^dagger$ and $hat a$ are the raising and lowering operators, and that they can be written in terms of $hat p_x$ and $hat x$, but how is this the "energy" basis? What does that even mean?
quantum-mechanics energy hilbert-space harmonic-oscillator hamiltonian
quantum-mechanics energy hilbert-space harmonic-oscillator hamiltonian
edited 2 hours ago
Qmechanic♦
98.8k121781090
asked 6 hours ago
matryoshka
337316
edited 2 hours ago
Qmechanic♦
98.8k121781090
asked 6 hours ago
matryoshka
337316
edited 2 hours ago
Qmechanic♦
98.8k121781090
edited 2 hours ago
Qmechanic♦
98.8k121781090
edited 2 hours ago
Qmechanic♦
98.8k121781090
98.8k121781090
asked 6 hours ago
matryoshka
337316
asked 6 hours ago
matryoshka
337316
asked 6 hours ago
matryoshka
337316
337316
1
Eigenstates of $hat a^dagger hat a$ are energy eigenstates, i.e. have definite energy if $hat H$ is the Hamiltonian.
– ZeroTheHero
5 hours ago
@ZeroTheHero What do you mean by "if $hat H$ is the Hamiltonian"? Isn't $hat H$ always the Hamiltonian?
– matryoshka
4 hours ago
2
maybe I phrased it wrong. What I meant is that $hat H$ doesn't necessarily have the form you gave.
– ZeroTheHero
4 hours ago
add a comment |Â
1
Eigenstates of $hat a^dagger hat a$ are energy eigenstates, i.e. have definite energy if $hat H$ is the Hamiltonian.
– ZeroTheHero
5 hours ago
@ZeroTheHero What do you mean by "if $hat H$ is the Hamiltonian"? Isn't $hat H$ always the Hamiltonian?
– matryoshka
4 hours ago
2
maybe I phrased it wrong. What I meant is that $hat H$ doesn't necessarily have the form you gave.
– ZeroTheHero
4 hours ago
1
1
Eigenstates of $hat a^dagger hat a$ are energy eigenstates, i.e. have definite energy if $hat H$ is the Hamiltonian.
– ZeroTheHero
5 hours ago
Eigenstates of $hat a^dagger hat a$ are energy eigenstates, i.e. have definite energy if $hat H$ is the Hamiltonian.
– ZeroTheHero
5 hours ago
@ZeroTheHero What do you mean by "if $hat H$ is the Hamiltonian"? Isn't $hat H$ always the Hamiltonian?
– matryoshka
4 hours ago
@ZeroTheHero What do you mean by "if $hat H$ is the Hamiltonian"? Isn't $hat H$ always the Hamiltonian?
– matryoshka
4 hours ago
2
2
maybe I phrased it wrong. What I meant is that $hat H$ doesn't necessarily have the form you gave.
– ZeroTheHero
4 hours ago
maybe I phrased it wrong. What I meant is that $hat H$ doesn't necessarily have the form you gave.
– ZeroTheHero
4 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
...how is this the "energy" basis? What does that even mean?
Any of our observable operators in their own eigenbasis are diagonal, where the diagonal entries are the eigenvalues.$^*$
We can see this is true. Let $|psi_irangle$ be the eigenvector such that $H|psi_irangle=E_i|psirangle$. Then the Hamiltonian in its own eigenbasis is:
$$[H]_m,n=langlepsi_m|H|psi_nrangle=langlepsi_m|E_n|psi_nrangle=E_nlanglepsi_m|psi_nrangle$$
Since the eigenvectors are orthonormal:
$$[H]_m,n=delta_m,nE_n$$
Which means that the Hamiltonian is diagonal in its own eigenbasis basis.
Notice how this doesn't depend on what $H$ actually is. If you want to work with your specific example (I'll leave the work to you):
$$langlepsi_m|hbaromegaleft(a^dagger a+frac 12right)|psi_nrangle=delta_m,nhbaromegaleft(n+frac12right)=delta_m,nE_n$$
Therefore, the expression you give must be the Hamiltonian is it's own eigenbasis.
$^*$In treating our operators like matrices, in general an operator in some basis tells us the following information. Each column of the operator tells us how the corresponding basis vector transforms upon multiplication by that operator. Therefore, it makes sense that an operator in its own eigenbasis is diagonal, because the eigenvectors are the basis vectors, and the resulting transformation of each basis vector corresponds to just multiplying them by the corresponding eigenvalue.
answered 3 hours ago
Aaron Stevens
6,0792830
add a comment |Â
up vote
1
down vote
It's the energy basis because the eigenstates of $hat H$ are the number of excited particles in a given state. The first term in your equation is also known as $n$ and represents the total number of particles in the nth state.
