A peculiarity in uncertainty principle

[ShayanJ]

Consider two hermitian operators A and B.
Imagine a system is in state |\psi\rangle,then we have:
<br /> \langle \psi|[A,B]|\psi\rangle=\langle \psi|AB-BA|\psi\rangle=B^{\dagger}A^{\dagger}|\psi\rangle-BA|\psi\rangle=BA|\psi\rangle-BA|\psi\rangle=0<br />
This just seems a little strange,for example we have [p,x]=-i \hbar but the above statement says that \langle [p,x] \rangle=0 which is strange because the average of a non-zero constant has become zero!
And also it does not give the uncertainty relations we know!
What's wrong?
Thanks

 

[George Jones]

See

https://www.physicsforums.com/showthread.php?t=122063

 

[cattlecattle]

Consider two hermitian operators A and B.
Imagine a system is in state |\psi\rangle,then we have:
<br /> \langle \psi|[A,B]|\psi\rangle=\langle \psi|AB-BA|\psi\rangle=B^{\dagger}A^{\dagger}|\psi\rangle-BA|\psi\rangle=BA|\psi\rangle-BA|\psi\rangle=0<br />
This just seems a little strange,for example we have [p,x]=-i \hbar but the above statement says that \langle [p,x] \rangle=0 which is strange because the average of a non-zero constant has become zero!
And also it does not give the uncertainty relations we know!
What's wrong?
Thanks

At your second equal sign, on the left, you have an expected value (complex number), on the right, you have a ket.

 

[K^2]

\langle \psi|AB-BA|\psi\rangle=B^{\dagger}A^{\dagger}|\psi\rangle-BA|\psi\rangle

Yeah, that's not even remotely right. For starters, \small \langle \psi|AB \neq B^\dagger A^\dagger|\psi\rangle. They don't even belong to the same space. More importantly, AB-BA is still an operator. Either the whole thing operates to the right, \small |\psi &#039;\rangle = (AB-BA)|\psi\rangle and you need to find \small \langle \psi |\psi &#039;\rangle, or the whole thing operates to the left with similar fallout.

Try to keep track of what sort of object you are dealing with. Kets are vectors, bras are dual vectors, and operators are rank-2 tensors that map one vector and one dual vector to field of complex numbers. The addition and subtraction operations can only be performed on the same type of object. You can add two kets or two bras, but you can't add a ket to a bra, because they belong to different spaces.

 

[ShayanJ]

Really sorry guys,I wasn't careful in wrting those
let me correct my question
Consider two hermitian operators A and B and a system in state |a\rangle which is an eigenstate of A with eigenvalue \lambda
So we have:
<br /> \langle a|[A,B]|a\rangle=\langle a|AB|a\rangle-\langle a |BA|a\rangle=(A^{\dagger}|a\rangle)^{\dagger}B|a \rangle-\lambda \langle a |B|a\rangle=(A|a \rangle)^{\dagger} B | a \rangle-\lambda \langle a |B|a \rangle=\lambda \langle a|B|a \rangle - \lambda \langle a|B|a \rangle=0<br />

This means that the uncertainty relation for A and B when the system is in an eigenstate of A has zero at the right side and if the two operators are p and x and the system is in a momentum eigen state we should have:
<br /> \Delta p \Delta x \geq 0<br />
And becuse we have a momentum eigen state,\Delta p=0 and so,there is no restriction on \Delta x which seems wrong becuase uncertainty principle should be right even in eigenstates(at least i think so!)
Sorry again
and thanks

 

[Fredrik]

Really sorry guys,I wasn't careful in wrting those
let me correct my question
Consider two hermitian operators A and B and a system in state |a\rangle which is an eigenstate of A with eigenvalue \lambda
So we have:
<br /> \langle a|[A,B]|a\rangle=\langle a|AB|a\rangle-\langle a |BA|a\rangle=(A^{\dagger}|a\rangle)^{\dagger}B|a \rangle-\lambda \langle a |B|a\rangle=(A|a \rangle)^{\dagger} B | a \rangle-\lambda \langle a |B|a \rangle=\lambda \langle a|B|a \rangle - \lambda \langle a|B|a \rangle=0<br />

This means that the uncertainty relation for A and B when the system is in an eigenstate of A has zero at the right side and if the two operators are p and x and the system is in a momentum eigen state we should have:
<br /> \Delta p \Delta x \geq 0<br />
And becuse we have a momentum eigen state,\Delta p=0 and so,there is no restriction on \Delta x which seems wrong becuase uncertainty principle should be right even in eigenstates(at least i think so!)
Sorry again
and thanks

Did you read George Jones's post? When [A,B] is proportional to the identity operator (and not zero), the left-hand side is non-zero. So the calculation "shows" that 0≠0. The problem is that the operators don't actually have eigenstates. Most of the time, it's harmless to pretend that they do, but this is one of the times when it's not. Edit: Maybe that's not the main problem. The rigged Hilbert space formalism can assign a meaning to |x> and |p>, but there are still issues with the domains of the operators. I don't have time to think this through right now.

 

[K^2]

Oh, I got as far as showing that the above result is non-zero if \small |a\rangle is not an eigen state of A. However, that's not good enough. The above result seems to show that \small \langle [p,x]\rangle is zero with \small |\psi \rangle = e^{ikx}, but it should be \small i\hbar for any \small |\psi \rangle. I think I see where the trouble is, but it would suggest that there were things about operator algebra that I was not aware of.

 

[ShayanJ]

Did you read George Jones's post? When [A,B] is proportional to the identity operator (and not zero), the left-hand side is non-zero. So the calculation "shows" that 0≠0. The problem is that the operators don't actually have eigenstates. Most of the time, it's harmless to pretend that they do, but this is one of the times when it's not. Edit: Maybe that's not the main problem. The rigged Hilbert space formalism can assign a meaning to |x> and |p>, but there are still issues with the domains of the operators. I don't have time to think this through right now.

