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4

$begingroup$

I'm a bit embarrassed because there seems to be an obvious thing that I'm missing, but I can't see what it is.

Consider $[mathcal{H},A]=B$, where $A$ and $B$ are some operators and $mathcal{H}$ is the Hamiltonian. Let's look at the expectation value of $mathcal{H}A$, and let's use energy eigenstates.

$$langle psi|mathcal{H}A|psirangle=langle psi|B+Amathcal{H}|psirangle=langle psi|B|psirangle+langle psi|Amathcal{H}|psirangle=langle Brangle+E langle Arangle.$$

But, can't I also write

$$langle psi|mathcal{H}A|psirangle=left(langle psi|mathcal{H}right) left(A|psirangleright)=E langle psi|A|psirangle=E langle Arangle,$$

where I've made use of $mathcal{H}| psi rangle=E |psirangle Rightarrow langle psi|mathcal{H}=E langle psi|?$

I don't see what's wrong with my argument, but I feel like something is wrong because I get different answers.

quantum-mechanics

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asked 8 hours ago

PtheguyPtheguy

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is all meaningless. There is no way to expect that the domains of A and B contain the eigenvectors of H.
  $endgroup$
  – DanielC
  8 hours ago
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
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  @DanielC That is a bit harsh, given that we could perfectly well postulate that we're in a finite-dimensional space of states where there are no domain issues.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

I'm a bit embarrassed because there seems to be an obvious thing that I'm missing, but I can't see what it is.

Consider $[mathcal{H},A]=B$, where $A$ and $B$ are some operators and $mathcal{H}$ is the Hamiltonian. Let's look at the expectation value of $mathcal{H}A$, and let's use energy eigenstates.

$$langle psi|mathcal{H}A|psirangle=langle psi|B+Amathcal{H}|psirangle=langle psi|B|psirangle+langle psi|Amathcal{H}|psirangle=langle Brangle+E langle Arangle.$$

But, can't I also write

$$langle psi|mathcal{H}A|psirangle=left(langle psi|mathcal{H}right) left(A|psirangleright)=E langle psi|A|psirangle=E langle Arangle,$$

where I've made use of $mathcal{H}| psi rangle=E |psirangle Rightarrow langle psi|mathcal{H}=E langle psi|?$

I don't see what's wrong with my argument, but I feel like something is wrong because I get different answers.

quantum-mechanics

share|cite|improve this question

asked 8 hours ago

PtheguyPtheguy

28222 silver badges1010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is all meaningless. There is no way to expect that the domains of A and B contain the eigenvectors of H.
  $endgroup$
  – DanielC
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielC That is a bit harsh, given that we could perfectly well postulate that we're in a finite-dimensional space of states where there are no domain issues.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I'm a bit embarrassed because there seems to be an obvious thing that I'm missing, but I can't see what it is.

Consider $[mathcal{H},A]=B$, where $A$ and $B$ are some operators and $mathcal{H}$ is the Hamiltonian. Let's look at the expectation value of $mathcal{H}A$, and let's use energy eigenstates.

$$langle psi|mathcal{H}A|psirangle=langle psi|B+Amathcal{H}|psirangle=langle psi|B|psirangle+langle psi|Amathcal{H}|psirangle=langle Brangle+E langle Arangle.$$

But, can't I also write

$$langle psi|mathcal{H}A|psirangle=left(langle psi|mathcal{H}right) left(A|psirangleright)=E langle psi|A|psirangle=E langle Arangle,$$

where I've made use of $mathcal{H}| psi rangle=E |psirangle Rightarrow langle psi|mathcal{H}=E langle psi|?$

I don't see what's wrong with my argument, but I feel like something is wrong because I get different answers.

quantum-mechanics

share|cite|improve this question

asked 8 hours ago

PtheguyPtheguy

28222 silver badges1010 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

PtheguyPtheguy

28222 silver badges1010 bronze badges

share|cite|improve this question

asked 8 hours ago

PtheguyPtheguy

28222 silver badges1010 bronze badges

asked 8 hours ago

PtheguyPtheguy

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asked 8 hours ago

PtheguyPtheguy

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2 Answers
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5

$begingroup$

Let's assume both $A$ and $B$ are Hermitian operators, since otherwise trying to take their expectation value doesn't make sense at all.

1. If $H$ and $A$ do not commute, then their product $HA$ is not an observable, since $(HA)^dagger = A^dagger H^dagger = AH neq HA$, so $HA$ is not Hermitian. So it is questionable what you're trying to compute here in the first place from a physical viewpoint.




2. Your argument is not wrong, it simply shows that $langle Brangle = 0$ for all eigenstates of $H$. In light of Ehrenfest's theorem ($frac{mathrm{d}}{mathrm{d}t}langle Arangle propto langle [H,A]rangle$ for not explicitly time-dependent $A$), this is not surprising: The eigenstates of the Hamiltonian are stationary states, so their expectation value for the commutator of an observable with the Hamiltonian needs to be zero, otherwise the expectation value (and thus the state) would not be stationary.