So, $hat H |nrangle = hbaromegaleft(hat a^daggerhat a+frac12right) |nrangle = hbaromega left(n+frac12right) |nrangle$
edited 3 hours ago
Aaron Stevens
6,0792830
answered 5 hours ago
JQK
787312
add a comment |Â
up vote
0
down vote
This is a good question, because what is going on here is a loose use of terminology. If you want strict terminology, then indeed when we write an operator such as $hata^dagger hata$ we have not adopted any particular basis but simply written down the operator. The correct use of the phrase "in the energy basis" would mean writing out the matrix elements $langle E_n | hatH | E_mrangle$. Then you would have a matrix representing the Hamiltonian in the basis $ E_n rangle $. The loose terminology here is drawing our attention to the fact that by manipulating raising and lowering operators it is possible to find out much that we might like to know, such as the energy levels and the effect of other operators on energy eigenstates, without needing to discover how the energy eigenstates can be written in terms of position or some other quantity such as momentum.
answered 2 hours ago
Andrew Steane
3666
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
...how is this the "energy" basis? What does that even mean?
Any of our observable operators in their own eigenbasis are diagonal, where the diagonal entries are the eigenvalues.$^*$
We can see this is true. Let $|psi_irangle$ be the eigenvector such that $H|psi_irangle=E_i|psirangle$. Then the Hamiltonian in its own eigenbasis is:
$$[H]_m,n=langlepsi_m|H|psi_nrangle=langlepsi_m|E_n|psi_nrangle=E_nlanglepsi_m|psi_nrangle$$
Since the eigenvectors are orthonormal:
$$[H]_m,n=delta_m,nE_n$$
Which means that the Hamiltonian is diagonal in its own eigenbasis basis.
Notice how this doesn't depend on what $H$ actually is. If you want to work with your specific example (I'll leave the work to you):
$$langlepsi_m|hbaromegaleft(a^dagger a+frac 12right)|psi_nrangle=delta_m,nhbaromegaleft(n+frac12right)=delta_m,nE_n$$
Therefore, the expression you give must be the Hamiltonian is it's own eigenbasis.
$^*$In treating our operators like matrices, in general an operator in some basis tells us the following information. Each column of the operator tells us how the corresponding basis vector transforms upon multiplication by that operator. Therefore, it makes sense that an operator in its own eigenbasis is diagonal, because the eigenvectors are the basis vectors, and the resulting transformation of each basis vector corresponds to just multiplying them by the corresponding eigenvalue.
answered 3 hours ago
Aaron Stevens
6,0792830
add a comment |Â
up vote
2
down vote
accepted
...how is this the "energy" basis? What does that even mean?
Any of our observable operators in their own eigenbasis are diagonal, where the diagonal entries are the eigenvalues.$^*$
We can see this is true. Let $|psi_irangle$ be the eigenvector such that $H|psi_irangle=E_i|psirangle$. Then the Hamiltonian in its own eigenbasis is:
$$[H]_m,n=langlepsi_m|H|psi_nrangle=langlepsi_m|E_n|psi_nrangle=E_nlanglepsi_m|psi_nrangle$$
Since the eigenvectors are orthonormal:
$$[H]_m,n=delta_m,nE_n$$
Which means that the Hamiltonian is diagonal in its own eigenbasis basis.
Notice how this doesn't depend on what $H$ actually is. If you want to work with your specific example (I'll leave the work to you):
$$langlepsi_m|hbaromegaleft(a^dagger a+frac 12right)|psi_nrangle=delta_m,nhbaromegaleft(n+frac12right)=delta_m,nE_n$$
Therefore, the expression you give must be the Hamiltonian is it's own eigenbasis.
$^*$In treating our operators like matrices, in general an operator in some basis tells us the following information. Each column of the operator tells us how the corresponding basis vector transforms upon multiplication by that operator. Therefore, it makes sense that an operator in its own eigenbasis is diagonal, because the eigenvectors are the basis vectors, and the resulting transformation of each basis vector corresponds to just multiplying them by the corresponding eigenvalue.
answered 3 hours ago
Aaron Stevens
6,0792830
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
...how is this the "energy" basis? What does that even mean?