I was in a rush while writing my last post,so I forgot to read that.
And now,after reading it,It seems that is just beyond me,I don't understand much but I know the following:
<br /> \frac{\hbar}{i} \frac{\partial}{\partial x}(e^{ikx})=\frac{\hbar}{i}ik e^{ikx}=\hbar k e^{ikx} \Rightarrow Pe^{ikx}=\hbar k e^{ikx}<br />
!
But well,I know,Sometimes in mathematics things aren't as you think they clearly are!I just want to know how?

Another point is that,unfortunately,the proof that \langle a|[A,B]|a\rangle=0 where |a \rangle is one of A's eigenstates is presented in one of the textbooks of quantum mechanics and that was the reason I posted this thread.

Any way,this is just very interesting and I sigh for not having enough knowledge in math for understanding the discussion.

And K^2,could you explain the trouble you see?
You know,It just seems a little mysterious to me

Thanks all guys

 

[George Jones]

If I get some time tomorrow, I'll try to work out an illustrative example.

Another point is that,unfortunately,the proof that \langle a|[A,B]|a\rangle=0 where |a \rangle is one of A's eigenstates is presented in one of the textbooks of quantum mechanics and that was the reason I posted this thread.

Which book?

In the proof given by the book, are A and B canonically conjugate?

 

[cattlecattle]

Oh, I got as far as showing that the above result is non-zero if \small |a\rangle is not an eigen state of A. However, that's not good enough. The above result seems to show that \small \langle [p,x]\rangle is zero with \small |\psi \rangle = e^{ikx}, but it should be \small i\hbar for any \small |\psi \rangle. I think I see where the trouble is, but it would suggest that there were things about operator algebra that I was not aware of.

Almost all proofs of 0=1 involve symbolic manipulation of infinity. This is no exception.
Intuitively, when in eigenstate of P, particle is in definite momentum, but <p|X|p>, the value of position is totally indefinite, so intuitively it's not well defined.
To see this more rigorously, consider eigenstate of |p>, whose x-basis function is plane wave e^{ipx/\hbar}. The expected value of position is then
<br /> \langle p | X | p \rangle = \int e^{-ipx/\hbar} x e^{ipx/\hbar}dx=\int x dx<br />
which is not well defined. So not surprisingly, this proof again involves infinity.

 

[dextercioby]

This is the famous problem of assuming there's no real mathematics behind the bra-ket formalism. For the recond, there is.

If A and B are canonically conjugate, i.e. there exists a complex separable Hilbert space \mathcal{H} and a dense everywhere invariant subset \displaystyle{\bar{\bf{\Phi}}=\mathcal{H}} included in both domains of A and B, such as

[A,B] = \hat{1}_{|\bar{\bf{\Phi}}} ,

then A and B must necessarily NOT have any common eigenvector, even if both A and B have a purely discrete spectrum (for the record, in QM there's never the case in which both A and B have purely discrete spectrum: position & momentum, angle & angular momentum, time and & Hamiltonian).

The illuminating article on this is written by David Gieres with the reference quant-ph/9907069v2 on arxiv.

 

[Fredrik]

<br /> \frac{\hbar}{i} \frac{\partial}{\partial x}(e^{ikx})=\frac{\hbar}{i}ik e^{ikx}=\hbar k e^{ikx} \Rightarrow Pe^{ikx}=\hbar k e^{ikx}<br />

Right, this shows that $e^{ikx}$ is an eigenfunction of $-i\hbar\, d/dx$. However, a state vector is a member of the system's Hilbert space, and the Hilbert space in this case is defined in the following (rather complicated) way. Let V be the set of square integrable functions from ℝ into ℂ. For all f,g in V, define f+g by (f+g)(x)=f(x)+g(x) for all x. For all f in V and all c in ℂ, define cf by (cf)(x)=c(f(x)) for all x. This turns V into a vector space. Now define, for all f,g in V, $\langle f,g\rangle=\int f(x)^*g(x) dx$. The map $(f,g)\mapsto\langle f,g\rangle$ is a semi-inner product on V. So we have turned V into a semi-inner product space. The reason why I call this a semi-inner product rather than an inner product is that <f,f>=0 doesn't imply f=0. For example, if f is the function such that f(1)=1 and f(x)=0 for all x≠0, then we have f≠0 and <f,f>=0. To proceed we define f and g to be equivalent if $\langle f-g,f-g\rangle=0$. Let's use the notation [f] for the set of members of V that are equivalent to f. [f] is called an equivalence class. Each f in G belongs to exactly one equivalence class. So now we define $L^2(\mathbb R)$ as the set of all equivalence classes. To turn this into a vector space, we define [f]+[g]=[f+g] and c[f]=[cf]. (It's not obvious that these definitions make sense, but it's not too hard to show that they do). Now we can finally define an inner product by $\langle[f],[g]\rangle=\langle f,g\rangle$. This turns $L^2(\mathbb R)$ into an inner product space, and that inner product happens to be a Hilbert space. This is proved in books on functional analysis (or integration theory).

So where does $e^{ikx}$ fit into all of this? It's not a member of $L^2(\mathbb R)$ or a member of V. It's a member of a larger vector space that has V as a subspace. The operator $-i\hbar\, d/dx$ has eigenfunctions in that larger space, but not in V. A state vector is a member of $L^2(\mathbb R)$, i.e. it's an equivalence class of members of V. A wavefunction is a member of V. $e^{ikx}$ isn't even that.

I like to write George Jones's calculation like this: Suppose that [A,B]=cI, where I is the identity operator and c is a complex number.
\begin{align}1 =\frac 1 c \langle a| cI|a\rangle =\frac 1 c \langle a|[A,B]|a\rangle =\frac 1 c \big(\langle a|AB|a\rangle - \langle a|BA|a\rangle\big) =\frac 1 c\big(a^*\langle a|B|a\rangle-a\langle a|B|a\rangle\big)=0.
\end{align} I used that a is real in the last step.

 

[ShayanJ]

If I get some time tomorrow, I'll try to work out an illustrative example.


Which book?

In the proof given by the book, are A and B canonically conjugate?

Quantum Physics by Stephen Gasiorowicz
I don't remember the edition but I think its dropped in later ones.

