# Another derivation of canonical position-momentum commutator relation

I am reading Martin Plenio's lecture notes on Quantum Mechanics from the Imperial College. In page 63 he wants to prove the relation $$[\hat{x},\hat{p}]=i\hbar$$ via $$\langle x | [\hat{x},\hat{p}]|\psi\rangle=\langle x | \hat{x}\hat{p}-\hat{p}\hat{x}|\psi\rangle \overset{?}{=} \langle x | i\hbar|\psi\rangle.\tag{1}$$

I have proven that $$\langle x | \hat{p}|\psi\rangle = \frac{\hbar}{i}\frac{\partial}{\partial x}\langle x |\psi\rangle.$$

From (1) I get $$\langle x | \hat{x}\hat{p}-\hat{p}\hat{x}|\psi\rangle = x \langle x | \hat{p}|\psi\rangle - \langle x | \frac{\hbar}{i}\frac{\partial}{\partial x}\left(\int dx\ x|x\rangle\langle x|\psi\rangle\right),$$

where $$\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial x}$$ and $$\hat{x} = \int dx\ x |x\rangle\langle x|.$$

If I could get the bra $\langle x |$ into the integral then I would get $$x \langle x | \hat{p}|\psi\rangle - \frac{\hbar}{i}\frac{\partial}{\partial x}\left( x\langle x|\psi\rangle\right),$$

which would lead to the desired result. However I don't see how getting the bra into the integral would be possible. Any help is appreciated (I have seen the proof in other questions here, but they take $|\psi\rangle$ as an eigenvector of $\hat{x}$). How could I make progress with this way of proving the relation? Any help is very much appreciated!

