Understanding the Commutator

Elekko

Hi!

In my textbook the explanation of the expectation value in general covering any observable $$Q$$ is:

$$\overline{Q} = \int \Psi^\ast (x,t) \hat{Q}\Psi(x,t) dx$$

Then they define the commutator as:

$$[\hat{A},\hat{B}] = \hat{A}\hat{B}-\hat{B}\hat{A}$$

Now for position and momentum they give directly it as:

$$[\hat{x},\hat{p}] = i\hbar$$

Now my question is: how is that possible? I know what the operators of the observables for both position and momentum are, but I do not understand why the commutator can be derived trough the general form of EXPECTATION value?? Expectation value is a value but the commutator is an operator on operators??
Can anyone help me with this confusion? Im new to quantum physics. Thanx

Related Quantum Physics News on Phys.org

tiny-tim

Homework Helper
Welcome to PF!

Hi Elekko! Welcome to PF! (write "itex" rather than "tex", and it won't keep starting a new line )
$$[\hat{x},\hat{p}] = i\hbar$$

… Expectation value is a value but the commutator is an operator on operators??
Are you asking how the LHS is an operator, but the RHS is only a number?

It's because the RHS is really $i\hbar I$, where I is the identity operator (the operator that converts anything to itself) … we don't usually bother to write it. Fredrik

Staff Emeritus
Gold Member
Now my question is: how is that possible? I know what the operators of the observables for both position and momentum are, but I do not understand why the commutator can be derived trough the general form of EXPECTATION value?? Expectation value is a value but the commutator is an operator on operators??
Can anyone help me with this confusion? Im new to quantum physics. Thanx
I don't understand your questions. You may want to clarify them a bit. The words "through the general form of expectation value" are especially confusing. Have you seen a derivation of a commutator that involves an expectation value? What do you mean by "operator on operators"? It's just an operator. The "product" AB is defined by (AB)f=A(Bf) where f is a square-integrable function from $\mathbb R^3$ to $\mathbb C$. So the "product" AB is just the composition $A\circ B$. The sum of two operators is defined by (A+B)f=Af+Bf. So [A,B]f=A(Bf)-B(Af).

mikeph

The commutator is not defined from the expectation value, the expectation value necessarily depends on the particular wavefunction whereas the commutator is defined as a property of a pair of operators, regardless of any particular wavefunction.

nonequilibrium

Well, in a sense the commutator itself is indeed an operator on operators, since the input is two operators and it gives a new operator. More exactly if $\mathcal B$ is the space of operators, then the commutator is a function $\mathcal B \times \mathcal B \to \mathcal B : (\hat A, \hat B) \mapsto \hat A \hat B - \hat B \hat A$.

As for the OP's initial question, it seems he's under the impression that the identity $[\hat x , \hat p ] = i\hbar$ was derived using expectation values; he probably thought so because the book seemed to imply it by introducing commutators after introducing expectation values and immediately stating "one can see that $[\hat x , \hat p ] = i\hbar$". I'm not sure if I interpreted the OP correctly though, but if so: no worries, the identity has nothing to do with expectation values. To get the "$[\hat x , \hat p ] = i\hbar$" identity, one writes $[\hat x , \hat p ]$ out in terms of the actual operators (i.e. derivatives) and then applying that to a general $\psi$ one can prove, using some algebra, that this is equal to $i\hbar \psi$.

Elekko

Re: Welcome to PF!

Hi all,

Hi Elekko! Welcome to PF! (write "itex" rather than "tex", and it won't keep starting a new line )

Are you asking how the LHS is an operator, but the RHS is only a number?

It's because the RHS is really $i\hbar I$, where I is the identity operator (the operator that converts anything to itself) … we don't usually bother to write it. So this means that in the case of momentum and position, the "identity" operator is: $\Psi[itex] ??? tiny-tim Science Advisor Homework Helper Hi Elekko! (remember to close your latex with a / ) So this means that in the case of momentum and position, the "identity" operator is: [itex]\Psi$ ???
nooo the identity operator is I

the effect of I on ψ is ψ​

Elekko

Well, in a sense the commutator itself is indeed an operator on operators, since the input is two operators and it gives a new operator. More exactly if $\mathcal B$ is the space of operators, then the commutator is a function $\mathcal B \times \mathcal B \to \mathcal B : (\hat A, \hat B) \mapsto \hat A \hat B - \hat B \hat A$.