And yes,that's after proving the general relation below:
<br /> (\Delta A)^2 (\Delta B)^2 \geq \frac{1}{4} \langle i [A,B] \rangle^2<br />

Thanks for the complete explanation Fredrik

 

[K^2]

So where does $e^{ikx}$ fit into all of this? It's not a member of $L^2(\mathbb R)$ or a member of V. It's a member of a larger vector space that has V as a subspace. The operator $-i\hbar\, d/dx$ has eigenfunctions in that larger space, but not in V. A state vector is a member of $L^2(\mathbb R)$, i.e. it's an equivalence class of members of V. A wavefunction is a member of V. $e^{ikx}$ isn't even that.

Alright... All of that makes sense. But surely, we can define an operator $\small P$, such that $\small P|[f(x)]\rangle = |[-i\hbar\frac{\partial}{\partial x}f(x)]\rangle$. If I now define $\small |k\rangle = |[e^{ikx}]\rangle$, then $\small |k\rangle$ is trivially eigen vectors of $\small P$. I can define $\small X$ in a similar manner, and we are right back to where we started. $\small [X,P]=i\hbar I$ and the derivation Shyan showed can be applied to $\small \langle k|[P,X]|k\rangle$ to show that it's zero, while, $\small \langle k|i\hbar I|k\rangle=i\hbar$.

Where's the catch?

 

[ShayanJ]

Alright... All of that makes sense. But surely, we can define an operator $\small P$, such that $\small P|[f(x)]\rangle = |[-i\hbar\frac{\partial}{\partial x}f(x)]\rangle$. If I now define $\small |k\rangle = |[e^{ikx}]\rangle$, then $\small |k\rangle$ is trivially eigen vectors of $\small P$. I can define $\small X$ in a similar manner, and we are right back to where we started. $\small [X,P]=i\hbar I$ and the derivation Shyan showed can be applied to $\small \langle k|[P,X]|k\rangle$ to show that it's zero, while, $\small \langle k|i\hbar I|k\rangle=i\hbar$.

Where's the catch?

Here's the catch
In QM we deal with square integrable functions because if we can't normalize them,we will have none sense results like infinite probabilities.

<br /> \int_{-\infty}^{\infty} e^{-ikx}e^{ikx} dx=\infty<br />
which shows that e^{ikx} is not square integrable so the momentum operator doesn't have an eigenstate in the space we're working in.

 

[Fredrik]

Alright... All of that makes sense. But surely, we can define an operator $\small P$, such that $\small P|[f(x)]\rangle = |[-i\hbar\frac{\partial}{\partial x}f(x)]\rangle$. If I now define $\small |k\rangle = |[e^{ikx}]\rangle$, then $\small |k\rangle$ is trivially eigen vectors of $\small P$.
...
Where's the catch?

I would prefer not to use the ket notation in most of those places. I would write something like this (using units such that $\hbar=1$):
\begin{align}
P[f] &=[-if']\\
u_p(x) &=e^{ipx}\\
|p\rangle &=[u_p]\\
P|p\rangle &= P[u_p]=[-i u_p']=[p u_p]=p[u_p]=p|p\rangle.
\end{align} Here's the catch: I defined the equivalence relation on the semi-inner product space of square-integrable functions, and u_p isn't a member of that space (it's not square integrable), so it's not a member of any of the equivalence classes. This means that the notation $[u_p]$ doesn't make sense.

On the other hand, you can always say "**** the Hilbert space, I'm just going to work with the semi-inner product space instead". Actually, if you want to include plane waves ($e^{ipx}$), you will have to work with some larger vector space that has that semi-inner product space as a subspace. I'm not sure what the appropriate choice is. Let's try the vector space of all functions from ℝ into ℂ. Denote this space by W, and the subspace of square-integrable functions by V. Now you can define everything the way it's usually done in introductory courses. $\hat p f(x)=-if'(x)$, $\hat x f(x)=xf(x)$, and so on. But you have to very carefully consider the domains of the operators you define.

For example, if f is in V, there's no guarantee that $\hat x f$ is in V. This means that the domain of $\hat x$ can't be the entire subspace V, if it's to be considered a linear operator on V. Because of problems like this, books on functional analysis don't require that a "linear operator on V" is defined on V. They only require that it's defined on a dense subspace of V. But they still require that the range is a subset of V.

 

[K^2]

and u_p isn't a member of that space (it's not square integrable)

Uhu. Now I take my particle, and I place it in a box of length L. The definitions of operators did not change. The eigen functions of P are still complex Bloch waves. (Albeit, with discrete k.) None of the algebra changed. The [X,P] commutator is exactly the same, and the problem persists.

Hilbert space - check. Square-integrable - check. What's the catch?

 

[ShayanJ]

There is still a point here.
We can't tell that the assumption that A has an eigenstate,causes the paradox because even if P doesn't have an eigenstate,one still can find an operator that does and do the calculation for that operator and its eigenstate and agian arrive at the paradox.So I think sth else should be
wrong.
I've been trying to understand it via reading the thread that Goerge suggested but I can't follow the discussion because I don't know enough math.
Can somebody present the final result of that thread in a simple language?
thanks

 

[Fredrik]

Uhu. Now I take my particle, and I place it in a box of length L. The definitions of operators did not change. The eigen functions of P are still complex Bloch waves. (Albeit, with discrete k.) None of the algebra changed. The [X,P] commutator is exactly the same, and the problem persists.

Hilbert space - check. Square-integrable - check. What's the catch?

I'm not sure about this. I've been saying (incorrectly) in other threads that the box potential is just a way of saying that the Hilbert space is now $L^2(\text{box})$ instead of $L^2(\mathbb R)$, but now that I think about it, that seems very wrong. Our wavefunctions aren't just zero or undefined outside of the box. They are required to got to zero as we approach the edge of the box. So if the Hilbert space is either of those two spaces I just mentioned, then the energy eigenstates only span a proper subspace of it. We should probably view one of those subspaces as the system's Hilbert space*. In that case, plane waves aren't included.

*) ...or rather, as the system's semi-inner product space, which can be used to define the system's Hilbert space.

Maybe you had periodic boundary conditions in mind, instead of a box potential. I guess that would solve the problem with momentum eigenvectors, but not position eigenvectors.