• Please clean up your notation. How do you prove something "via" (1), when (1) is just some number instead of an actual equation? What does "from (1) I get" mean? You also need to clean up your usage of $x$, which is overloaded as both an integration variable and a free variable. Your fourth equation looks basically like "x = \int f(x) dx$" which is nonsensical. Commented Sep 5, 2016 at 4:04 • @knzhou I have made things clearer. The notation is from Plenio's notes. Commented Sep 5, 2016 at 4:24 • I'm thinking part of the problem you're having his that you have$x$both inside the integral as a dummy variable and outside. Try using:$$\hat{x} = \int \operatorname{d}y\ y |y\rangle\langle y|,$$ and $$\langle x|y\rangle = \delta(x-y),$$ and things might be clearer. Commented Sep 5, 2016 at 4:28 ## 2 Answers Although I like knzhou's answer, I'd write it slightly differently. Start with the identtity $${\cal I} = \int |x'\rangle \langle x'| dx' ,$$ which one then inserts into the second expression in (1) $$\int \langle x| (\hat{x}\hat{p}-\hat{p}\hat{x})(|x'\rangle \langle x'|) |\psi\rangle dx'$$ Next we define the wave function$\langle x'|\psi\rangle = \psi(x')$, so that we get $$\int \langle x|x'\rangle (\hat{x}\hat{p}-\hat{p}\hat{x})\psi(x') dx'$$ Then we get the Dirac delta which removes the integral $$\int \delta(x-x') (\hat{x}\hat{p}-\hat{p}\hat{x})\psi(x') dx' = (\hat{x}\hat{p}-\hat{p}\hat{x})\psi(x) .$$ Express the operators in terms of their definitions and then evaluate the derivatives $$x (-i\hbar)\partial_x \psi(x) - x (-i\hbar)\partial_x [x\psi(x)] = i\hbar \psi(x) .$$ The result is the same as$\langle x|i\hbar|\psi\rangle$. Still pendantic, but not painfully so. :-) • Elegant, cheers! Commented Sep 5, 2016 at 5:09 • This is indeed another way of getting at the answer, but it's important to note that the$\hat{x}$you're using is not the same as the$\hat{x}$I'm using. In your answer,$\hat{x}$is a differential operator that acts on functions, while in mine,$\hat{x}$is just a regular operator that acts on kets. Commented Sep 5, 2016 at 5:11 • (Of course, the two pictures are equivalent! The former is just the latter, specialized to position space.) Commented Sep 5, 2016 at 5:14 • @knzhou: point taken, but as you say, they are equivalent. Commented Sep 5, 2016 at 6:01 The problem is that several$x$'s are being mixed up. Let me rewrite part of your first result making the distinction clear. Suppose we want to simplify the quantity $$\langle x | \hat{p} \hat{x} | \psi \rangle.$$ We first use the identity $$1 = \int dx' |x' \rangle \langle x'|$$ and insert this 'factor of$1$' into the expression, to get $$\langle x | \hat{p} \hat{x} | \psi \rangle = \langle x | \hat{p} \hat{x} \left( \int dx' | x' \rangle \langle x' | \psi \rangle \right).$$ Here,$x'$is an integration variable. Next, we act with the$\hat{x}$operator to get $$\langle x | \hat{p} \left( \int dx' x' | x' \rangle \langle x' | \psi \rangle \right).$$ Now we need to act with the$\hat{p}$operator. As you stated in your post, this operator differentiates the coefficient of$| x' \rangle$, giving $$\langle x | \left( \int dx' \frac{\partial}{\partial x'} \left( x' \langle x' | \psi \rangle \right) | x' \rangle \right).$$ Finally, we act with the bra$\langle x| $, using the result $$\langle x | x' \rangle = \delta(x-x').$$ This simplifies the integral to $$\int dx' \delta(x-x') \frac{\partial}{\partial x'} \left( x' \langle x' | \psi \rangle \right) = \frac{\partial}{\partial x} \left( x \langle x | \psi \rangle \right).$$ From this point onward, the rest of the derivation is ordinary calculus. In practice, you probably wouldn't want to be as pedantic as I'm being above (this is a really slow way of doing it), but mixing up the$x$'s can be very confusing the first time around. Let me explain the action of$\hat{p}$. Start from the expression you know, $$\langle x| \hat{p} | \psi \rangle = \frac{\hbar}{i} \frac{\partial}{\partial x} \langle x | \psi \rangle.$$ This equation is saying that the$|x \rangle$component of$\hat{p} | \psi \rangle$is the right-hand side. That is,$\hat{p} |\psi \rangle$itself is $$\hat{p} | \psi \rangle = \frac{\hbar}{i} \int dx \frac{\partial}{\partial x} \left( \langle x | \psi \rangle\right) |x \rangle.$$ Note that here, I'm not using$x'$notation, because there's only one kind of$x$involved. Some people like to always put a prime for an integration variable, though. What this equation is saying is that$\hat{p}$acts on$|\psi \rangle$by differentiating the$|x \rangle$component of$|\psi \rangle$. However, this is totally independent of the fact that$|\psi \rangle$is the state of your system; this is simply how$\hat{p}$acts on all vectors. (In fact, this is a common way to define$\hat{p}$in the first place.) This is exactly what we did above, where the$|x'\rangle$component of our vector in parentheses was$x' \langle x' | \psi \rangle$. • Hi, it is not very clear to me why$\hat{p}$enters the integral and acts only on the coefficient of$|x'\rangle$. Thank you for your excellent answer. Commented Sep 5, 2016 at 4:47 • @VladimirVargas Yeah, I made a jump there. I'll edit in a justification. Commented Sep 5, 2016 at 4:50 • Oh, I see! Maybe practice will give me that third eye to see things more quickly. Although now I'm not sure that the notes I'm reading are the best for a quantum mechanics beginner. Thank you very much. Commented Sep 5, 2016 at 5:08 • @VladimirVargas Yeah, usually people start with flippiefanus's picture. (Note, however, that his expression$\hat{p} = -i\hbar \partial_x\$ only works if all functions are position-space coefficients, which can be confusing if you want to, say, switch between position space and momentum space.) Commented Sep 5, 2016 at 5:13
• With a little practice, though, you'll be able to do this as fast as you can write. It looks like there are a lot of rules, but I already used almost every rule there is in my answer! There's not too much to learn beyond that. Commented Sep 5, 2016 at 5:13
The text passage is a discussion about a derivation of the canonical position-momentum commutator relation in quantum mechanics. It includes a question about how to proceed with a specific step in the proof.
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# Another derivation of canonical position-momentum commutator relation