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, so if my argument isn't wrong, then let's consider the following: At $t=0$, $langle A rangle=a_0$ is known. What is the time dependence of $langle A rangle$? On the one hand, $langle psi(0)|U(t)^{dagger}AU(t)|psi(0)rangle = a_0$, since we can apply my argument above to each term if we expand the temporal unitary operator. On the other hand, $d/dt langle A rangle propto langle [H,A] rangle$. So, if my argument above is correct, then the expectation value doesn't change and we should have $[H,A]=0$, but the problem says $A$ and $H$ don't commute. How do we resolve this?
  $endgroup$
  – Ptheguy
  7 hours ago
  
  
  

  
  
  
  


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  @Ptheguy I do not understand your question. $langle [H,A]rangle = 0 $ for all eigenstates of $H$ does not imply $[H,A] = 0$.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, thank you for that clarification. That was one point I was missing. Still, if my argument in the original post is correct, then I get that the expectation value of an observable $A$ is time independent. That is, I arrive at $langle psi(t)|A|psi(t)rangle = langle psi(0)|A|psi(0)rangle$. Is this correct? Wouldn't the expectation value change over time since the states are evolving?
  $endgroup$
  – Ptheguy
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ptheguy Yes - which is entirely correct for an eigenstate of the Hamiltonian, since these states are eigenstates of the time evolution and do not change (except for a phase)! You seem to be confusing yourself by thinking that this applies to all states - it does not, you explicitly restricted yourself to eigenstates of the Hamiltonian in your question.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  YES! Thank you. This now makes sense again.
  $endgroup$
  – Ptheguy
  7 hours ago
  

  
  
  
  

add a comment
 | 

0

$begingroup$

$$leftlangle Brightrangle =leftlangle left[H,Aright]rightrangle=Eleftlangle Arightrangle-leftlangle Arightrangle E=0$$

share|cite|improve this answer

answered 8 hours ago

olegoleg

39411 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean, this does answer the question but really would be a lot better with a sentence or two explaining why it does.
  $endgroup$
  – jacob1729
  5 hours ago
  

  
  
  
  

add a comment
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5

$begingroup$

Let's assume both $A$ and $B$ are Hermitian operators, since otherwise trying to take their expectation value doesn't make sense at all.

1. If $H$ and $A$ do not commute, then their product $HA$ is not an observable, since $(HA)^dagger = A^dagger H^dagger = AH neq HA$, so $HA$ is not Hermitian. So it is questionable what you're trying to compute here in the first place from a physical viewpoint.




2. Your argument is not wrong, it simply shows that $langle Brangle = 0$ for all eigenstates of $H$. In light of Ehrenfest's theorem ($frac{mathrm{d}}{mathrm{d}t}langle Arangle propto langle [H,A]rangle$ for not explicitly time-dependent $A$), this is not surprising: The eigenstates of the Hamiltonian are stationary states, so their expectation value for the commutator of an observable with the Hamiltonian needs to be zero, otherwise the expectation value (and thus the state) would not be stationary.

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

76k1818 gold badges139139 silver badges356356 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, so if my argument isn't wrong, then let's consider the following: At $t=0$, $langle A rangle=a_0$ is known. What is the time dependence of $langle A rangle$? On the one hand, $langle psi(0)|U(t)^{dagger}AU(t)|psi(0)rangle = a_0$, since we can apply my argument above to each term if we expand the temporal unitary operator. On the other hand, $d/dt langle A rangle propto langle [H,A] rangle$. So, if my argument above is correct, then the expectation value doesn't change and we should have $[H,A]=0$, but the problem says $A$ and $H$ don't commute. How do we resolve this?
  $endgroup$
  – Ptheguy
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ptheguy I do not understand your question. $langle [H,A]rangle = 0 $ for all eigenstates of $H$ does not imply $[H,A] = 0$.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, thank you for that clarification. That was one point I was missing. Still, if my argument in the original post is correct, then I get that the expectation value of an observable $A$ is time independent. That is, I arrive at $langle psi(t)|A|psi(t)rangle = langle psi(0)|A|psi(0)rangle$. Is this correct? Wouldn't the expectation value change over time since the states are evolving?
  $endgroup$
  – Ptheguy
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ptheguy Yes - which is entirely correct for an eigenstate of the Hamiltonian, since these states are eigenstates of the time evolution and do not change (except for a phase)! You seem to be confusing yourself by thinking that this applies to all states - it does not, you explicitly restricted yourself to eigenstates of the Hamiltonian in your question.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  YES! Thank you. This now makes sense again.
  $endgroup$
  – Ptheguy
  7 hours ago
  

  
  
  
  

add a comment
 | 

5

$begingroup$

Let's assume both $A$ and $B$ are Hermitian operators, since otherwise trying to take their expectation value doesn't make sense at all.