Anyway, I'm not so sure that non-existence of eigenvectors is really the issue in the 0=1 "proof". The other thread talks mainly about the domains of the operators, so that's probably the real problem here. The commutator [x,p]=i has to be interpreted as in dextercioby's post (in this thread) to make sense.

 

[ShayanJ]

I asked a professor
He said that the problem is,as I understood,that when [A,B]=cI then non of the operators has normalizable eigenvectors.
In fact you can find no A and B for which both [A,B]=cI and \langle a | a \rangle=1,\langle b | b \rangle =1 are true where a and b are eigenvectors of A and B

But I will be more happy if I see a proof of the latter

 

[Dickfore]

The problem with x and p is that they have continuous spectra, so their eigenvectors are elements of a rigged Hilbert space. You may get rid of your confusion by the following artifice:

Imagine you impose periodic boundary conditions, so that x and x + L are physically equivalent points. The consequence of this is that momentum acquires only a dicrete set of eigenvalues:
<br /> k_n = \frac{2\pi n}{L}, \ n = 0, \pm 1, \pm 2, \ldots<br />
and the corresponding eigenfunctions are \langle x \vert k_n \rangle = L^{-1/2} e_{i k_n x}, normalized by the condition:
<br /> \int_{0}^{L} \langle k_m \vert x \rangle \, \langle x \vert k_n \rangle \, dx = \delta_{m,n}<br />
Now, let us expand an arbitrary state ket in the basis of (discrete) momentum eigenvectors. This is nothin more than a Fourier series:
<br /> \vert \psi \rangle = \sum_{n = -\infty}^{\infty} a_n \, \vert k_n \rangle, \ a_n = \langle k_n \vert \psi \rangle = L^{-1/2} \, \int_{0}^{L} dx e^{-i k_n x} \langle x \vert \psi \rangle<br />
Then, the commutator is:
<br /> \langle k_m \vert [ x, p ] \vert \psi \rangle \hbar \, \sum_{n} a_n (k_n - k_m) \langle k_m \vert x \vert k_n \rangle<br />
The matrix element of the position operator between discrete momentum eigenstates is:
<br /> \langle k_m \vert x \vert k_n \rangle = \frac{1}{L} \, \int_{0}^{L} dx x \, e^{-i (k_m - k_n)} = \left\lbrace \begin{array}{lc}<br /> \frac{1}{2} &amp; m = n \\<br /> <br /> \frac{i}{k_m - k_n} &amp; m \neq n<br /> \end{array}\right.<br />

Thus, the commuator matrix element is
<br /> \langle k_m \vert [x, p] \vert \psi \rangle = -i \hbar \sum_{n \neq m} a_n<br />
If you take into account the following identity
<br /> \sum_{n = -\infty}^{\infty} e^{i \frac{2 \pi n x}{L}} = L \sum_{n = -\infty}^{\infty} \delta(x - n L)<br />
you get the following expression:
<br /> [x, p] \vert \psi \rangle= i \hbar \left( \vert \psi \rangle - L \sum_{n = -\infty}^{\infty} e^{-\frac{i}{\hbar} n L p} \vert x = 0 \rangle \right)<br />
where the p on the r.h.s. is an operator. Thus, the commutator has changed.

 

[ShayanJ]

Then, the commutator is:
<br /> \langle k_m \vert [ x, p ] \vert \psi \rangle \hbar \, \sum_{n} a_n (k_n - k_m) \langle k_m \vert x \vert k_n \rangle<br />
The matrix element of the position operator between discrete momentum eigenstates is:
<br /> \langle k_m \vert x \vert k_n \rangle = \frac{1}{L} \, \int_{0}^{L} dx x \, e^{-i (k_m - k_n)} = \left\lbrace \begin{array}{lc}<br /> \frac{1}{2} &amp; m = n \\<br /> <br /> \frac{i}{k_m - k_n} &amp; m \neq n<br /> \end{array}\right.<br />

Thus, the commuator matrix element is
<br /> \langle k_m \vert [x, p] \vert \psi \rangle = -i \hbar \sum_{n \neq m} a_n<br />
If you take into account the following identity
<br /> \sum_{n = -\infty}^{\infty} e^{i \frac{2 \pi n x}{L}} = L \sum_{n = -\infty}^{\infty} \delta(x - n L)<br />
you get the following expression:
<br /> [x, p] \vert \psi \rangle= i \hbar \left( \vert \psi \rangle - L \sum_{n = -\infty}^{\infty} e^{-\frac{i}{\hbar} n L p} \vert x = 0 \rangle \right)<br />
where the p on the r.h.s. is an operator. Thus, the commutator has changed.

I don't understand what you've written in this part.
Looks like some symbols aren't in their place.
Could you write them again and also explain?

And the other point is that,even with this,we can find other operators with the same illness.Can you cure them the same way too?

Thanks

 

[dextercioby]

The idea is that they are not cured. It all stems from the x operator being defined the way it is, namely as a mere multiplication by the real variable also denoted by x. This operator has a purely continuous spectrum, its 'eigenvectors' do not fit in the Hilbert space, they are delta Dirac distributions.

 

[K^2]

Dickfore, what you've written is consistent with results I got numerically earlier by testing X and P operators in discrete space with periodic boundary conditions. The [X,P] operator ended up having zeroes on diagonal, which immediately tells you it has zero expectation in eigen vectors of X operator. I've separately verified that it has zero expectations for eigen vectors of P as well. So in finite, discrete space, Shyan's proof works, and it is consistent with how we expect commutator to work based on your explanation. So there is nothing strange going on there.

I guess, I just expected the operators to behave very similar in continuous space and discrete space with enough points. It's very interesting to see that there is a qualitative change. I wonder if that says anything about how much we can trust the Lattice calculations.

 

[George Jones]

I am working on a long post that gives an example that uses fairly elementary, but somewhat lengthy, calculations to illustrate the problem with domains. If my wife has work (marking) to do tonight, then I might finish the post tonight; if my wife wants to watch a movie, then I won't finish tonight.