I am reading Martin Plenio's lecture notes on Quantum Mechanics from the Imperial College. In page 63 he wants to prove the relation $$[\hat{x},\hat{p}]=i\hbar$$ via $$\langle x | [\hat{x},\hat{p}]|\psi\rangle=\langle x | \hat{x}\hat{p}-\hat{p}\hat{x}|\psi\rangle \overset{?}{=} \langle x | i\hbar|\psi\rangle.\tag{1}$$

I have proven that $$\langle x | \hat{p}|\psi\rangle = \frac{\hbar}{i}\frac{\partial}{\partial x}\langle x |\psi\rangle.$$

From (1) I get $$\langle x | \hat{x}\hat{p}-\hat{p}\hat{x}|\psi\rangle = x \langle x | \hat{p}|\psi\rangle - \langle x | \frac{\hbar}{i}\frac{\partial}{\partial x}\left(\int dx\ x|x\rangle\langle x|\psi\rangle\right),$$

where $$\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial x}$$ and $$\hat{x} = \int dx\ x |x\rangle\langle x|.$$

If I could get the bra $\langle x |$ into the integral then I would get $$x \langle x | \hat{p}|\psi\rangle - \frac{\hbar}{i}\frac{\partial}{\partial x}\left( x\langle x|\psi\rangle\right),$$

which would lead to the desired result. However I don't see how getting the bra into the integral would be possible. Any help is appreciated (I have seen the proof in other questions here, but they take $|\psi\rangle$ as an eigenvector of $\hat{x}$). How could I make progress with this way of proving the relation? Any help is very much appreciated!