Any of our observable operators in their own eigenbasis are diagonal, where the diagonal entries are the eigenvalues.$^*$
We can see this is true. Let $|psi_irangle$ be the eigenvector such that $H|psi_irangle=E_i|psirangle$. Then the Hamiltonian in its own eigenbasis is:
$$[H]_m,n=langlepsi_m|H|psi_nrangle=langlepsi_m|E_n|psi_nrangle=E_nlanglepsi_m|psi_nrangle$$
Since the eigenvectors are orthonormal:
$$[H]_m,n=delta_m,nE_n$$
Which means that the Hamiltonian is diagonal in its own eigenbasis basis.
Notice how this doesn't depend on what $H$ actually is. If you want to work with your specific example (I'll leave the work to you):
$$langlepsi_m|hbaromegaleft(a^dagger a+frac 12right)|psi_nrangle=delta_m,nhbaromegaleft(n+frac12right)=delta_m,nE_n$$
Therefore, the expression you give must be the Hamiltonian is it's own eigenbasis.
$^*$In treating our operators like matrices, in general an operator in some basis tells us the following information. Each column of the operator tells us how the corresponding basis vector transforms upon multiplication by that operator. Therefore, it makes sense that an operator in its own eigenbasis is diagonal, because the eigenvectors are the basis vectors, and the resulting transformation of each basis vector corresponds to just multiplying them by the corresponding eigenvalue.
answered 3 hours ago
Aaron Stevens
6,0792830
...how is this the "energy" basis? What does that even mean?
Any of our observable operators in their own eigenbasis are diagonal, where the diagonal entries are the eigenvalues.$^*$
We can see this is true. Let $|psi_irangle$ be the eigenvector such that $H|psi_irangle=E_i|psirangle$. Then the Hamiltonian in its own eigenbasis is:
$$[H]_m,n=langlepsi_m|H|psi_nrangle=langlepsi_m|E_n|psi_nrangle=E_nlanglepsi_m|psi_nrangle$$
Since the eigenvectors are orthonormal:
$$[H]_m,n=delta_m,nE_n$$
Which means that the Hamiltonian is diagonal in its own eigenbasis basis.
Notice how this doesn't depend on what $H$ actually is. If you want to work with your specific example (I'll leave the work to you):
$$langlepsi_m|hbaromegaleft(a^dagger a+frac 12right)|psi_nrangle=delta_m,nhbaromegaleft(n+frac12right)=delta_m,nE_n$$
Therefore, the expression you give must be the Hamiltonian is it's own eigenbasis.
$^*$In treating our operators like matrices, in general an operator in some basis tells us the following information. Each column of the operator tells us how the corresponding basis vector transforms upon multiplication by that operator. Therefore, it makes sense that an operator in its own eigenbasis is diagonal, because the eigenvectors are the basis vectors, and the resulting transformation of each basis vector corresponds to just multiplying them by the corresponding eigenvalue.
answered 3 hours ago
Aaron Stevens
6,0792830
edited 3 hours ago
answered 3 hours ago
Aaron Stevens
6,0792830
answered 3 hours ago
Aaron Stevens
6,0792830
answered 3 hours ago
Aaron Stevens
6,0792830
6,0792830
add a comment |Â
add a comment |Â
up vote
1
down vote
It's the energy basis because the eigenstates of $hat H$ are the number of excited particles in a given state. The first term in your equation is also known as $n$ and represents the total number of particles in the nth state.
So, $hat H |nrangle = hbaromegaleft(hat a^daggerhat a+frac12right) |nrangle = hbaromega left(n+frac12right) |nrangle$
edited 3 hours ago
Aaron Stevens
6,0792830
answered 5 hours ago
JQK
787312
add a comment |Â
up vote
1
down vote
It's the energy basis because the eigenstates of $hat H$ are the number of excited particles in a given state. The first term in your equation is also known as $n$ and represents the total number of particles in the nth state.
So, $hat H |nrangle = hbaromegaleft(hat a^daggerhat a+frac12right) |nrangle = hbaromega left(n+frac12right) |nrangle$
edited 3 hours ago
Aaron Stevens
6,0792830
answered 5 hours ago
JQK
787312
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's the energy basis because the eigenstates of $hat H$ are the number of excited particles in a given state. The first term in your equation is also known as $n$ and represents the total number of particles in the nth state.
So, $hat H |nrangle = hbaromegaleft(hat a^daggerhat a+frac12right) |nrangle = hbaromega left(n+frac12right) |nrangle$
edited 3 hours ago
Aaron Stevens
6,0792830
answered 5 hours ago
JQK
787312
It's the energy basis because the eigenstates of $hat H$ are the number of excited particles in a given state. The first term in your equation is also known as $n$ and represents the total number of particles in the nth state.