 

[ShayanJ]

I don't know much math compared to other ones who have posted here.
I see you guys are really good in this stuff.
But let me give it a try too.
I've tried to figure out what are two general operators which satisfy [A,B]=cI and both are hermitian.
I found the following(Maybe not the most general but general enough)
<br /> A=a \frac{d}{d \alpha}+f(\alpha) \\<br /> B=b \alpha<br />
Now consider:
<br /> A \psi(\alpha)=\lambda \psi(\alpha) \Rightarrow \psi(\alpha)=A e^{- \int \frac{f(\alpha)-\lambda}{a} d\alpha}<br />
You see that the above function isn't normalizable.
We also have:
<br /> B \psi(\alpha)=\lambda \psi(\alpha) \Rightarrow b \alpha \psi(\alpha)=\lambda \psi(\alpha)
Which can't be true unless \alpha is a constant which can't be.
So when [A,B]=cI and both A and B are hermitian,One of them will not have normalizable eigenstates and the other doesn't have eigenstates at all and the assumption of their existence is the wrong part of the proof.
I will be happy to hear corrections.
Thanks all

 

[dextercioby]

[...]
<br /> A=a \frac{d}{d \alpha}+f(\alpha) \\<br /> B=b \alpha<br />
Now consider:
<br /> A \psi(\alpha)=\lambda \psi(\alpha) \Rightarrow \psi(\alpha)=A e^{- \int \frac{f(\alpha)-\lambda}{a} d\alpha}<br />
You see that the above function isn't normalizable.
[...]

Take f(\alpha) = e^{-\alpha^2} with \alpha\in\mathbb{R}

 

[ShayanJ]

I guess you meant:
<br /> f(\alpha)=2ca \alpha+\lambda \Rightarrow \psi(\alpha)=A e^{-c \alpha^2}<br />
Which is normalizable.
Well,I think I'll wait for the answer!

 

[dextercioby]

Yes, I thought of the convergent exponential in the shape of a gaussian. So you can find potentials for which the wavefunction is a well-behaved mathematical object.

 

[ShayanJ]

So we have concluded that the problem is NOT the assumption of the existence of A's and B's normalizable eigenstates?

 

[ShayanJ]

In Quantum mechanics:Concepts and applications by Nouredine Zettili (2nd edition),one can find the following:
<br /> | \psi \rangle = \int d^3 r | \vec{r} \rangle \langle \vec{r} | \psi \rangle = \int d^3 r \psi(\vec{r}) | \vec{r} \rangle<br />
With \langle \vec{r} | \psi \rangle = \psi(\vec{r})
I tried to write the paradox with this convention:
<br /> <br /> 1=\langle a | a \rangle=\frac{1}{c} \langle a | cI | a \rangle=\frac{1}{c} [ \int d^3 r a^*(\vec{r}) (cI) \int d^3 r a(\vec{r}) ] | \vec{r} \rangle=\frac{1}{c} [ \int d^3 r a^*(\vec{r}) [A,B] \int d^3 r a(\vec{r}) ] | \vec{r} \rangle=\frac{1}{c} [ \int d^3 r a^*(\vec{r}) AB \int d^3 r a(\vec{r})-\int d^3 r a^*(\vec{r}) BA \int d^3 r a(\vec{r}) ] | \vec{r} \rangle<br /> <br />
I'm not sure how to continue but I think it may help.So I give it to more capable hands.

 

[ShayanJ]

Yeah,again me.
This time i asked a professor of pure mathematics.
He said the problem is that [A,B] | \psi \rangle =c |\psi \rangle doesn't mean [A,B]=cI
I think the reason is that the domain of I is the whole space but the domain of [A,B] is more restricted so the two operators only have the same prescription but different domains so they're not equal.

 

[Dickfore]

Yeah,again me.
This time i asked a professor of pure mathematics.
He said the problem is that [A,B] | \psi \rangle =c |\psi \rangle doesn't mean [A,B]=cI
I think the reason is that the domain of I is the whole space but the domain of [A,B] is more restricted so the two operators only have the same prescription but different domains so they're not equal.

I think the simplest way to put it is that \langle a \vert a \rangle and \langle a \vert B \vert a \rangle, where \vert a \rangle is an eigenvector of A formally do not exist (they diverge, are infinite)!
Calculate the M.E. of this commutation relation between two different eigenstates:
<br /> \langle a \vert [A, B] \vert a&#039; \rangle = (a - a&#039;) \, \langle a \vert B \vert a&#039; \rangle = c \langle a \vert a&#039; \rangle<br />
When a&#039; \neq a, the r.h.s. is zero, so we conclude \langle a \vert B \vert a&#039; \rangle = 0, \ a&#039; \neq a. But, if we take the limit a&#039; \rightarrow a, the l.h.s. is an indeterminate form of the type 0 \cdot \infty, whereas the r.h.s. is c \, \delta(a - a&#039;), according to the orthonormalization condition for continuous spectra.

This was the motivation behind making the eigenvalues of at least one of the operators A or B, discrete (in my case, the momentum operator p), because then you may normalize everything to a finite norm.

 

[ShayanJ]

I think the simplest way to put it is that \langle a \vert a \rangle and \langle a \vert B \vert a \rangle, where \vert a \rangle is an eigenvector of A formally do not exist (they diverge, are infinite)!
Calculate the M.E. of this commutation relation between two different eigenstates:
<br /> \langle a \vert [A, B] \vert a&#039; \rangle = (a - a&#039;) \, \langle a \vert B \vert a&#039; \rangle = c \langle a \vert a&#039; \rangle<br />
When a&#039; \neq a, the r.h.s. is zero, so we conclude \langle a \vert B \vert a&#039; \rangle = 0, \ a&#039; \neq a. But, if we take the limit a&#039; \rightarrow a, the l.h.s. is an indeterminate form of the type 0 \cdot \infty, whereas the r.h.s. is c \, \delta(a - a&#039;), according to the orthonormalization condition for continuous spectra.

This was the motivation behind making the eigenvalues of at least one of the operators A or B, discrete (in my case, the momentum operator p), because then you may normalize everything to a finite norm.

Ok,but can you prove \langle a | B | a \rangle=\infty and \langle a | a \rangle=\infty for all A,B and |a \rangle satisfying the conditions?