As for the OP's initial question, it seems he's under the impression that the identity $[\hat x , \hat p ] = i\hbar$ was derived using expectation values; he probably thought so because the book seemed to imply it by introducing commutators after introducing expectation values and immediately stating "one can see that $[\hat x , \hat p ] = i\hbar$". I'm not sure if I interpreted the OP correctly though, but if so: no worries, the identity has nothing to do with expectation values. To get the "$[\hat x , \hat p ] = i\hbar$" identity, one writes $[\hat x , \hat p ]$ out in terms of the actual operators (i.e. derivatives) and then applying that to a general $\psi$ one can prove, using some algebra, that this is equal to $i\hbar \psi$.
So here, it means that when we derive the commutator for momentum and position to $i\hbar \psi$ the $\psi$ is always applied and it can be seen as and identity operator as tiny-tim mentioned?

Elekko

Hi Elekko! (remember to close your latex with a / )

nooo the identity operator is I

the effect of I on ψ is ψ​
Wow you're quick in replying.
So let me see if I have got it right:

the expectation value $\overline{Q} = \int \Psi^\ast (x,t) \hat{Q}\Psi(x,t) dx$ has nothing to do with to derive the commutator.
The commutator of momentum and position is simply $i\hbar$ by using the observables in the definition of commutator. And $\psi$ can be applied in case of showing it in general?

mikeph

Post #8 doesn't seem correct, (though post #9 does)

To find the commutator, evaluate [AB-BA](phi). If you get zero, then the operators commute, otherwise you will get X(phi) where X is the commutator operator, and you can read off what X is. It is always an operator.

For position and momentum,

[x,p](phi) = iħ*phi ≡ X(phi)

Note the brackets are still used on the right, because the commutator is not simply a number "iħ", it is an operator, equivalent to "multiply argument by iħ".

In this sense, [x,p] is not strictly iħ, it is iħ*, but the multiplication sign is often omitted.

tiny-tim

Homework Helper
The commutator of momentum and position is simply $i\hbar$ by using the observables in the definition of commutator.
See what MikeyW says.
And $\psi$ can be applied in case of showing it in general?
I have no idea what you mean. ψ is any wavefunction

any operator equation A = B means "Aψ = Bψ for any ψ"​

Elekko

I think I got it now guys. Thank you!
Tiny-tim, nevermind. It seems that I don't understand why we even do have to apply $\psi$ on the commutator to begin with. Its just simply xp-px where x = x, p = -ihd/dx

Fredrik

Staff Emeritus
Gold Member
I think I got it now guys. Thank you!
Tiny-tim, nevermind. It seems that I don't understand why we even do have to apply $\psi$ on the commutator to begin with. Its just simply xp-px where x = x, p = -ihd/dx
Right, but unless you have xp-px act on a function, how are you going to prove that [x,p]=iħ?

Last edited:

Elekko

Right, but unless you have xp-px act on a function, how are you going to prove that [x,p]=iħ?
Thanks!
I got it now!
In case of quantum mechanics we use the waveFUNCTION $\psi$ in order to prove the commutator.
I appreciate for all your help guys!

vanhees71

Gold Member
To make things really clear: The commutation relation between position and momentum opertor (in the same direction) follows from the fact that, by definition and as in classical mechanics, the (canonical) momentum of a system is defined by it's operation as a generator of spacial translations.

In the position-space representation, which sometimes is called "wave mechanics", but in fact it's just a representation of quantum theory in the position eigenbasis, the operators thus read
$$\hat{x} \psi(x)=x \psi(x), \quad \hat{p} \psi(x)=-\mathrm{i} \hbar \mathrm{d}_x \psi(x).$$
The commutation relations are found by applying these operators in different orders
$$(\hat{x} \hat{p}-\hat{p} \hat{x})\psi(x)=-\mathrm{i} \hbar [x \mathrm{d}_x \psi(x) - \mathrm{d}_x (x \psi(x))]=\mathrm{i} \hbar \psi(x).$$
This means that for any wave function the commutator acts as by simply multiplying the wave function by $\mathrm{i} \hbar$, and this is written in operator form as
$$[\hat{x},\hat{p}]=\mathrm{i} \hbar \mathbb{1}.$$

Fredrik

Staff Emeritus
Gold Member
I suspect that most people who are new to this don't fully understand how results like $\hat p\hat x\psi(x)=\frac{d}{dx}(x\psi(x))$ follow from the definitions, so I will explain that. I will only consider the 1-dimensional case here, so our wavefunctions are functions from $\mathbb R$ into $\mathbb C$. First note that when $\psi$ denotes such a function, $\psi(x)$ denotes a complex number in its range. So $\psi(x)$ never denotes a function. It's always just a number. This means that an operator like $\hat x$ or $\hat p$ acts on $\psi$, never on $\psi(x)$.