1. If $H$ and $A$ do not commute, then their product $HA$ is not an observable, since $(HA)^dagger = A^dagger H^dagger = AH neq HA$, so $HA$ is not Hermitian. So it is questionable what you're trying to compute here in the first place from a physical viewpoint.




2. Your argument is not wrong, it simply shows that $langle Brangle = 0$ for all eigenstates of $H$. In light of Ehrenfest's theorem ($frac{mathrm{d}}{mathrm{d}t}langle Arangle propto langle [H,A]rangle$ for not explicitly time-dependent $A$), this is not surprising: The eigenstates of the Hamiltonian are stationary states, so their expectation value for the commutator of an observable with the Hamiltonian needs to be zero, otherwise the expectation value (and thus the state) would not be stationary.

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

76k1818 gold badges139139 silver badges356356 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, so if my argument isn't wrong, then let's consider the following: At $t=0$, $langle A rangle=a_0$ is known. What is the time dependence of $langle A rangle$? On the one hand, $langle psi(0)|U(t)^{dagger}AU(t)|psi(0)rangle = a_0$, since we can apply my argument above to each term if we expand the temporal unitary operator. On the other hand, $d/dt langle A rangle propto langle [H,A] rangle$. So, if my argument above is correct, then the expectation value doesn't change and we should have $[H,A]=0$, but the problem says $A$ and $H$ don't commute. How do we resolve this?
  $endgroup$
  – Ptheguy
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ptheguy I do not understand your question. $langle [H,A]rangle = 0 $ for all eigenstates of $H$ does not imply $[H,A] = 0$.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, thank you for that clarification. That was one point I was missing. Still, if my argument in the original post is correct, then I get that the expectation value of an observable $A$ is time independent. That is, I arrive at $langle psi(t)|A|psi(t)rangle = langle psi(0)|A|psi(0)rangle$. Is this correct? Wouldn't the expectation value change over time since the states are evolving?
  $endgroup$
  – Ptheguy
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ptheguy Yes - which is entirely correct for an eigenstate of the Hamiltonian, since these states are eigenstates of the time evolution and do not change (except for a phase)! You seem to be confusing yourself by thinking that this applies to all states - it does not, you explicitly restricted yourself to eigenstates of the Hamiltonian in your question.
  $endgroup$
  – ACuriousMind♦
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  YES! Thank you. This now makes sense again.
  $endgroup$
  – Ptheguy
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Let's assume both $A$ and $B$ are Hermitian operators, since otherwise trying to take their expectation value doesn't make sense at all.

1. If $H$ and $A$ do not commute, then their product $HA$ is not an observable, since $(HA)^dagger = A^dagger H^dagger = AH neq HA$, so $HA$ is not Hermitian. So it is questionable what you're trying to compute here in the first place from a physical viewpoint.




2. Your argument is not wrong, it simply shows that $langle Brangle = 0$ for all eigenstates of $H$. In light of Ehrenfest's theorem ($frac{mathrm{d}}{mathrm{d}t}langle Arangle propto langle [H,A]rangle$ for not explicitly time-dependent $A$), this is not surprising: The eigenstates of the Hamiltonian are stationary states, so their expectation value for the commutator of an observable with the Hamiltonian needs to be zero, otherwise the expectation value (and thus the state) would not be stationary.

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

76k1818 gold badges139139 silver badges356356 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

76k1818 gold badges139139 silver badges356356 bronze badges

answered 8 hours ago

ACuriousMind♦ACuriousMind

76k1818 gold badges139139 silver badges356356 bronze badges

answered 8 hours ago

ACuriousMind♦ACuriousMind

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0

$begingroup$

$$leftlangle Brightrangle =leftlangle left[H,Aright]rightrangle=Eleftlangle Arightrangle-leftlangle Arightrangle E=0$$

share|cite|improve this answer

answered 8 hours ago

olegoleg

39411 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean, this does answer the question but really would be a lot better with a sentence or two explaining why it does.
  $endgroup$
  – jacob1729
  5 hours ago
  

  
  
  
  

add a comment
 | 

0

$begingroup$

$$leftlangle Brightrangle =leftlangle left[H,Aright]rightrangle=Eleftlangle Arightrangle-leftlangle Arightrangle E=0$$

share|cite|improve this answer

answered 8 hours ago

olegoleg

39411 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean, this does answer the question but really would be a lot better with a sentence or two explaining why it does.
  $endgroup$
  – jacob1729
  5 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

$$leftlangle Brightrangle =leftlangle left[H,Aright]rightrangle=Eleftlangle Arightrangle-leftlangle Arightrangle E=0$$

share|cite|improve this answer

answered 8 hours ago

olegoleg

39411 silver badge88 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

olegoleg

39411 silver badge88 bronze badges

answered 8 hours ago

olegoleg

39411 silver badge88 bronze badges

answered 8 hours ago

olegoleg

39411 silver badge88 bronze badges