# A peculiarity in uncertainty principle

1. Dec 16, 2012

### ShayanJ

Consider two hermitian operators A and B.
Imagine a system is in state $|\psi\rangle$,then we have:
$\langle \psi|[A,B]|\psi\rangle=\langle \psi|AB-BA|\psi\rangle=B^{\dagger}A^{\dagger}|\psi\rangle-BA|\psi\rangle=BA|\psi\rangle-BA|\psi\rangle=0$
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

2. Dec 16, 2012

### George Jones

Staff Emeritus
3. Dec 16, 2012

### cattlecattle

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

4. Dec 16, 2012

### K^2

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, $\small |\psi '\rangle = (AB-BA)|\psi\rangle$ and you need to find $\small \langle \psi |\psi '\rangle$, 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.

5. Dec 16, 2012

### 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:
$\langle a|[A,B]|a\rangle=\langle a|AB|a\rangle-\langle a |BA|a\rangle=(A^{\dagger}|a\rangle)^{\dagger}B|a \rangle-\lambda \langle a |B|a\rangle=(A|a \rangle)^{\dagger} B | a \rangle-\lambda \langle a |B|a \rangle=\lambda \langle a|B|a \rangle - \lambda \langle a|B|a \rangle=0$

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:
$\Delta p \Delta x \geq 0$
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

6. Dec 16, 2012

### Fredrik

Staff Emeritus
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.

Last edited: Dec 16, 2012
7. Dec 16, 2012

### 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.

8. Dec 16, 2012

### ShayanJ

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:
$\frac{\hbar}{i} \frac{\partial}{\partial x}(e^{ikx})=\frac{\hbar}{i}ik e^{ikx}=\hbar k e^{ikx} \Rightarrow Pe^{ikx}=\hbar k e^{ikx}$
!!!
But well,I know,Sometimes in mathematics things aren't as you think they clearly are!I just wanna 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

9. Dec 16, 2012

### George Jones

Staff Emeritus
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?

10. Dec 16, 2012

### cattlecattle

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
$\langle p | X | p \rangle = \int e^{-ipx/\hbar} x e^{ipx/\hbar}dx=\int x dx$
which is not well defined. So not surprisingly, this proof again involves infinity.

11. Dec 16, 2012

### 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.

Last edited: Dec 16, 2012
12. Dec 16, 2012

### Fredrik

Staff Emeritus
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.

13. Dec 16, 2012

### ShayanJ

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:
$(\Delta A)^2 (\Delta B)^2 \geq \frac{1}{4} \langle i [A,B] \rangle^2$

Thanks for the complete explanation Fredrik

Last edited: Dec 16, 2012
14. Dec 16, 2012

### K^2

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?

15. Dec 16, 2012

### ShayanJ

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.

$\int_{-\infty}^{\infty} e^{-ikx}e^{ikx} dx=\infty$
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.

16. Dec 16, 2012

### Fredrik

Staff Emeritus
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 up 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.

17. Dec 17, 2012

### K^2

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?

18. Dec 17, 2012

### 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

19. Dec 17, 2012

### Fredrik

Staff Emeritus
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.

20. Dec 18, 2012

### ShayanJ

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

21. Dec 18, 2012

### 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:
$$k_n = \frac{2\pi n}{L}, \ n = 0, \pm 1, \pm 2, \ldots$$
and the corresponding eigenfunctions are $\langle x \vert k_n \rangle = L^{-1/2} e_{i k_n x}$, normalized by the condition:
$$\int_{0}^{L} \langle k_m \vert x \rangle \, \langle x \vert k_n \rangle \, dx = \delta_{m,n}$$
Now, let us expand an arbitrary state ket in the basis of (discrete) momentum eigenvectors. This is nothin more than a Fourier series:
$$\vert \psi \rangle = \sum_{n = -\infty}^{\infty} a_n \, \vert k_n \rangle, \ a_n = \langle k_n \vert \psi \rangle = L^{-1/2} \, \int_{0}^{L} dx e^{-i k_n x} \langle x \vert \psi \rangle$$
Then, the commutator is:
$$\langle k_m \vert [ x, p ] \vert \psi \rangle \hbar \, \sum_{n} a_n (k_n - k_m) \langle k_m \vert x \vert k_n \rangle$$
The matrix element of the position operator between discrete momentum eigenstates is:
$$\langle k_m \vert x \vert k_n \rangle = \frac{1}{L} \, \int_{0}^{L} dx x \, e^{-i (k_m - k_n)} = \left\lbrace \begin{array}{lc} \frac{1}{2} & m = n \\ \frac{i}{k_m - k_n} & m \neq n \end{array}\right.$$

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

Last edited: Dec 18, 2012
22. Dec 18, 2012

### ShayanJ

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

23. Dec 18, 2012

### 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.

24. Dec 18, 2012

### 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.

25. Dec 18, 2012

### George Jones

Staff Emeritus
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.