• Please clean up your notation. How do you prove something "via" (1), when (1) is just some number instead of an actual equation? What does "from (1) I get" mean? You also need to clean up your usage of $x$, which is overloaded as both an integration variable and a free variable. Your fourth equation looks basically like "x = \int f(x) dx$" which is nonsensical. Commented Sep 5, 2016 at 4:04 • @knzhou I have made things clearer. The notation is from Plenio's notes. Commented Sep 5, 2016 at 4:24 • I'm thinking part of the problem you're having his that you have$x$both inside the integral as a dummy variable and outside. Try using:$$\hat{x} = \int \operatorname{d}y\ y |y\rangle\langle y|,$$ and $$\langle x|y\rangle = \delta(x-y),$$ and things might be clearer. Commented Sep 5, 2016 at 4:28 ## 2 Answers Although I like knzhou's answer, I'd write it slightly differently. Start with the identtity $${\cal I} = \int |x'\rangle \langle x'| dx' ,$$ which one then inserts into the second expression in (1) $$\int \langle x| (\hat{x}\hat{p}-\hat{p}\hat{x})(|x'\rangle \langle x'|) |\psi\rangle dx'$$ Next we define the wave function$\langle x'|\psi\rangle = \psi(x')$, so that we get $$\int \langle x|x'\rangle (\hat{x}\hat{p}-\hat{p}\hat{x})\psi(x') dx'$$ Then we get the Dirac delta which removes the integral $$\int \delta(x-x') (\hat{x}\hat{p}-\hat{p}\hat{x})\psi(x') dx' = (\hat{x}\hat{p}-\hat{p}\hat{x})\psi(x) .$$ Express the operators in terms of their definitions and then evaluate the derivatives $$x (-i\hbar)\partial_x \psi(x) - x (-i\hbar)\partial_x [x\psi(x)] = i\hbar \psi(x) .$$ The result is the same as$\langle x|i\hbar|\psi\rangle$. Still pendantic, but not painfully so. :-) • Elegant, cheers! Commented Sep 5, 2016 at 5:09 • This is indeed another way of getting at the answer, but it's important to note that the$\hat{x}$you're using is not the same as the$\hat{x}$I'm using. In your answer,$\hat{x}$is a differential operator that acts on functions, while in mine,$\hat{x}$is just a regular operator that acts on kets. Commented Sep 5, 2016 at 5:11 • (Of course, the two pictures are equivalent! The former is just the latter, specialized to position space.) Commented Sep 5, 2016 at 5:14 • @knzhou: point taken, but as you say, they are equivalent. Commented Sep 5, 2016 at 6:01 The problem is that several$x$'s are being mixed up. Let me rewrite part of your first result making the distinction clear. Suppose we want to simplify the quantity $$\langle x | \hat{p} \hat{x} | \psi \rangle.$$ We first use the identity $$1 = \int dx' |x' \rangle \langle x'|$$ and insert this 'factor of$1$' into the expression, to get $$\langle x | \hat{p} \hat{x} | \psi \rangle = \langle x | \hat{p} \hat{x} \left( \int dx' | x' \rangle \langle x' | \psi \rangle \right).$$ Here,$x'$is an integration variable. Next, we act with the$\hat{x}$operator to get $$\langle x | \hat{p} \left( \int dx' x' | x' \rangle \langle x' | \psi \rangle \right).$$ Now we need to act with the$\hat{p}$operator. As you stated in your post, this operator differentiates the coefficient of$| x' \rangle$, giving $$\langle x | \left( \int dx' \frac{\partial}{\partial x'} \left( x' \langle x' | \psi \rangle \right) | x' \rangle \right).$$ Finally, we act with the bra$\langle x| $, using the result $$\langle x | x' \rangle = \delta(x-x').$$ This simplifies the integral to $$\int dx' \delta(x-x') \frac{\partial}{\partial x'} \left( x' \langle x' | \psi \rangle \right) = \frac{\partial}{\partial x} \left( x \langle x | \psi \rangle \right).$$ From this point onward, the rest of the derivation is ordinary calculus. In practice, you probably wouldn't want to be as pedantic as I'm being above (this is a really slow way of doing it), but mixing up the$x$'s can be very confusing the first time around. Let me explain the action of$\hat{p}$. Start from the expression you know, $$\langle x| \hat{p} | \psi \rangle = \frac{\hbar}{i} \frac{\partial}{\partial x} \langle x | \psi \rangle.$$ This equation is saying that the$|x \rangle$component of$\hat{p} | \psi \rangle$is the right-hand side. That is,$\hat{p} |\psi \rangle$itself is $$\hat{p} | \psi \rangle = \frac{\hbar}{i} \int dx \frac{\partial}{\partial x} \left( \langle x | \psi \rangle\right) |x \rangle.$$ Note that here, I'm not using$x'$notation, because there's only one kind of$x$involved. Some people like to always put a prime for an integration variable, though. What this equation is saying is that$\hat{p}$acts on$|\psi \rangle$by differentiating the$|x \rangle$component of$|\psi \rangle$. However, this is totally independent of the fact that$|\psi \rangle$is the state of your system; this is simply how$\hat{p}$acts on all vectors. (In fact, this is a common way to define$\hat{p}$in the first place.) This is exactly what we did above, where the$|x'\rangle$component of our vector in parentheses was$x' \langle x' | \psi \rangle$. • Hi, it is not very clear to me why$\hat{p}$enters the integral and acts only on the coefficient of$|x'\rangle$. Thank you for your excellent answer. Commented Sep 5, 2016 at 4:47 • @VladimirVargas Yeah, I made a jump there. I'll edit in a justification. Commented Sep 5, 2016 at 4:50 • Oh, I see! Maybe practice will give me that third eye to see things more quickly. Although now I'm not sure that the notes I'm reading are the best for a quantum mechanics beginner. Thank you very much. Commented Sep 5, 2016 at 5:08 • @VladimirVargas Yeah, usually people start with flippiefanus's picture. (Note, however, that his expression$\hat{p} = -i\hbar \partial_x\$ only works if all functions are position-space coefficients, which can be confusing if you want to, say, switch between position space and momentum space.) Commented Sep 5, 2016 at 5:13
• With a little practice, though, you'll be able to do this as fast as you can write. It looks like there are a lot of rules, but I already used almost every rule there is in my answer! There's not too much to learn beyond that. Commented Sep 5, 2016 at 5:13
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The text passage is a discussion about a derivation of the canonical position-momentum commutator relation in quantum mechanics. It includes a question about how to proceed with a specific step in the proof.

TYPE: SINGLE_QUESTION