# Prove that 0=1 in quantum mechanics

1. Aug 6, 2010

### Petar Mali

Of course every prove of this type have some mistake.

If nobody will see I will post solution!

$$\hat{A}$$, $$\hat{B}$$ - linear, hermitian operator which commutator is

$$[\hat{A},\hat{B}]=i\hbar\hat{I}$$

$$\hat{I}$$ - unit operator
Eigen problems of operators are:

$$\hat{A}|\psi \rangle=a|\psi \rangle$$

$$\hat{B}|\psi \rangle=b|\psi \rangle$$

$$|\psi \rangle$$ - normalized eigen vector

$$\langle \psi|\psi\rangle=1$$

Look now

$$[\hat{A},\hat{B}]=i\hbar\hat{I}$$

Take a sandwich of right and left side with vector $$|\psi\rangle$$

$$\langle \psi|[\hat{A},\hat{B}]|\psi\rangle=i\hbar\langle \psi|\hat{I}|\psi\rangle$$

$$\frac{1}{i\hbar}\langle \psi|(\hat{A}\hat{B}-\hat{B}\hat{A})|\psi \rangle=1$$

$$\frac{a}{i\hbar}(\langle \psi|\hat{B}|\psi\rangle-\langle \psi|\hat{B}|\psi\rangle)=1$$

$$0=1$$

Where is mistake? :D

2. Aug 6, 2010

### Petr Mugver

If

$$A|a>=a|a>$$

consider the state ket

$$|\psi>=e^{-ibB}|a>$$

where b is a real number. Then

$$A|\psi>=Ae^{-ibB}|a>=\left(e^{-ibB}A+[A,e^{-ibB}]\right)|a> = \left(e^{-ibB}A-ib[A,B]e^{-ibB}\right)|a> = (a+b)e^{-ibB}|a>$$

In other words

$$e^{-ibB}|a>=|a+b>$$

and this is true for every $$a,b\in\mathbb{R}$$. Thus A and B have continuous spectrum, and you can't normalize $$<a|a>=1$$. Rather

$$<a|a'>=\delta(a-a')$$

So you get something like

$$-ia<a|B|a'>=\delta(a-a')$$

and this is not 1 = 0 when you put a = a'.

Last edited: Aug 6, 2010
3. Aug 6, 2010

No, it's far simpler than that. If $A \lvert\psi\rangle = a \lvert\psi\rangle$, and $B \lvert\psi\rangle = b \lvert\psi\rangle$, then
$$[A,B] \lvert\psi\rangle = AB \lvert\psi\rangle - BA \lvert\psi\rangle = ab \lvert\psi\rangle - ba \lvert\psi\rangle = 0 \ne i\hbar I \lvert\psi\rangle.$$
Hence $[A,B] \ne i\hbar I$.

4. Aug 6, 2010

### Petr Mugver

You simply proved that, since A and B do not commute, they cannot be simultaneously diagonalized. Petar Mali didn't use the fact that psi is also an eigenvector of B.

5. Aug 6, 2010

### Petar Mali

If $$[\hat{A},\hat{B}]=i\hbar\hat{I}$$

then $$\langle \psi|\psi\rangle\ne1$$

because operators $$\hat{A}, \hat{B}$$ have continual spectrum!

6. Aug 6, 2010

### Petr Mugver

Yeah we solved the arcane!

7. Aug 6, 2010

### George Jones

Staff Emeritus
8. Aug 6, 2010

### George Jones

Staff Emeritus
9. Aug 6, 2010

### Fwiffo

$$<A><B>\ne<AB>$$

10. Aug 6, 2010

### jostpuur

Do you know how to prove this claim?

11. Aug 6, 2010

### jostpuur

If

$$A = \left(\begin{array}{cccc} A_{11} & 0 & 0 & \cdots \\ 0 & A_{22} & 0 & \cdots \\ 0 & 0 & A_{33} & \cdots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right)$$

and

$$B = \left(\begin{array}{cccc} B_{11} & B_{12} & B_{13} & \cdots \\ B_{21} & B_{22} & B_{23} & \cdots \\ B_{31} & B_{32} & B_{33} & \cdots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right)$$

then

$$AB-BA = \left(\begin{array}{cccc} 0 & B_{12}(A_{11} - A_{22}) & B_{13}(A_{11}-A_{33}) & \cdots \\ B_{21}(A_{22}-A_{11}) & 0 & B_{23}(A_{22}-A_{33}) & \cdots \\ B_{31}(A_{33}-A_{11}) & B_{32}(A_{33}-A_{22}) & 0 & \cdots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right)$$

so it is impossible to find $$A,B\in\mathbb{C}^{\mathbb{N}\times\mathbb{N}}$$ such that $A$ is diagonalizable and $AB-BA=1$.

On the other hand $M_x(-\partial_x) - (-\partial_x)M_x = 1$, where $\partial_x$ is the differentation operator and $M_x$ is the multiplication operator $(M_x\psi)(x)=x\psi(x)$. The claim about continuous spectrum and non-existence of (normalizable) eigenvectors seems to be correct.

