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Under a change of basis i.e., transforming from one orthonormal $\{|\phi_n\rangle\}$ base to another $\{|\phi^\prime_n\rangle\}$ (when looked from a passive point of view) implies that the state doesn't change but only the "components" change: $$|\psi\rangle=\sum\limits_{n}\langle\phi_n|\psi\rangle|\phi_n\rangle=\sum\limits_{n}\langle\phi^\prime_n|\psi\rangle|\phi^\prime_n\rangle$$ where $|\phi^\prime_n\rangle=U|\phi_n\rangle$, $U$ being an unitary operator which ensures the ortonormality of two bases. Using $$\langle\phi_n|\psi\rangle\to\langle\phi^\prime_n|\psi\rangle=\langle\phi_n|U^{-1}|\psi\rangle\tag{A}$$ one can also say that "net result'' of a basis transformation as an active transformation where the base is left unchanged but any state $|\psi\rangle$ changes to $U^{-1}|\psi\rangle$. In two different bases, an operator $A$ can be expanded as: $$A=\sum\limits_{m}\sum\limits_{n}|\phi_m\rangle\langle\phi_m|A|\phi_n\rangle\langle\phi_n|=\sum\limits_{m}\sum\limits_{n}|\phi^\prime_m\rangle\langle\phi^\prime_m|A|\phi^\prime_n\rangle\langle\phi^\prime_n|.$$ Using $$\langle\phi_m|A|\phi_n\rangle\to\langle\phi^\prime_m|A|\phi^\prime_n\rangle=\langle\phi_m|U^{-1}AU|\phi_n\rangle.\tag{1}$$ Hence, the result of the basis transformation, is an active transformation where the operator changes as $A\to U^{-1}AU$ but the base is left unchanged.

What happens to the various object under basis transformation? It is easy to see that for any $|\phi\rangle,|\psi\rangle$ in the Hilbert space, the inner products remain unchanged: $$\langle\psi|\phi\rangle\to \langle\psi| UU^{-1}\phi\rangle=\langle\psi|\phi\rangle.$$ Also $$\langle\psi|A|\phi\rangle\to \langle\psi|U(U^{-1}AU)|U^{-1}\phi\rangle=\langle\psi|A|\phi\rangle\tag{2}$$ i.e., objects such as $\langle\psi|A|\phi\rangle$ do not chnage.

Doesn't Eq.(2) contradict Eq.(1)? Eq.(2) says objects such as $\langle\psi|A|\phi\rangle$ do not change. But Eq.(1) says matrix elements of an operator changes under basis transformation. But Eq.(1) is just a special case of Eq.(2). What is wrong here with my understanding?

Note: I'm trying to compare active transformation with passive transformation. Either the basis changes or the system changes. Note that Eq.(A) and Eq.(1) says if the base is assumed to be fixed, then both the state and the operator must change. I'm reading this from Modern Quantum Mechanics by Sakurai, section 1.5

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  • $\begingroup$ Which picture are you in: Schrodinger or Heisenberg? $\endgroup$ – probably_someone Apr 25 '18 at 19:31
  • $\begingroup$ I'm trying to compare active transformation with passive transformation. Either the basis changes or the system changes. Note that Eq.(A) and Eq.(1) says if the base is assumed to be fixed, then both the state and the operator must change. I'm reading this from Modern Quantum Mechanics by Sakurai, section 1.5 @probably_someone $\endgroup$ – SRS Apr 25 '18 at 19:35
  • $\begingroup$ In (1) you are considering different states before ($|\phi_m>,|\phi_n>$) and after ($|\phi_m'>,|\phi_n'>$) the transformation, while in (2) you are using the same states ($|\phi>,|\psi>$)? $\endgroup$ – user8153 Apr 25 '18 at 19:46
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    $\begingroup$ May I ask what does "$\dotso \to \dotso$" mean to you, in words? I have the impression that you're using this notation with two different meanings. $\endgroup$ – pglpm Apr 25 '18 at 20:42
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My impression is that you're using the symbol "$\to$" with two different meanings in eqs (1) and (2).

In eqs (1) and (A), "$X \to Y$" means "As you change basis, you must abandon $X$ as the representative of your object, and use $Y$ instead". In this case a set of $n\times n$ numbers, that represent the operator $A$. You see that this is giving you a prescription for a passive transformation. The symbol "$\to$" doesn't really mean "gets mapped to", but "use ... instead of ...".

