Is the expectation value of this commutator zero?

[IFNT]

If I have $$H=p^2/2m+V(x)$$, $$|a'>$$ are energy eigenkets with eigenvalue $$E_{a'}$$, isn't the expectation value of $$[H,x]$$ wrt $$|a'>$$ not always 0? Don't I have that
$$<a'|[H,x]|a'> = <a'|(Hx-xH)|a'> = <a'|Hx|a'> - <a'|xH|a'> = 0$$?
But if I calculate the commutator, I get:
$$<a'|[H,x]|a'> = <a'|-i p \hbar / m|a'> \neq 0$$

I suddenly can't see where I did wrong...

 

[arkajad]

And what is for you the value of $$<a'|p|a'>$$?

 

[IFNT]

And what is for you the value of $$<a'|p|a'>$$?

I don't know, but do we know it is zero?

 

[arkajad]

Take two simple, exactly solvable, examples in one dimension:

1) free particle
2) harmonic oscillator

Calculate both values in each case. One which you think should be zero, and one which you think should not be zero.

Very easy and entertaining exercise.

 

[IFNT]

Take two simple, exactly solvable, examples in one dimension:

1) free particle
2) harmonic oscillator

Calculate both values in each case. One which you think should be zero, and one which you think should not be zero.

Very easy and entertaining exercise.

Now suppose I did two examples and they both give zero. So now I know it is true for all hamiltonians? nice proof

 

[arkajad]

I do not understand. So you calculated all four values (two for each of the cases) for each energy eigenstate, and all gave zero, 0=0?. So where is the problem? Can you be more precise?

 

[IFNT]

I do not understand. So you calculated all four values (two for each of the cases) for each energy eigenstate, and all gave zero, 0=0?. So where is the problem? Can you be more precise?

Hmm okay you do not understand, I hope some other on this forum will understand and answer.

 

[arkajad]

I do understand. For the harmonic oscillator you will get 0=0. No problem.

For free particle you will not get reasonable answer at all, except 0xinfinity, which can be any number, because energy eigenstates are not normalized there. But let's see, maybe someone will fall into your trap.

 

[The_Duck]

I think your math constitutes a proof that $$\left \langle a \right | p \left | a \right \rangle = 0$$ when "a" is a normalizable eigenstate of the Hamiltonian (I guess as arkajad points out you have to be careful about non-normalizable states).

This makes sense intuitively. Eigenstates of H are "stationary states": the probably density does not move around with time. For a bound (normalizable) state, this means that the expectation of the position of the particle is not changing, so we expect (hah) the expectation of momentum to be 0.

 

[arkajad]

Well, even for normalizable states you may have a surprise if you find that your energy eigenstate is not in the the dense domain on which the momentum operator is self-adjoint. These subtleties (anomalies?) are discussed in the old good von Neumann's book "Mathematical Foundations of Quantum Mechanics" (be careful because of some old-fashioned terminology, not in use today), but not too often in typical physics books.

 

[sweet springs]

Hi.

[H,x]=[p^2,x]/2m=(p^2x-xp^2)/2m
px-xp=h'/i
ppx-pxp=h'/i p
pxp-xpp=h'/i p
ppx-xpp=2h'/i p
[H,x]=[p^2,x]/2m=h'/i p/m

Energy eigenstate is decomposed to symmetric and antisymmetric states
|e>=|s>+|a>
[H,x] is antisymmetric, i.e. [H(x),x]=-[H(-x),-x]

<e|[H,x]|e>
= <s|[H,x]|s> + <a|[H,x]|a> whose integrand is odd function of x
+ <a|[H,x]|s> + <s|[H,x]|a> whose integrand is even function of x

So <e|[H,x]|e>=0 in case |s>=0 or |a>=0. Otherwise <e|[H,x]|e> can be different from zero and is a pure imaginary number.
It depends on how you choose V(x) in the problem. If V(x) is symmetric i.e.V(x)=V(-x), |s>=0 or |a>=0 so <e|[H,x]|e>=0.

Regards.

 

[The_Duck]

So <e|[H,x]|e>=0 in case |s>=0 or |a>=0. Otherwise <e|[H,x]|e> can be different from zero and is a pure imaginary number.