12. Aug 6, 2010

### Fwiffo

The problem is not the continuous spectrum, although what you (and jostpuur) wrote is certainly right. The (bigger) problem is that two operators which don't commute don't have the same eigenvectors. The line above is impossible unless A and B commute. This whole exercise could be done with a discrete spectrum with a few minor changes. Or am I missing something?

13. Aug 6, 2010

### jostpuur

Petar Mali didn't mean that the $|\psi\rangle$ would be the same on both of those lines.

It is only assumed that $A|\psi\rangle=a|\psi\rangle$, and nothing about $B$.

14. Aug 6, 2010

### George Jones

Staff Emeritus
Here is a proof of something close to this.

Set $\hbar = 1$ and assume

$$AB - BA = iI.$$

Multiplying the commutation relation by $B$ and reaaranging gives

$$\begin{equation*} \begin{split} AB - BA &= iI \\ AB^2 - BAB &= iB \\ AB^2 - B \left( BA + iI \right) &= iB \\ AB^2 - B^2 A &= 2iB. \end{split} \end{equation*}$$

By induction,

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

for every positive integer $n$. Consequently,

$$\begin{equation*} \begin{split} n\left\| B \right\|^{n-1} &=\left\| AB^n - B^n A \right\| \\ &\leq 2\left\| A \right\| \left\| B \right\|^n\\ n &\leq 2\left\| A \right\| \left\| B \right\| . \end{split} \end{equation*}$$

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!

15. Aug 6, 2010

### jostpuur

I think I pretty much answered my own question already. My calculation proves that if $A$ can be diagonalized so that it has eigenvectors in the Hilbert space, then $[A,B]\neq 1$.

At this moment I'm not 100% sure that the non-existence of eigenvectors in the Hilbert space would be the same thing as the spectrum being continuous, though. Rigorous definitions should be recalled from somewhere...

George Jones, notice that you are proving a different thing. You proved unboundedness, but for example

$$A = \left(\begin{array}{cccc} 1 & 0 & 0 & \cdots \\ 0 & 2 & 0 & \cdots \\ 0 & 0 & 3 & \cdots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right)$$

is diagonalized so that it has eigenvectors in the Hilbert space, despite the fact that it is unbounded.

16. Aug 6, 2010

### The_Duck

If you ignore the inconsistent and unnecessary assertion that $$|\psi\rangle$$ is an eigenvector of B, then there's no reason that $$|\psi\rangle$$ can't be normalizable. I think what the "proof" tells you, and what you'll find if you plug in some real operators with this property, is that the expression

$$\langle \psi|\hat{B}|\psi\rangle-\langle \psi|\hat{B}|\psi\rangle$$

is an indeterminate expression like $$\infty - \infty$$ which can't be simplified down to 0.

17. Aug 6, 2010

### Petr Mugver

I bet you can't find two hermitian operators with discrete spectrum, in any Hilbert space you like, with the commutation relation above...

18. Aug 6, 2010

### jostpuur

I repeat the original paradox with a "concrete" representation.

Operators are $M_x$ and $-i\hbar\partial_x$. Then $[M_x,-i\hbar\partial_x]=i\hbar\;\textrm{id}$.

We'll pick some eigenvector of $M_x$. $|\psi\rangle$! So that $M_x|\psi\rangle = x_0|\psi\rangle$ and $\langle\psi|\psi\rangle=1$. In position representation this means

$$\psi(x) = \frac{1}{\sqrt{\delta(0)}}\delta(x-x_0)$$

Let's sandwich the commutator between brackets then!

$$\langle\psi| [M_x,-i\hbar\partial_x] |\psi\rangle = \int\limits_{-\infty}^{\infty} \psi(x)^* \big( M_x (-i\hbar\partial_x) - (-i\hbar\partial_x)M_x\big)\psi(x) dx$$
$$= \frac{-i\hbar}{\delta(0)}\int\limits_{-\infty}^{\infty} \delta(x-x_0)\big(x \partial_x\delta(x-x_0) -\partial_x(x\delta(x-x_0))\big) dx = \cdots$$

Now there is two different ways to continue. We can substitute

$$\partial_x(x\delta(x-x_0)) = x\partial_x\delta(x-x_0) + \delta(x-x_0)$$

and then the calculation comes to end like this

$$\cdots = \frac{-i\hbar}{\delta(0)}\int\limits_{-\infty}^{\infty}\delta(x-x_0)\big(-\delta(x-x_0)\big)dx = i\hbar$$

But we can also substitute

$$x\delta(x-x_0) = x_0\delta(x-x_0)$$

Then the calculation comes to end like this

$$\cdots = \frac{-i\hbar}{\delta(0)} x_0\int\limits_{-\infty}^{\infty} \delta(x-x_0)\big(\partial_x\delta(x-x_0) -\partial_x\delta(x-x_0)\big)dx = 0$$

So $i\hbar = 0$. Did it look rigorous?

19. Aug 6, 2010

### The_Duck

Ah; you're right; for some reason I thought that e.g. position eigenstates were normalizable in the desired way but of course they aren't.

20. Aug 6, 2010

### Fwiffo

Good point. This by the way is nice proof that the commutation relation $$[A,B]$$ $$= \alpha I$$ is impossible in the discrete case where operators can be diagonalized.