Prove that 0=1 in quantum mechanics

[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

 

[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'.

 

[adriank]

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 $[A,B] \ne i\hbar I$.

 

[Petr Mugver]

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 $[A,B] \ne i\hbar I$.

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.

 

[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!

 

[Petr Mugver]

Yeah we solved the arcane!

 

[George Jones]

See

https://www.physicsforums.com/showthread.php?p=1813103#post1813103.

 

[George Jones]

See

https://www.physicsforums.com/showthread.php?p=1813103#post1813103.

Oops, that should be

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

 

[Fwiffo]

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

 

[jostpuur]

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

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

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

Do you know how to prove this claim?

 

[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.

 

[Fwiffo]

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

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

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?

 

[jostpuur]

Eigen problems of operators are:

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

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

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?

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

Take a closer look at the steps leading to the contradiction:

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

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

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

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

$$0=1$$

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

 

[George Jones]

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

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

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

Do you know how to prove this claim?

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!

 

[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.

 

[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.

 

[Petr Mugver]

...there's no reason that $$|\psi\rangle$$ can't be normalizable...

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

 

[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?

 

[The_Duck]

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

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.

 

[Fwiffo]

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

Take a closer look at the steps leading to the contradiction:



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

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.

 

[Petr Mugver]

The physical point about all this (see post above) is that the commutation relation we are talking about says that the one operator is the generator of translations for the other operator. Since you can translate eigenvalues by the continuum amount you like, theese eigenvalues must be... continuous!

 

[ismaili]

The physical point about all this (see post above) is that the commutation relation we are talking about says that the one operator is the generator of translations for the other operator. Since you can translate eigenvalues by the continuum amount you like, theese eigenvalues must be... continuous!

Good point!
This physical reason is elucidating.

This is pretty much similar to the trace paradox.
i.e. Using discrete basis and taking trace on both sides.

 

[strangerep]

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

[...]

Did it look rigorous?

Not even close. But you intended that post as a joke, right?

 

[jostpuur]

Not even close. But you intended that post as a joke, right?

Something like that. The intention was to repeat the same calculation that was done in the post #1. In my post #18 the calculation is done with delta functions and derivatives, so it is easy to see that the calculation is uncertain. In the post #1 it is done with brackets, so the calculation was camouflaged as rigorous.

 

[jostpuur]

There's one more point of view that can help bringing clarity to the paradox. If we approximate the real line as a discrete, by choosing some $\Delta x$ to denote the smallest step on the real line, then multiplication and differentiation operators are

$$M_x = \Delta x\left(\begin{array}{ccc|cccc} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \\ \cdots & -2 & 0 & 0 & 0 & 0 & \cdots \\ \cdots & 0 & -1 & 0 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 0 & 0 & 0 & 0 & \cdots \\ \cdots & 0 & 0 & 0 & 1 & 0 & \cdots \\ \cdots & 0 & 0 & 0 & 0 & 2 & \cdots \\ & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}\right)$$

$$\partial_x = \frac{1}{\Delta x} \left(\begin{array}{ccc|cccc} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \\ \cdots & -1 & 1 & 0 & 0 & 0 & \cdots \\ \cdots & 0 & -1 & 1 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 0 & -1 & 1 & 0 & \cdots \\ \cdots & 0 & 0 & 0 & -1 & 1 & \cdots \\ \cdots & 0 & 0 & 0 & 0 & -1 & \cdots \\ & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}\right)$$

The commutator looks like this

$$M_x\partial_x - \partial_x M_x = \left(\begin{array}{cc|cccc} \ddots & \vdots & \vdots & \vdots & \vdots & \\ \cdots & 0 & -1 & 0 & 0 & \cdots \\ \hline \cdots & 0 & 0 & -1 & 0 & \cdots \\ \cdots & 0 & 0 & 0 & -1 & \cdots \\ \cdots & 0 & 0 & 0 & 0 & \cdots \\ & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}\right)$$

That means that

$$[M_x,-i\hbar\partial_x] \approx i\hbar\;\textrm{id}$$

only approximately. If we take an eigenvector $|\psi\rangle$ such that $M_x|\psi\rangle=x_0|\psi\rangle$, then

$$\langle\psi|[M_x,-i\hbar\partial_x]|\psi\rangle = 0$$

is the right answer, because the commutator will translate $|\psi\rangle$ such that it becomes an eingenstate with eigenvalue $x_0-\Delta x$. The paradox is related to the orthogonality of the eigenstates corresponding to delta spikes at $x_0$ and $x_0-\Delta x$, when $\Delta x$ is considered as zero.

 

[jostpuur]

$$AB - BA = iI.$$

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 $n$.

This can be applied to the following identity too.

$$[A,e^{-ibB}] = \sum_{n=0}^{\infty} \frac{(-ib)^n}{n!}(AB^n - B^nA) = \sum_{n=1}^{\infty}\frac{(-ib)^n}{n!}niB^{n-1} = be^{-ibB}$$

Suddenly Peter Mugver's earlier post started to make sense.

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

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

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

I don't see what was the point of an intermediate step $-ib[A,B]e^{-ibB}$, but it's right anyway, I see.

 

[Petr Mugver]

I don't see what was the point of an intermediate step $-ib[A,B]e^{-ibB}$, but it's right anyway, I see.

It's a property of Commutators:

$$[A,f(B)]=[A,B]f'(B)$$

$$[f(A),B]=f'(A)[A,B]$$

 

[strangerep]

It's a property of Commutators:

$$[A,f(B)]=[A,B]f'(B)$$

$$[f(A),B]=f'(A)[A,B]$$

It's not actually a property of all commutators, but only if
[A,B] commutes with B (which is of course the case for the
Heisenberg algebra).

 

[Petr Mugver]

It's not actually a property of any commutators, but only if
[A,B] commutes with B (which is of course the case for the
Heisenberg algebra).

Yeah it's true... forgive me!