 

[George Jones]

Consider two hermitian operators A and B and a system in state |a\rangle which is an eigenstate of A with eigenvalue \lambda
So we have:
<br /> \langle a|[A,B]|a\rangle=\langle a|AB|a\rangle-\langle a |BA|a\rangle=(A^{\dagger}|a\rangle)^{\dagger}B|a \rangle-\lambda \langle a |B|a\rangle=(A|a \rangle)^{\dagger} B | a \rangle-\lambda \langle a |B|a \rangle=\lambda \langle a|B|a \rangle - \lambda \langle a|B|a \rangle=0<br />

There is still a point here.
We can't tell that the assumption that A has an eigenstate,causes the paradox because even if P doesn't have an eigenstate,one still can find an operator that does and do the calculation for that operator and its eigenstate and again arrive at the paradox.So I think sth else should be
wrong.
I've been trying to understand it via reading the thread that Goerge suggested but I can't follow the discussion because I don't know enough math.
Can somebody present the final result of that thread in a simple language?
thanks

There is a problem with the step

\left\langle a \vert AB \vert a\right\rangle =\left\langle Aa \vert Ba\right\rangle
To see the problem in terms of domains, let's work through in detail a fairly elementary example for which the eigenstates and eigenvalues above actually exist. In this example, subtleties with domains definitely come into play.

First, consider something even more elementary, real-valued functions of a single real variable. The domain of such a function f is the (sub)set of all real numbers x on which f is allowed to act. Suppose f is defined by f\left(x\right) = 1/x. The domain of f cannot be the set of all real numbers \mathbb{R}, but it can be any subset of \mathbb{R} that doesn't contain zero. Take the domain of f to be the set of all non-zero real numbers. Define g by g\left(x\right) = 1/x with domain the set of all positive real numbers. As functions, f \ne g, because f and g have different domains, i.e., it takes both a domain and an action to specify a function. As functions, f = g only when f and g have the same actions and the same domains.

Let the action of the momentum operator be given by P=-id/dx (for convenience, set \hbar =1). On what wave functions can P act, i.e, what is the domain, D_{P}, of P? Since P operates the Hilbert space H of square-integrable functions, D_{P} must be a subset of the set of square-integrable functions. The action of P has to give as output something that lives in the Hilbert space H, i.e., the output has to be square-integrable, and thus D_{P} must be subset of the set of square-integrable functions whose derivatives are also square-integrable. Already, we see that the domain of P cannot be all of the Hilbert space H.

As an observable, we want P to be self-adjoint, i.e, we want P=P^{\dagger }. As in the case of functions above, this means that the actions of P and P^{\dagger } must be the same, and that the domains (the states on which P and P^{\dagger } act) D_{P} and D_{P^{\dagger }} must be the same. For concreteness, take wave functions on the interval with endpoints x=0 and x=1. The adjoint of the momentum operator is defined by

<br /> \begin{align}<br /> \left\langle P^{\dagger }g \vert f\right\rangle &amp;=\left\langle g \vert Pf\right\rangle \\<br /> &amp;=-i\int_{0}^{1}g* \frac{df}{dx}dx \\<br /> &amp;=-i\left( \left[ g* f\right] _{0}^{1}-\int_{0}^{1}\frac{dg}{dx}* fdx\right) \\<br /> &amp; =-i\left[ g* f\right] _{0}^{1}+\left\langle Pg \vert f\right\rangle ,<br /> \end{align}<br />
where integration by parts has been used.

Consequently, the actions of P and P^{\dagger } are the same as long as the first term in the last line vanishes, i.e., as long as

<br /> \begin{align}<br /> 0 &amp;= g* \left( 1\right) f\left( 1\right) -g* \left( 0\right) f\left( 0\right) \\<br /> \frac{f\left( 1\right) }{f\left( 0\right) } &amp;= \frac{g* \left( 0\right) }{g* \left( 1\right) }<br /> \end{align}<br />
for non zero f\left( 0\right) and g\left( 1\right). Now, f\left( 1\right) /f\left( 0\right) is some complex number, say \lambda, so f\left( 1\right) = \lambda f\left( 0\right). Hence,

<br /> \begin{align}<br /> \lambda &amp; =\frac{g* \left( 0\right) }{g* \left( 1\right) }\\<br /> \lambda * &amp; =\frac{g\left( 0\right) }{g\left( 1\right)} \\<br /> g\left( 1\right) &amp; =\frac{1}{\lambda * }g\left( 0\right) .<br /> \end{align}<br />
From the relation \left\langle P^{\dagger }g \vert f\right\rangle =\left\langle g \vert Pf\right\rangle, we see that g\in D_{P^{\dagger }} and f\in D_{P}. These domains can be made to be the same if the same \lambda restrictions are placed on f and g, i.e, if \lambda =1/\lambda*, or \lambda* \lambda =1. This means that \lambda can be written as \lambda =e^{i\theta } with \theta real. Different choices of \theta correspond to different boundary conditions, with \theta =0 corresponding to the periodic boundary condition used by Dickfore above (with L=1). Let's use this choice, so that f is in D_{P^{\dagger }}=D_{P} if f is square-integrable, f&#039; is square-integrable, and f\left( 1\right) =f\left( 0\right). Note that this works even if 0=f\left( 0\right). With this choice, P=P^{\dagger }, and we can write \left\langle Pg \vert f\right\rangle =\left\langle g \vert Pf\right\rangle. We cannot use this when P (on either side) acts on a wave function h that is not in D_{P^{\dagger }}=D_{P}. This is the problem with the "proof" above.

In \left\langle a \vert AB \vert a\right\rangle, take A=P and B=X. Take f\left( x\right) =e^{2\pi ix}. Then, \left( Pf\right) \left( x\right) =2\pi f\left( x\right) and \left( PXf\right) \left( x\right) =P\left( xf\left( x\right) \right). However, h\left( x\right) =xf\left( x\right) =xe^{2\pi ix} does not satisfy the boundary condition h\left( 1\right) =h\left( 0\right), so h is not in the domain D_{P}, and we cannot just slide [/itex]A=P[/itex] to the left.

