A The meaning of the commutator for two operators

[SemM]

Hi, what is the true meaning and usefulness of the commutator in:

\begin{equation}
[T, T'] \ne 0
\end{equation}

and how can it be used to solve a parent ODE?

In a book on QM, the commutator of the two operators of the Schrödinger eqn, after factorization, is 1, and this commutation relation is used to solve the ODE. However, this can only work on that eqn, and not when a commutator is far more complex than 1, as in the given example of the Schrödinger eqn:

I have an ODE which has the following operators with the commutator:\begin{equation}
[T, T'] = -2i\hbar \frac{d}{dx}
\end{equation}

what does it really say about the ODE and it's properties? Can it be used to solve the ODE or re-write it in a different manner?The operators are:

T = $\bigg(i\hbar d/dx +\gamma)$
T' = $\bigg(-i\hbar d/dx +\gamma)$

Thanks!

 

[stevendaryl]

Hi, what is the true meaning and usefulness of the commutator in:

\begin{equation}
[T, T'] \ne 0
\end{equation}

[Some stuff deleted]

The operators are:

T = $\bigg(i\hbar d/dx +\gamma)$
T' = $\bigg(-i\hbar d/dx +\gamma)$

Thanks!

\gamma is just a constant? In that case, [T, T'] = 0 for those two operators.

 

[Nguyen Son]

Hi SenM,
As I know, when we have two operators act on a function, in general, we cannot change their order arbitrarily like our mood.
So $\hat{A}\hat{B}\psi(x)$ means that operator $\hat{B}$ must act on function $\psi(x)$ first, and then operator $\hat{A}$ acts on the result of $\hat{B}\psi(x)$ later, that's the rule.
For example, we have two operators $\hat{A}$ and $\hat{B}$, with
$$\hat{A} \psi(x)=a\psi(x)$$ $$\hat{B}\psi(x)=b\psi(x)$$ (just that function multiply by a real number $a,b$)
The formula of commutator of 2 operators is:
$$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$$
So their commutator is $$[\hat{A},\hat{B}]\psi(x)=(\hat{A}\hat{B}-\hat{B}\hat{A})\psi(x)=ab\psi(x)-ba\psi(x)=0$$
We say that they are "commute" with each other, so you can act $\hat{A}$ on $\psi(x)$ first or $\hat{B}$ on $\psi(x)$first, it doesn't matter, same result.
Another example, we have two operators $\hat{x}$ and $\hat{p}$ (momentum operator), with $$\hat{x}\psi(x)=x⋅\psi(x)$$ $$\hat{p}\psi(x)=-i\hbar\frac {\partial } {\partial x}\psi(x)$$
Their commutator is $$[\hat{x},\hat{p}]\psi(x)=(\hat{x}\hat{p}-\hat{p}\hat{x})\psi(x)=x⋅\left[-i\hbar\frac {\partial } {\partial x}\psi(x)\right]+i\hbar\frac {\partial } {\partial x}\left[x⋅\psi(x)\right]=i\hbar\psi(x)≠0$$
So they are "not commute" with each other, so if we have $\hat{x}\hat{p}\psi(x)$, we can't let $\hat{x}$ acts on $\psi(x)$ before $\hat{p}$, just try and you will see the different result!
So the calculation of "commutator" (just in this field, I don't know about commutator in abstract algebra like Group, Ring, Field,...) just let us know that the order of operators is arbitrary or not, which one can act first.
Too long example! Back to your operators $\hat{T}$ & $\hat{T'}$, I see that if $\gamma$ is a constant, they are "commute" with each other, can't be $-2i\hbar\frac {d} {dx}$!
Something wrong? Could you upload the whole question/exercise for us?

 

[SemM]

Something wrong? Could you upload the whole question/exercise for us?

Thanks Son!

Take:

$T = (ihd/dx + \gamma)$
$T' = (-ihd/dx + \gamma)$

Multiply out:

TT' = $(ihd/dx + \gamma) \times (-ihd/dx + \gamma) = h^2 d^2/dx^2 + ihd/dx \gamma (0) -\gamma ihd/dx + \gamma^2$
T'T = $(-ihd/dx + \gamma) \times (ihd/dx + \gamma) = h^2 d^2/dx^2 - ihd/dx \gamma (0) +\gamma ihd/dx + \gamma^2$

This yields :

$[TT'] = h^2 d^2/dx^2 -\gamma ihd/dx + \gamma^2 - (h^2 d^2/dx^2 +\gamma ihd/dx + \gamma^2)$

$[TT'] = -2\gamma ihd/dx$

Can you confirm that this is not wrong, or if it is why that minus doesn't disappear in the second non-vanishing term $-\gamma ihd/dx - (\gamma ihd/dx)$?

 

[BvU]

What is it that makes $ih\; {d\over dx} \gamma(0)$ disappear ? (bearing in mind the thing is an operator) ?

 

[Nguyen Son]

Sorry for my laziness
https://imgur.com/a/vF0NN
So $\left[\hat{T},\hat{T'}\right]=0$

 

[SemM]

What is it that makes $ih\; {d\over dx} \gamma(0)$ disappear ? (bearing in mind the thing is an operator) ?

No, this is the derivation of a constant $\gamma$ times 0. That makes 0.

 

[SemM]

Sorry for my laziness
https://imgur.com/a/vF0NN
So $\left[\hat{T},\hat{T'}\right]=0$

I am sorry, but I think you all have made a mistake here.