In eq. (2), "$X \to Y$" seems to mean "When you make the transformation, $X$ is mapped into $Y$ by this transformation". In this case $X$ is a scalar product, and it is invariant under unitary transformations. You see that this is telling you what the effect of an active transformation is. Here the symbol "$\to$" really means "gets mapped to", not "use ... instead of ...".

In neither case does the symbol "$\to$" have its standard functional meaning, say like $U\colon \mathcal{H} \to \mathcal{H}$ or $U\colon \lvert\phi\rangle \mapsto \lvert \phi'\rangle$.

I'm not completely sure, but my impression is that your conundrum lies there.

A "representation" in a given basis is just an isomorphism $P$ from your Hilbert space $\mathcal{H}$ to the complex vector space $\boldsymbol{\mathrm{C}}^n$: $$R\colon\mathcal{H} \to \boldsymbol{\mathrm{C}}^n\quad\text{by} \quad \lvert\phi\rangle \mapsto (x_i).$$ For the moment, forget about the "basis". A representation is simply a map that sends vectors in $\mathcal{H}$ to vectors in $\boldsymbol{\mathrm{C}}^n$, preserving the scalar product. I can specify such a map even without choosing a basis. For example, for a spin-1/2 system, $\lvert +z \rangle \mapsto (1,1)/\sqrt{2}$ and $\lvert +x \rangle \mapsto (0,1)$ together fully specify such a map. The "basis" are simply those vectors in $\mathcal{H}$ that are mapped to the special vectors $(1,0,\dotsc)$, $(0,1,\dotsc)$, etc. in $\boldsymbol{\mathrm{C}}^n$.

This isomorphism also induces another isomorphism $R_*$ between the space $L_{\mathcal{H}}$ of operators on $\mathcal{H}$ and the space of complex matrices $M_{n,n}(\boldsymbol{\mathrm{C}})$: $$R_*\colon L_{\mathcal{H}} \to M_{n,n}(\boldsymbol{\mathrm{C}}) \quad\text{by}\quad A \mapsto (Z_{ij}).$$ This is how the relationship between all these maps looks like:
commutative diagram

The same scheme holds if you replace "$A$" with "$U$" or any other linear operator on $\mathcal{H}$.

When we "work in a given basis", we have just chosen to move to $\boldsymbol{\mathrm{C}}^n$ and $M_{n,n}(\boldsymbol{\mathrm{C}})$ and to do our calculations there, using standard matrix multiplication. When we've finished we can go back to $\mathcal{H}$ by using the inverse map $R^{-1}$.

When we do a "change of basis" we are considering a new isomorphism $R'$ between $\mathcal{H}$ and $\boldsymbol{\mathrm{C}}^n$, with the induced one $R'_*$ on the space of linear operators. The relationship with the previous representation is
commutative diagram

Now when you consider a unitary operator $U$ – I'm not thinking of a change of basis: just a unitary operator acting on $\mathcal{H}$ – then this operator induces a "superoperator" $\hat{U}$ on the space of operators of $\mathcal{H}$: $$\hat{U}\colon L(\mathcal{H})\to L(\mathcal{H}) \quad\text{by}\quad A \mapsto U^{-1}A U.$$

Using a representation $R$ or $R'$, this superoperator can also be represented – though not as a simple matrix: it's a set of matrices multiplying by left and right, called Kraus matrices; see eg https://arxiv.org/abs/quant-ph/9906023. Here's the scheme:
commutative diagram

Now, when you write "$A \to U^{-1}A U$", you're considering the map $\hat{U}$ in the last diagram. You aren't considering any representation (but if you want you can also consider its representation $R_{**}(\hat{U})$ too, of course). This is an "active" transformation.

When you write "$A_{mn}\to (U^{-1}A U)_{mn}$", instead, you are considering the map $R'_* \circ R_*^{-1}$ in the second diagram, which brings you from one matrix representation $R_*(A)$ of $A$, to another, $R'_*(A)$. This is a "passive" transformation – a bad name, in my opinion, because it isn't a transformation at all from $\mathcal{H}$'s perspective.