 

[ShayanJ]

That's very good Goerge,thanks.
But sth that still bothers me is that you somewhere assumed one of the operators to be P.Doesn't that restrict the domain of your argument?I mean,that doesn't tell us that for every A and B for which [A,B]=cI,we can't switch the order,but just about P and X.And also that doesn't tell that there is no function that we can do such calculations on,but that there are some functions that are not suitable.So I think you should generalize your argument.(Or I'm wrong somewhere?)
Thanks again

 

[ShayanJ]

And a question,Why by having same \lambda restrictions on f and g,the domains become the same?
Can we tell that's because it makes the proportionality symmetrical between f and g?If yes,how this one affects the domains?
<br /> \lambda=\frac{f(1)}{f(0)}=\frac{g^*(0)}{g^*(1)}= \frac{f^*(0)}{f^*(1)}=\frac{1}{\lambda^*}<br />

 

[Dickfore]

Ok,but can you prove \langle a | B | a \rangle=\infty and \langle a | a \rangle=\infty for all A,B and |a \rangle satisfying the conditions?

Well, you certainly know that \langle a \vert a&#039; \rangle = \delta(a - a&#039;), so, from the commutator it would follow that:
<br /> \langle a \vert B \vert a&#039; \rangle = \frac{c}{a - a&#039;} \, \delta(a - a&#039;) \sim c \, \delta&#039;(a - a&#039;), \ a \rightarrow a&#039;<br />

EDIT:
Example:
<br /> \langle p \vert x \vert p&#039; \rangle = \frac{1}{2\pi \hbar} \, \int_{-\infty}^{\infty} x \, e^{-\frac{i}{\hbar}(p - p&#039;) x} \, dx<br />
<br /> = i \, \frac{\partial}{\partial p} \, \left( \frac{1}{2\pi} \, \int_{-\infty}^{\infty} e^{-\frac{i}{\hbar}(p - p&#039;) x} \, dx \right) = i \, \hbar \, \delta&#039;(p - p&#039;) \rightarrow i \, \hbar \ \delta&#039;(0), \ p \rightarrow p&#039;<br />

 

[ShayanJ]

Well, you certainly know that \langle a \vert a&#039; \rangle = \delta(a - a&#039;), so, from the commutator it would follow that:
<br /> \langle a \vert B \vert a&#039; \rangle = \frac{c}{a - a&#039;} \, \delta(0) \sim c \, \delta&#039;(a - a&#039;), \ a \rightarrow a&#039;<br />

EDIT:
Example:
<br /> \langle p \vert x \vert p&#039; \rangle = \frac{1}{2\pi \hbar} \, \int_{-\infty}^{\infty} x \, e^{-\frac{i}{\hbar}(p - p&#039;) x} \, dx<br />
<br /> = i \, \frac{\partial}{\partial p} \, \left( \frac{1}{2\pi} \, \int_{-\infty}^{\infty} e^{-\frac{i}{\hbar}(p - p&#039;) x} \, dx \right) = i \, \hbar \, \delta&#039;(p - p&#039;) \rightarrow i \, \hbar \ \delta&#039;(0)<br />

I mean can you tell that when [A,B]=cI,then one the operators has Continuous spectrum?

 

[Dickfore]

I mean can you tell that when [A,B]=cI,then one the operators has continues spectrum?

This is a hard question, but it may be true. I have not seen a proof. But, notice that, for angular momentum, \left[ J_x, J_y \right] = i \, J_z, so the commutator between two operators with a discrete spectrum is not the identity operator.

EDIT:
By the Baker-Hausdorf lemma:

<br /> f(\alpha) \equiv e^{\alpha A} B e^{-\alpha A}<br />
<br /> f(\alpha) = B + \alpha [A, B] + \frac{\alpha^2}{2} \, [A, [A, B]] + \ldots = B + \alpha \, c \, I<br />
so any eigenstate of B, is also an eigenstate of f. Maybe this can be used to prove that B has a continuous spectrum, but I don't know how.

 

[ShayanJ]

This is a hard question, but it may be true. I have not seen a proof. But, notice that, for angular momentum, \left[ J_x, J_y \right] = i \, J_z, so the commutator between two operators with a discrete spectrum is not the identity operator.

So I guess because the observables in QM are not so many,its possible to check all conjugate observables and see whether they have Continuous spectrum or not.
You agree?
Can someone list all conjugate observables?

 

[George Jones]

Some general results:

Set \hbar = 1 and assume

AB - BA = iI.

Multiplying the commutation relation by B and reaaranging gives

<br /> \begin{equation*}<br /> \begin{split}<br /> AB - BA &amp;= iI \\<br /> AB^2 - BAB &amp;= iB \\<br /> AB^2 - B \left( BA + iI \right) &amp;= iB \\<br /> AB^2 - B^2 A &amp;= 2iB.<br /> \end{split}<br /> \end{equation*}<br />

By induction,

AB^n - B^n A = niB^{n-1}

for every positive integer n. Consequently,

<br /> \begin{equation*}<br /> \begin{split}<br /> n\left\| B \right\|^{n-1} &amp;=\left\| AB^n - B^n A \right\| \\<br /> &amp;\leq 2\left\| A \right\| \left\| B \right\|^n\\<br /> n &amp;\leq 2\left\| A \right\| \left\| B \right\| .<br /> \end{split}<br /> \end{equation*}<br />

Because this is true for every n, at least one of A and B must be unbounded. Say it is A. Then, by the Hellinger-Toeplitz theorem, if A is self-adjoint, the domain of physical observable A cannot be all of Hilbert space!

 

[ShayanJ]

Ok Goerge,So let's say D(A)=S \subset H \mbox{ and } D(B)=H
Then we have:
<br /> D([A,B])=D(AB-BA)=D(AB) \cap D(BA)=\{ \psi \epsilon H | B \psi \epsilon S\} \cap \{ \psi \epsilon S | A \psi \epsilon H\}=\{\psi|B \psi \epsilon S\} \cap S<br />

I don't think this helps because we have D([A,B]) \subset D(A) \subset D(B) so we still can choose a state which is in the domain of all three operators.This helps only if you can prove there is no eigenstate of A in that subspace.(Is that what you mean?)