$\hat x$ is defined by specifying a set of functions that will be its domain, and then specifying $\hat x\psi$ for all $\psi$ in that set. Since $\hat x\psi$ is a function with domain $\mathbb R$, the way to do this is to specify $\hat x\psi(x)$ for all $x\in\mathbb R$. (Note that $\hat x\psi(x)$ means $(\hat x\psi)(x)$. $\hat x$ acts on $\psi$ and $\hat x\psi$ acts on $x$).

The domain of $\hat x$ is the set of all square-integrable $\psi:\mathbb R\to\mathbb C$ such that the map $x\mapsto x\psi(x)$ is square-integrable. For each $\psi$ in that set, we define
$$\hat x\psi(y)=y\psi(y),$$ for all $y\in\mathbb R.$ (Note that it never matters what variable symbol is used in a "for all" statement. I chose to use y instead of x because I wanted to make it clear that we don't have to use the same symbol that's used in the notation for the position operator).

$\hat p$ can be defined without even mentioning a variable (like x) that represents a real number. The domain of $\hat p$ is the set of all square-integrable $\psi:\mathbb R\to\mathbb C$ such that $\psi'$ is square-integrable. For all such $\psi$, we define
$$\hat p\psi=-i\psi'.$$ I like to use units such that $\hbar=1$. That's why I don't include $\hbar$ explicitly on the right-hand side.
The "product" $\hat p\hat x$ is defined as follows: For all $\psi:\mathbb R\to\mathbb C$ in the domain of $\hat x$, such that $\hat x\psi$ is in the domain of $\hat p$, we define
$$(\hat p\hat x)\psi=\hat p(\hat x\psi).$$ This definition is the reason why we don't need to insert parentheses when we write things like $\hat p\hat x\psi$.

The other product $\hat x\hat p$ is defined similarly. The domain of the commutator is the intersection of the domains of $\hat p\hat x$ and $\hat x\hat p$. Note that this means that the identity operator that we don't write out on the right-hand side of $[\hat x,\hat p]=i\hbar$ is the identity operator on that set, not the identity operator on the full set of square-integrable functions.

Now, let's use the definitions to evaluate $\hat p\hat x\psi(x)$, where $\psi$ is an arbitrary member of the domain of $[x,p]$.
$$\hat p\hat x\psi(x)=\hat p(\hat x\psi)(x)=-i(\hat x\psi)'(x)=-i\frac{d}{dx}(x\psi(x)).$$ The notation on the right means "-i times the derivative of the map $y\mapsto y\psi(y)$ evaluated at x". The definition of $\hat x$ tells us that that map is precisely what we denote by $\hat x\psi$, so the right-hand side above means $-i(\hat x\psi)'(x)$.

The rest of the calculation that proves the identity $[x,p]=i$ goes like this:
\begin{align}&=-i\big(\psi(x)+x\psi'(x)\big)=-i\big(\psi(x)+\hat x(\psi')(x)\big)=-i\big(\psi(x)+\hat x(i\hat p\psi)(x)\big)\\ &=-i\psi(x)+\hat x\hat p\psi(x)=(-i+\hat x\hat p)\psi(x). \end{align} I used that $\hat x$ is linear here. I haven't proved that, but it's easy to do, using the definition of $\hat x$.

Elekko

I suspect that most people who are new to this don't fully understand how results like $\hat p\hat x\psi(x)=\frac{d}{dx}(x\psi(x))$ follow from the definitions, so I will explain that. I will only consider the 1-dimensional case here, so our wavefunctions are functions from $\mathbb R$ into $\mathbb C$. First note that when $\psi$ denotes such a function, $\psi(x)$ denotes a complex number in its range. So $\psi(x)$ never denotes a function. It's always just a number. This means that an operator like $\hat x$ or $\hat p$ acts on $\psi$, never on $\psi(x)$.