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  • $\begingroup$ Agreed. Exactly what I am saying! On the one hand OP is talking about how the matrix elements of an operator change w.r.t to basis transformations on the other hand he is looking at the expectation value. $\endgroup$ – sagittarius_a Apr 26 '18 at 6:07
  • $\begingroup$ Agreed. But I think the problem is that the OP keeps on using this "$\to$" (also in the comments to your answer) without realizing that he/she's using it inconsistently. $\endgroup$ – pglpm Apr 26 '18 at 6:57
  • $\begingroup$ @sagittarius_a I've added some diagrams trying to explain the two changes of your answer. $\endgroup$ – pglpm Apr 26 '18 at 8:10
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    $\begingroup$ Absolutely. It’s not all the OP’s fault either, $\to$ is used horribly inconsistently all the time. $\endgroup$ – knzhou Apr 26 '18 at 8:12
  • $\begingroup$ @knzhou Absolutely agree. Also the terms "passive transformation" and "active transformation" are confusing and don't help, in my opinion. $\endgroup$ – pglpm Apr 26 '18 at 8:15
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I think that you're mixing up change of basis with coordinate transformations.

one can also say that "net result'' of a basis transformation as an active transformation where the base is left unchanged but any state |ψ⟩ changes to U−1|ψ⟩.

A basis transformation does not affect states. It just affects what coordinates are used to represent the state. So if you have some projection operator P that goes from state space to coordinate space, then you can create a new projection operator PU$^{-1}$ that takes states to their new coordinates. It's meaningful to talk about U$^{-1}$ right-acting on P, but not so much to talk about it left-acting on |ψ⟩. What you are calling the "passive" transformation seems to be P -> PU: instead of taking the coordinates as being the coefficients of |ψ$_n$⟩, you're taking them to be the coefficients of U|ψ$_n$⟩. You're then separating out U and calling that the "active" transformation.

In the original basis, getting the coordinates consists of |ψ⟩ -> P|ψ⟩. After the basis transformation, it consists of |ψ⟩ -> PU|ψ⟩. Instead of conceptualizing the coordinate change as replacing P with PU, you're conceptualizing it as replacing |ψ⟩ with U|ψ⟩. The former is dealing only with coordinates, the second is actually changing states and is thus not a "basis transformation" but just a plain old unitary transformation.

Taking PUA|ψ⟩ gives the result of applying A, then taking the new coordinates. That is, it represent "apply A, change basis, take coordinates". If you want instead "apply U, apply A, take coordinates", that's PAU|ψ⟩. And to get from PUA|ψ⟩ to PAU|ψ⟩, you replace A with $U^{-1}AU$. That's why A is changing: you're trying to get A applied to the "change basis state", rather than the changed basis of (A applied to the state). If you do all of your operations with the old states, then apply the new projection operator, A doesn't change.

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There is no contradiction. If you want, you can define your operator $A$ by what is does to the basis states in the system. Usually you have what is called a "partition of unity", i.e.,

$$ \mathrm{id} = \sum_n | \phi_m \rangle \langle \phi_m |, $$

is the identity map. Your second equation then states that

$$ A = \mathrm{id} ~A~ \mathrm{id} $$

There is no unique partition of unity. So you can also take $$ \mathrm{id} = \sum_n | \phi_m ' \rangle \langle \phi_m' |, $$

The coefficients $\langle \phi_m | A | \phi_m\rangle$ tell you how $A$ acts w.r.t to the basis spanned by $|\phi_m\rangle$. But you can re-express them in any basis you want. The trick is always to insert $\mathrm{id}$, i.e.,

$$ \langle \phi_m' | A | \phi_m'\rangle = \langle \phi_m' | \mathrm{id}~ A~\mathrm{id} | \phi_m'\rangle $$

If you take your partition of unity from the $|\phi_m\rangle$ basis you get your first equation.

EDIT

Let me go through this in detail. If you want, your equation (1) tells you how to relate the matrix elements of $A$ in one basis to the matrix elements in another basis. Let me do this for you

$$\begin{align} \langle \phi_m' | A |\phi_n' \rangle &= \langle \phi_m' | \mathrm{id}~ A ~\mathrm{id} |\phi_n' \rangle \\ & = \sum_{pq} \langle \phi_m' | \phi_p \rangle \langle{\phi_p} | A |\phi_q \rangle \langle \phi_q | \phi_n' \rangle \end{align}$$

Now I have to say, the unitary matrix you have given is not the most general one. Typically, a given state will rotate into a multitude of new ones. Using our good old trick:

$$ |\phi_n '\rangle = \mathrm{id}~|\phi_n '\rangle = \sum_q \langle \phi_q|\phi_n '\rangle |\phi_q\rangle = \sum_q \mathcal{U}_{nq} |\phi_q\rangle, $$ where we have defined $\mathcal{U}_{nq} = \langle \phi_q|\phi_n '\rangle $. Can you take it from here?

OK.. let me continue. To make the connection with Sakurai you choose your basis such that $\mathcal{U}_{nq}= \delta_{nq} \mathcal{U}_{nn}$. This means we now have

$$\begin{align} \langle \phi_m' | A |\phi_n' \rangle &= \langle{\phi_m} | \mathcal{U}^{-1}_{mm} A \mathcal{U}_{nn}|\phi_n \rangle \\ &= \langle{\phi_m} | \mathcal{U}^{-1} A \mathcal{U}|\phi_n \rangle \end{align}$$

The general case

Since you where asking, in the general case we have

$$\begin{align} \langle \phi_m' | A |\phi_n' \rangle & = \sum_{pq} \langle \phi_m' | \phi_p \rangle \langle{\phi_p} | A |\phi_q \rangle \langle \phi_q | \phi_n' \rangle \\ & = \sum_{pq} \mathcal{U}^{-1}_{mp} \langle{\phi_p} | A |\phi_q \rangle \mathcal{U}_{qn} \end{align}$$

This you recognize as matrix multiplication and so

$$\begin{align} \langle \phi_m' | A |\phi_n' \rangle = \langle{\phi_m} | \mathcal{U}^{-1} A \mathcal{U}|\phi_n \rangle \end{align}$$

if we write this general $\mathcal{U}$ as

$$ \mathcal{U} = \sum_{qn} ~ \mathcal{U}_{qn} | \phi_q \rangle \langle \phi_n | $$

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  • $\begingroup$ I still don't get it. Is Eq.(2) in my question correct? If yes, let me choose $|\psi\rangle=|\phi_m\rangle$ and $|\phi\rangle=|\phi_n\rangle$ in Eq.(2). With these substitutions, Eq.(2) doesn't give back Eq.(1). Rather it says $\langle\phi_m|A|\phi_n\rangle\to \langle\phi_m|A|\phi_n\rangle$. That's the contradiction I'm talking about. $\endgroup$ – SRS Apr 25 '18 at 20:19
  • $\begingroup$ But these are two different things you are talking about. On the one hand you are talking about the expectation value of $A$ on the other hand you are talking about the matrix elements of $A$ w.r.t some basis. $\endgroup$ – sagittarius_a Apr 25 '18 at 20:43
  • $\begingroup$ I'm not saying your last equation is wrong. In fact, it looks correct. But it contradicts with Sakurai. Assuming your last equation to be more general, how do states and operators transform under it i.e., what are the analogs of Eq.(A) and Eq.(1)? $\endgroup$ – SRS Apr 26 '18 at 5:35
  • $\begingroup$ How do states and operators transform i.e., what are the analogues of Eq.(A) and Eq.(1) acc. to the transformation you wrote? $\endgroup$ – SRS Apr 26 '18 at 6:08
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    $\begingroup$ @SRS You keep writing "$\dotso \to\dotso$", and it seems to me you're using this notation inconsistently. What does "$\dotso \to\dotso$" mean to you? $\endgroup$ – pglpm Apr 26 '18 at 7:00
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Equation (1) should be

$\langle\phi_m|A|\phi_n\rangle =\langle\phi^\prime_mU|A|U^{-1}\phi^\prime_n\rangle=\langle\phi_m^\prime|UAU^{-1}|\phi_n^\prime\rangle=\langle\phi^\prime_m|A^\prime|\phi^\prime_n\rangle. \tag{1}$

The basis transform also changes $A$ into $A^\prime$.

Equation (2) is incorrect, as is the comment immediately following this post.

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  • $\begingroup$ This cannot be correct since this would imply $| \phi_n' \rangle = | \phi_n \rangle$ $\endgroup$ – sagittarius_a Apr 25 '18 at 20:16
  • $\begingroup$ Why should it be $A^\prime$? Under a passive transformation, basis vectors change. Not the abstract operators. Please look at Sakurai's Eq. 1.5.10 @my2cts $\endgroup$ – SRS Apr 25 '18 at 20:17

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