I feel like IFNT's simple manipulations show that for "nice" eigenstates of H the expectation of p must be 0 no matter whether the state is even or odd or neither. Can you find a bound eigenstate |e> of any Hamiltonian with a nonzero value for <e|[H,x]|e> ?

 

[sweet springs]

Hi.
I here show another condition for <p>=0. By partial integaration,
<p>=h'/i ∫Psi*(x) Psi'(x)dx =h'/i [Psi*(x) Psi(x)] at both the infinity - h'/i ∫Psi*'(x)Psi(x)dx
If wave function Psi(x) is real and Psi(x)=0 at x=plus and minus infinity,
<p>=h'/2i [Psi(x) Psi(x)] at both the infinity = 0.
Regards.

 

[AndersonMD]

I don't understand why you think $$[H,x] = 0$$ in general. Why should this be the case, position and momentum operators don't commute. If I remember correctly this means that you can't simultaneously have a basis |a'> which obeys the eigenvalue rules for both p and x.

 

[sweet springs]

Hi.

Can you find a bound eigenstate |e> of any Hamiltonian with a nonzero value for <e|[H,x]|e> ?

Actually I do not have examples to show you with me. In return, can you show me proof that <a|[H,x]|s> + <s|[H,x]|a> = 0 or <a|p|s> + <s|p|a> = 2Re<a|p|s> = 0 where |e>=|a>+|s> ?
Regards

PS
|e>=|a>+|s> may hold in case of non symmetric potential e.g. V(x)=ax+bx^2.

 

[sweet springs]

Hi.
For a famous particle-in-a-box which is on the train of speed v, <p>=mv.　Can it be the case that <p>≠0　？
Regards.

 

[arkajad]

The canonical commutation relation

px-xp=-ih'

does not really hold for a particle in a box. A whole line is needed.

 

[strangerep]

If I have $$H=p^2/2m+V(x)$$, $$|a'>$$ are energy eigenkets with eigenvalue $$E_{a'}$$, isn't the expectation value of $$[H,x]$$ wrt $$|a'>$$ not always 0? Don't I have that
$$<a'|[H,x]|a'> = <a'|(Hx-xH)|a'> = <a'|Hx|a'> - <a'|xH|a'> = 0$$?
But if I calculate the commutator, I get:
$$<a'|[H,x]|a'> = <a'|-i p \hbar / m|a'> \neq 0$$

I suddenly can't see where I did wrong...

Adding to all the other answers in this thread...

This seems like a variation on Dirac's famous "0 = 1" paradox: if Q,P are the
usual canonically-conjugate operators satisfying [Q,P] = ih, we can (naively) write:

$$\langle q|\,[Q,P]\,|q\rangle ~=~ i\hbar \langla q|I|q\rangle ~=~ i\hbar \langle q|q\rangle$$

$$\frac{1}{i \hbar} \Big( \langle q|QP|q\rangle - \langle q|PQ|q\rangle \Big) ~=~ \langle q|q \rangle ~.$$

In the first term, let Q act on the bra; in the second, let Q act on the ket:

$$\frac{1}{i\hbar} \Big( q \langle q|P|q\rangle - q \langle q|P|q\rangle \!\!\Big) ~=~ \langle q|q \rangle ~~,~~\Longrightarrow 0 = 1 ~(!)$$

The resolution of this paradox in standard quantum theory is that the
spectra of both operators is continuous, so the inner products in the above
are meaningless. You can't validly manipulate adjoints as above for this case.
This is a consequence of another theorem that for canonically conjugate
operators as above, at least one of the operators must be unbounded.
(Proof can be found in Reed & Simon.) Another theorem (Hellinger-Toeplitz)
then says this means that Q and P cannot both have the whole Hilbert space
as their domain.

This is (one of) the primary motivations why people move to rigged Hilbert spaces.
(Anyone using the Dirac bra-ket formalism in this context is secretly using
rigged Hilbert spaces whether they realize it or not.)

For a gentle introduction, try Ballentine section 1.4

Also see quant-ph/0502053 by Rafael de la Madrid. It also gives a gentle
introduction, and also covers the example of the 1D rectangular barrier
potential -- which might shed some light more directly relevant to the original
post in this thread.

HTH.

PS: I wish there was a sticky thread about this... :-)

 

[sweet springs]

Hi.