 

[dextercioby]

This thread really contains useful information so I'll add my bit (as usually from the historical perspective).

The commutation relations initially due to Born and Jordan 1925 written for infinite matrices (1925 was the year of matrix mechanics) were reinterpreted in terms of Hilbert spaces with the suitable example l^2 (R).

In 1926 Schrödinger coined wave mechanics, a new form of quantum mechanics, based not on infinite sequences of numbers and infinite matrices, but on functions, a little different than 'ordinary functions. Without knowledge of Hilbert space methods (in 1926 most of Hilbert space methods did not exist !), Schrödinger proved the equivalence of these 2 new 'mechanics' in a formal manner. Today, one knows that matrix mechanics and wave mechanics are essentially the same, because l^2(R) and L^2(R) are isomorphic Hilbert spaces, a particular case of a general theorem which says that any 2 complex separable Hilbert spaces are isomorphic.

In terms of wave functions, Schrödinger re-wrote the fundamental commutation relations of Born and Jordan and 'found a solution', i.e. what we call today a representation of the Lie algebra of the Heisenberg group in terms of essentially self-adjoint operators on a complex separable Hilbert space.

In 1927 in his ground-breaking paper on group theory in quantum mechanics, H. Weyl re-wrote the commutation relations in terms of what we call today <Weyl unitaries>, i.e. semigroups of unitary operators, having the position and momentum coordinates as generators.

1928-1929: In his famous series of papers in the Proc. Nat. Acad. Sci., M. Stone in the US proves the famous <Stone theorem> which sets Weyl's work on rigorous mathematical terms. Stone even formulates the first version of the Stone-von Neumann's theorem and sketches a proof of it.

1929: In Germany John von Neumann was already working on the overall theory of Hilbert spaces and his interest in quantum mechanics through his friend and compatriot E. Wigner led him to apply his results in the theory of quantum mechanics and he particularly chose Weyl's 1927 paper and Stone's 1929 result. He formulated and proved a completely rigorous version of Stone's theorem in terms of Weyl's unitaries which said that the 1926 solution of the Born-Jordan commutation relations found by Erwin Schrödinger was essentially unique.

In the 1950's von Neumann's work was obtained as a particular case of the general theory of unitary (and also projective unitary) representations of locally compact Lie grous developed by G.W. Mackey.

Having gone through most of the work on the fundamental commutation relations for coordinate and momentum, I don't recall any rigorous mathematical result regarding the spectrum of the 2 operators in a general setting.

I don't think there's a case in which one can imagine a physical situation in which both the coordinate and the momentum operators both have a completely discrete (perhaps unbounded) spectrum and obey the commutation relation on a dense everywhere subset of a complex separable Hilbert space.

P.S.

0. Born and Jordan initially wrote the commutation relations in terms of frequencies, not position and momentum.

1. Position and momentum can be replaced by orbital angular momentum and azimuthal angle or by Hamiltonian and time operators.

2. If one can prove that at least 1 of the 2 operators in the commutator must have a part of its spectrum absolutely continuous, then one must bring in the theory of rigged Hilbert spaces.

 

[dextercioby]

This is a completion to post #42 by George above. The proof he quoted was similar to the one in

Wielandt, H. - Über die Unbeschränktheit der Operatoren der Quantenmechanik (Math. Annalen, 1949, S.21).

 

[ShayanJ]

I think this article helps.
For one thing,we have,for one of the eigenvectors of an operator having only continuous part in its spectrum:
<br /> \langle a | a&#039; \rangle = \delta(a-a&#039;)<br />
So we have:
<br /> \infty=\langle a | a \rangle=\frac{1}{c}\langle a | cI | a \rangle =\frac{1}{c}\langle a |c[A,B]| a \rangle =\frac{1}{c} (a-a)\langle a |B| a \rangle \Rightarrow \langle a |B| a \rangle =\frac{\infty}{0}=\infty
We just have to prove that the conditions stated,imply that the spectrum of the operators is purely continuous.
And about \langle a |B|a \rangle being infinite.consider:
<br /> \int _{-\infty} ^{\infty} e^{-ikx} x e^{ikx} dx= \frac{1}{2} x^2 ] _{-\infty} ^{\infty}= \infty - \infty<br />
which is not infinite.

 

[ShayanJ]

There is another proof for \langle a |B|a \rangle being infinite.
[A,B] |\psi \rangle=c | \psi \rangle \Rightarrow \langle \psi | [A,B] | \psi \rangle=c
Now if A | \psi \rangle=\lambda | \psi \rangle:
<br /> \langle \psi|AB|\psi \rangle=\lambda \langle \psi |B|\psi\rangle \\<br /> \langle \psi|BA|\psi \rangle=\lambda \langle \psi |B|\psi\rangle <br />
But:
<br /> \langle \psi|AB|\psi \rangle-\langle \psi |BA|\psi\rangle=\langle \psi|[A,B]|\psi \rangle \Rightarrow (\lambda-\lambda) \langle \psi|B|\psi \rangle=c \Rightarrow \langle \psi|B|\psi\rangle=\frac{c}{0} \Rightarrow \langle \psi |B|\psi \rangle=\infty<br />

So I think this should be the problem in the proof.
I don't know how,but looks like whenever we have [A,B]=cI for two self-adjoint operators and A|\psi \rangle=\lambda |\psi\rangle,then \langle \psi |B|\psi \rangle=\infty.

 

[Jorriss]

I don't know how,but looks like whenever we have [A,B]=cI for two self-adjoint operators and A|\psi \rangle=\lambda |\psi\rangle,then \langle \psi |B|\psi \rangle=\infty.

[A,B]=cI can only be realized if A, B are infinite dimensional. A trace is not defined on an infinite dimensional vector space (as far as I know).

 

[ShayanJ]

[A,B]=cI can only be realized if A, B are infinite dimensional. A trace is not defined on an infinite dimensional vector space (as far as I know).

Well,we're OK with that,They ARE infinite dimensional!

 

[dextercioby]

[...]A trace is not defined on an infinite dimensional vector space (as far as I know).

It is. Just think of von Neumann's density operator (trace class). Tr(rho) is perfectly well defined.