The sequence of operation of:

$ihd/dx \times \gamma \ne \gamma \times ihd/dx$

The former yields the derivation of the constant $\gamma$ which gives 0. The second gives the $\gamma i h$ times derivative operator, which can then act on some function $\psi$
Please check! That is the whole core of this post.

 

[BvU]

No, this is the derivation of a constant $\gamma$ times 0. That makes 0.

Where did that (0) come from ?

Have you a good idea of what a commutator is ?

 

[SemM]

Where did that (0) come from ?

Have you a good idea of what a commutator is ?

Sorry for confusion, that zero is just to say that that term is 0. Please see my previous post.

 

[BvU]

It is not. The derivative of something times a constant is that constant times the derivative.

 

[SemM]

It is not. The derivative of something times a constant is that constant times the derivative.

But the function $\psi$ is multiplied to the operators at last, because it is on the right. Doesn't the position (left or right) supersede the nature of what is multiplied to the operator?

 

[BvU]

Correct. So ${d\over dx} \gamma\Psi - \gamma {d\over dx }\Psi = 0$

 

[SemM]

Correct. So ${d\over dx} \gamma\Psi - \gamma {d\over dx }\Psi = 0$

What you write here I have no problem with. However, I am pretty sure that what happens in the brackets would be first "to decide"? Then outside the brackets ($\psi$) comes later?

 

[BvU]

Uclear what you mean. $T, T', TT', T'T$ and $[T,T']$ are all operators.
I am referring to the cross terms in your commutators: unlike you think, they cancel. Post #13 () should be clear to you, but I suspect it isn't, given your reply in #14.

 

[SemM]

Uclear what you mean. $T, T', TT', T'T$ and $[T,T']$ are all operators.
I am referring to the cross terms in your commutators: unlike you think, they cancel. Post #13 () should be clear to you, but I suspect it isn't, given your reply in #14.

This was actually taught to me by a Professor in QM, that :

$\gamma d/dx \ne d/dx \gamma$

because the latter is zero

 

[BvU]

Goes to show that even a professorate isn't adequate protection -- but I suspect the poor chap is misquoted somehow.

Ask any mathematician (or math student): With gamma constant, $$ {d\over dx }\Bigl \{ \gamma [{\rm something}] \Bigr \} = \Biggl \{ {d\over dx }\gamma\Biggr \} [{\rm something}] + \gamma {d\over dx } [{\rm samething}] = 0 +\gamma {d\over dx } [{\rm samething}] = \gamma {d\over dx } [{\rm samething}]$$

 

[stevendaryl]

This was actually taught to me by a Professor in QM, that :

$\gamma d/dx \ne d/dx \gamma$

because the latter is zero

That is not correct. \frac{d}{dx} \gamma \neq 0. Why not?

What's confusing is that an expression such as \gamma is ambiguous:

• It might be interpreted as a function, whose value is constant.
• It might be interpreted as an operator on functions, that returns a multiple of the original function.

An operator is something that takes a function and returns a different function.

Commutators work on operators. When you write a product of two operators A B the meaning of that is NOT A acting on B. The meaning is that A B is a new operator whose action on a function f(x) is given by:

A B f(x) = A (B f(x))

First, you operate using B. The result is a new function. Then you let A operate on that.

So the meaning of \frac{d}{dx} \gamma is NOT "take the derivative of the constant function \gamma". As an operator, it is defined by:

\frac{d}{dx} \gamma f(x) = \frac{d}{dx} (\gamma f(x)) = \gamma \frac{d}{dx} f(x) = \gamma \frac{df}{dx}

First, you let \gamma operate on f(x) to produce a constant multiple of the original function, \gamma f(x). Then you let \frac{d}{dx} operate on that.

So the result of letting \frac{d}{dx} \gamma act on f(x) is to produce \gamma \frac{df}{dx}. Which is not zero.

As an operator equation, we can write:

\frac{d}{dx} \gamma = \gamma \frac{d}{dx}

In other words, as operators, \gamma and \frac{d}{dx} commute.

 

[SemM]

As an operator equation, we can write:

\frac{d}{dx} \gamma = \gamma \frac{d}{dx}

In other words, as operators, \gamma and \frac{d}{dx} commute.

This was completely new to me!

So, when $\gamma$ is a constant, $\frac{d}{dx} \gamma \ne 0$ , when $\frac{d}{dx}$ acts on some function f?

 

[SemM]

Goes to show that even a professorate isn't adequate protection -- but I suspect the poor chap is misquoted somehow.

I think he has been misquoted. I am probably to blame, not him.

 

[SemM]

Ask any mathematician (or math student): With gamma constant, $$ {d\over dx }\Bigl \{ \gamma [{\rm something}] \Bigr \} = \Biggl \{ {d\over dx }\gamma\Biggr \} [{\rm something}] + \gamma {d\over dx } [{\rm samething}] = 0 +\gamma {d\over dx } [{\rm samething}] = \gamma {d\over dx } [{\rm samething}]$$

Does this mean that this accounts ALSO for variables (which I wouldn't think so). So, this would be correct:

$xd/dx \psi \ne d/dx x \psi$

right?

 

[Nguyen Son]

Sure, because $$\frac{d}{dx}(x\psi)=x\frac{d\psi}{dx}+\psi\frac{dx}{dx}=x\frac{d\psi}{dx}+\psi \neq x\frac{d\psi}{dx}$$

 

[SemM]

Sure, because $$\frac{d}{dx}(x\psi)=x\frac{d\psi}{dx}+\psi\frac{dx}{dx}=x\frac{d\psi}{dx}+\psi \neq x\frac{d\psi}{dx}$$

Thanks!

 

[Nguyen Son]

You are welcome