The canonical commutation relation

px-xp=-ih'

does not really hold for a particle in a box. A whole line is needed.

Please see wiki "Particle in a box" where QM is applied. I see it holds there.

Regards.

 

[arkajad]

Wikipedia is not to be considered as a Bible in such questions involving rather subtle tricky mathematical concepts (unbounded, symmetric but not self-adjoint operators).

You should rather consult Reed and Simon, "Methods of Mathematical Physics", Vol 2, pp. 141-143.

You are dealing with unbounded operators and trying to formally manipulate them. This can easily lead to paradoxes. From [p,x]=-ih' you easily
get

exp(ipa/h') x exp(-ipa/h')= x+a

and this will lead you out of the box for a big enough.

See also the discussion in the chapter "C. Driven particle in a box. Quantum particle" in http://eprints.whiterose.ac.uk/1374/1/weigerts28.pdf".

 

[George Jones]

http://arxiv.org/abs/quant-ph/0103153

is another related article.

 

[sweet springs]

Hi.
Huum... I just can smell difficulties in particle-in-a-box problem. Thank you for your suggesitions.

Referring to my second last post, I do not care about specific models in explaining my ground-and-moving-train case.
Harmonics, well of finite depth or any not controversial model whose <p> is zero in its center of mass system is OK.
Regards.

 

[arkajad]

http://arxiv.org/abs/quant-ph/0103153

is another related article.

Thanks for this link - very useful.

 

[sweet springs]

Hi.

QUOTE=arkajad;2867907

For free particle you will not get reasonable answer at all, except 0xinfinity, which can be any number, because energy eigenstates are not normalized there. But let's see, maybe someone will fall into your trap.

For V(x)=0 free particle, all the momentum eigenstates are also energy eigenstates i.e. stationary states. Expectation value of momentum of these states can take any real numbers.
Regards.

 

[sweet springs]

Hi.

Adding to all the other answers in this thread...

This seems like a variation on Dirac's famous "0 = 1" paradox: if Q,P are the
usual canonically-conjugate operators satisfying [Q,P] = ih, we can (naively) write:

$$\langle q|\,[Q,P]\,|q\rangle ~=~ i\hbar \langla q|I|q\rangle ~=~ i\hbar \langle q|q\rangle$$

$$\frac{1}{i \hbar} \Big( \langle q|QP|q\rangle - \langle q|PQ|q\rangle \Big) ~=~ \langle q|q \rangle ~.$$

In the first term, let Q act on the bra; in the second, let Q act on the ket:

$$\frac{1}{i\hbar} \Big( q \langle q|P|q\rangle - q \langle q|P|q\rangle \!\!\Big) ~=~ \langle q|q \rangle ~~,~~\Longrightarrow 0 = 1 ~(!)$$

<q'|[Q,P]|q>=<q'|(ih')|q>=ih' delta(q'-q)
Whereas
<q'|[Q,P]|q>=<q'|QP|q>-<q'|PQ|q>=(q'-q)<q'|P|q>
Writing down formula of wave function in momentum space and change order of integration and differentiation results,
=-ih' (q'-q) delta'(q'-q)

Differentiating formula x delta(x) = 0, we get delta(x)+x delta'(x) = 0 that shows both the results are same. There seems to be no paradox.

Regards.

 

[sweet springs]

Hi. Coming back to original post

If I have $$H=p^2/2m+V(x)$$, $$|a'>$$ are energy eigenkets with eigenvalue $$E_{a'}$$, isn't the expectation value of $$[H,x]$$ wrt $$|a'>$$ not always 0? Don't I have that
$$<a'|[H,x]|a'> = <a'|(Hx-xH)|a'> = <a'|Hx|a'> - <a'|xH|a'> = 0$$?
But if I calculate the commutator, I get:
$$<a'|[H,x]|a'> = <a'|-i p \hbar / m|a'> \neq 0$$

I suddenly can't see where I did wrong...

From formula d/dt<ψ|A|ψ> = 1/ih'<ψ| [H, A] |ψ>,
d/dt<e|A|e> = 1/ih'<e| [H, A] |e> for energy eigenstate |e> whose eigenvalue is e. In case <e|A|e> is finite, d/dt<e|A|e> =0 and d/dt<e|p|e> =0 if we take x for A.

Regards.