So, $hat H |nrangle = hbaromegaleft(hat a^daggerhat a+frac12right) |nrangle = hbaromega left(n+frac12right) |nrangle$
edited 3 hours ago
Aaron Stevens
6,0792830
answered 5 hours ago
JQK
787312
edited 3 hours ago
Aaron Stevens
6,0792830
edited 3 hours ago
Aaron Stevens
6,0792830
edited 3 hours ago
Aaron Stevens
6,0792830
6,0792830
answered 5 hours ago
JQK
787312
answered 5 hours ago
JQK
787312
answered 5 hours ago
JQK
787312
787312
add a comment |Â
add a comment |Â
up vote
0
down vote
This is a good question, because what is going on here is a loose use of terminology. If you want strict terminology, then indeed when we write an operator such as $hata^dagger hata$ we have not adopted any particular basis but simply written down the operator. The correct use of the phrase "in the energy basis" would mean writing out the matrix elements $langle E_n | hatH | E_mrangle$. Then you would have a matrix representing the Hamiltonian in the basis $ E_n rangle $. The loose terminology here is drawing our attention to the fact that by manipulating raising and lowering operators it is possible to find out much that we might like to know, such as the energy levels and the effect of other operators on energy eigenstates, without needing to discover how the energy eigenstates can be written in terms of position or some other quantity such as momentum.
answered 2 hours ago
Andrew Steane
3666
add a comment |Â
up vote
0
down vote
This is a good question, because what is going on here is a loose use of terminology. If you want strict terminology, then indeed when we write an operator such as $hata^dagger hata$ we have not adopted any particular basis but simply written down the operator. The correct use of the phrase "in the energy basis" would mean writing out the matrix elements $langle E_n | hatH | E_mrangle$. Then you would have a matrix representing the Hamiltonian in the basis $ E_n rangle $. The loose terminology here is drawing our attention to the fact that by manipulating raising and lowering operators it is possible to find out much that we might like to know, such as the energy levels and the effect of other operators on energy eigenstates, without needing to discover how the energy eigenstates can be written in terms of position or some other quantity such as momentum.
answered 2 hours ago
Andrew Steane
3666
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is a good question, because what is going on here is a loose use of terminology. If you want strict terminology, then indeed when we write an operator such as $hata^dagger hata$ we have not adopted any particular basis but simply written down the operator. The correct use of the phrase "in the energy basis" would mean writing out the matrix elements $langle E_n | hatH | E_mrangle$. Then you would have a matrix representing the Hamiltonian in the basis $ E_n rangle $. The loose terminology here is drawing our attention to the fact that by manipulating raising and lowering operators it is possible to find out much that we might like to know, such as the energy levels and the effect of other operators on energy eigenstates, without needing to discover how the energy eigenstates can be written in terms of position or some other quantity such as momentum.
answered 2 hours ago
Andrew Steane
3666
This is a good question, because what is going on here is a loose use of terminology. If you want strict terminology, then indeed when we write an operator such as $hata^dagger hata$ we have not adopted any particular basis but simply written down the operator. The correct use of the phrase "in the energy basis" would mean writing out the matrix elements $langle E_n | hatH | E_mrangle$. Then you would have a matrix representing the Hamiltonian in the basis $ E_n rangle $. The loose terminology here is drawing our attention to the fact that by manipulating raising and lowering operators it is possible to find out much that we might like to know, such as the energy levels and the effect of other operators on energy eigenstates, without needing to discover how the energy eigenstates can be written in terms of position or some other quantity such as momentum.
answered 2 hours ago
Andrew Steane
3666
answered 2 hours ago
Andrew Steane
3666
answered 2 hours ago
Andrew Steane
3666
answered 2 hours ago
Andrew Steane
3666
3666
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f438970%2fwhat-is-the-hamiltonian-in-the-energy-basis-for-a-simple-harmonic-oscillator%23new-answer', 'question_page');
);
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
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
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
1
Eigenstates of $hat a^dagger hat a$ are energy eigenstates, i.e. have definite energy if $hat H$ is the Hamiltonian.
– ZeroTheHero
5 hours ago
@ZeroTheHero What do you mean by "if $hat H$ is the Hamiltonian"? Isn't $hat H$ always the Hamiltonian?
– matryoshka
4 hours ago
2
maybe I phrased it wrong. What I meant is that $hat H$ doesn't necessarily have the form you gave.
– ZeroTheHero
4 hours ago