$\hat x$ is defined by specifying a set of functions that will be its domain, and then specifying $\hat x\psi$ for all $\psi$ in that set. Since $\hat x\psi$ is a function with domain $\mathbb R$, the way to do this is to specify $\hat x\psi(x)$ for all $x\in\mathbb R$. (Note that $\hat x\psi(x)$ means $(\hat x\psi)(x)$. $\hat x$ acts on $\psi$ and $\hat x\psi$ acts on $x$).

The domain of $\hat x$ is the set of all square-integrable $\psi:\mathbb R\to\mathbb C$ such that the map $x\mapsto x\psi(x)$ is square-integrable. For each $\psi$ in that set, we define
$$\hat x\psi(y)=y\psi(y),$$ for all $y\in\mathbb R.$ (Note that it never matters what variable symbol is used in a "for all" statement. I chose to use y instead of x because I wanted to make it clear that we don't have to use the same symbol that's used in the notation for the position operator).

$\hat p$ can be defined without even mentioning a variable (like x) that represents a real number. The domain of $\hat p$ is the set of all square-integrable $\psi:\mathbb R\to\mathbb C$ such that $\psi'$ is square-integrable. For all such $\psi$, we define
$$\hat p\psi=-i\psi'.$$ I like to use units such that $\hbar=1$. That's why I don't include $\hbar$ explicitly on the right-hand side.
The "product" $\hat p\hat x$ is defined as follows: For all $\psi:\mathbb R\to\mathbb C$ in the domain of $\hat x$, such that $\hat x\psi$ is in the domain of $\hat p$, we define
$$(\hat p\hat x)\psi=\hat p(\hat x\psi).$$ This definition is the reason why we don't need to insert parentheses when we write things like $\hat p\hat x\psi$.

The other product $\hat x\hat p$ is defined similarly. The domain of the commutator is the intersection of the domains of $\hat p\hat x$ and $\hat x\hat p$. Note that this means that the identity operator that we don't write out on the right-hand side of $[\hat x,\hat p]=i\hbar$ is the identity operator on that set, not the identity operator on the full set of square-integrable functions.

Now, let's use the definitions to evaluate $\hat p\hat x\psi(x)$, where $\psi$ is an arbitrary member of the domain of $[x,p]$.
$$\hat p\hat x\psi(x)=\hat p(\hat x\psi)(x)=-i(\hat x\psi)'(x)=-i\frac{d}{dx}(x\psi(x)).$$ The notation on the right means "-i times the derivative of the map $y\mapsto y\psi(y)$ evaluated at x". The definition of $\hat x$ tells us that that map is precisely what we denote by $\hat x\psi$, so the right-hand side above means $-i(\hat x\psi)'(x)$.

The rest of the calculation that proves the identity $[x,p]=i$ goes like this:
\begin{align}&=-i\big(\psi(x)+x\psi'(x)\big)=-i\big(\psi(x)+\hat x(\psi')(x)\big)=-i\big(\psi(x)+\hat x(i\hat p\psi)(x)\big)\\ &=-i\psi(x)+\hat x\hat p\psi(x)=(-i+\hat x\hat p)\psi(x). \end{align} I used that $\hat x$ is linear here. I haven't proved that, but it's easy to do, using the definition of $\hat x$.
That's a perfect explanation.
Thank you very much!

Fredrik

Staff Emeritus
Gold Member
That's a perfect explanation.
Thank you very much!
You're welcome. Thanks for letting me know that you read it and understood it.

Here's a slightly different version that doesn't even use the d/dx notation. I will not include all the information about domains and stuff this time. Here I denotes the function that takes x to x. To avoid confusion, the identity operator will be denoted by 1 instead of I. We define two operators, Q and D.

Qf=If (This means that for all real numbers x, Qf(x)=(If)(x)=I(x)f(x)=xf(x), so the Q we have defined here is the position operator).
Df=f' (We define the momentum operator by P=-iD, so D=iP).

\begin{align}
DQf(x) &=(If)'(x)=I'(x)f(x)+I(x)f'(x) =f(x)+xf'(x)=f(x)+QDf(x) =(f+QDf)(x) \\
DQf &=f+QDf=(1+QD)f\\
DQ &= 1+QD\\
\left[Q,P\right] &= \left[Q,-iD\right]=-i\left[Q,D\right]=-i(-1)=i.
\end{align}

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving