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I'm having trouble understanding why is $\psi(\Lambda^{-1}p')$ the correct wave function in the Lorentz transformed frame $p' = \Lambda p$.

Suppose the state in frame $O$ is given by

$$ |\Psi\rangle = \int dp\, \psi(p)\, |p\rangle $$

then in frame $O'$ with $p' = \Lambda p$, the same state has representation

$$ |\Psi\rangle = \int dp'\, \psi'(p')\, |p'\rangle $$

where $|p'\rangle = U(\Lambda) |p\rangle =|\Lambda p\rangle$ and $\psi'(q') = \langle q' | \Psi \rangle$ (all quantities unprimed $p, q, . . .$ are in frame $O$, all quantities primed $p', q', . . .$ are in frame $O'$). Calculating now

$$ \langle q' | \Psi \rangle = \langle q | U^{\dagger}(\Lambda)|\Psi\rangle = \int dp\, \psi(p)\, \langle q | U^{\dagger}(\Lambda)\, |p\rangle = \int dp\, \psi(p)\, \delta(q - \Lambda^{-1} p) = \psi(\Lambda q) $$

where the last step follows since $q - \Lambda^{-1} p = 0$ gives $p = \Lambda q$. So all in all $\psi'(q') = \psi(\Lambda q)$ but instead we know the correct answer is $\psi'(q') = \psi(\Lambda^{-1} q')$. Where is the mistake in the reasoning?

Note: similar questions have been asked before but none addresses the issue I'm raising, namely, how to convince oneself that the function space transformation $\psi(\Lambda^{-1} q')$ arises in the way shown above.

konstam56
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    Related (possible duplicate?): http://physics.stackexchange.com/q/95837/ – Kyle Kanos Feb 16 '15 at 14:15
  • Thanks, it's related (as are many others as I noted) but I'm afraid it does not address the issue I raised. I am interested in seeing in detail how the said transformation on a wave function arises due to a passive basis transformation $|p \rangle \mapsto |p' \rangle$ in the Hilbert space, i.e. as a result of calculating the representation $\langle p' | \Psi \rangle$ of the wave function in the new basis. – konstam56 Feb 17 '15 at 03:03
  • You seem to mix something up here - if this is QFT, there is no wavefunction, and also, no momentum basis in the way you wrote it, since the Hilbert space of QFT is always the entire Fock space. – ACuriousMind Feb 18 '15 at 15:15
  • No I am not mixing up---this is relativistic QM, not QFT. – konstam56 Feb 19 '15 at 02:41

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For the mass $m>0$, spin $j$ representation of the (universal cover of the) Poincaré group, the theory of induced representations give the transformation law $$(u(A,a)\phi)(p) = e^{ip\cdot a} D^{(j)}(A_pAA_{\Lambda(A^{-1})p})\phi(\Lambda(A^{-1})p),\qquad \forall\phi\in L^2(\Omega_m^+,\text d\Omega_m^+)\otimes\mathbb C^{2j+1}$$ where $(A,a)\in SL_2(\mathbb C)\ltimes\mathbb R^n$, $D^{(j)}$ is the spin-$j$ irreducible representation of the little group $SU(2)$ (recall that I have assumed $m>0$ here for definiteness), $p\mapsto A_p$ is any section from the orbit $\Omega_m^+$ to Lorentz transformations in $SL_2(\mathbb C)$ (the exact details of these objects come from the Mackey-Wigner theory of induced representations) and $\text d\Omega_m^+$ is the invariant measure.

For spin $j=0$, this transformation law reduces to $$(u(A,a)\phi)(p) = e^{ip\cdot a}\phi(\Lambda(A^{-1})p),\qquad p\in\Omega_m^+,$$ and if we neglect the action of the translation semigroup this further reduces to $$(u(A,0)\phi)(p) = \phi(\Lambda(A^{-1})p),\qquad p\in\Omega_m^+.$$ Taking a Fourier transform to go from momentum space to position space, the transposed action is seen to be $$(u(A,0)\psi)(x) = \psi(\Lambda(A^{-1})x),\qquad x\in\mathbb R^4,$$ which is the special case of the more general relation $$(u(A,a)\psi)(x) = \psi(\Lambda(A^{-1})(x-a)),\qquad x\in\mathbb R^4.$$

Phoenix87
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  • Thanks, this is informative but I'm afraid does not answer the question: how does the said transformation on the wave function arise in the Hilbert space as a result of a passive basis change? That is, how to arrive at the first equation, and in particular $\phi(\Lambda(A^{-1}) p)$, in the first place? – konstam56 Feb 17 '15 at 03:16
  • As I have said in the answer, that requires the theory of induced representations (and in postocular the little group method). The derivation of that formula is not hard but quite lengthy and technical, so I guess this is why in most of the books (e.g. Weinberg) this is just motivated on physical grounds. I suggest you then have a look at (Wigner 1939). – Phoenix87 Feb 17 '15 at 09:04
  • Let me try to explain the problem from a different perspective. I am already assuming that we have Wigner's (1939) result, that we know how the boost $U(\Lambda)$ acts on a basis vector in the Hilbert space [1],

    $$ U(\Lambda) |p, \sigma \rangle = \sum_{\sigma'} |\Lambda p, \sigma'\rangle D_{\sigma' \sigma}(W(\Lambda, p)) $$

    Take a boost with no rotation, D = \mathbb{1}, then the basis vector transforms

    $$ U(\Lambda) |p, \sigma \rangle = |\Lambda p, \sigma\rangle $$

     – konstam56 Feb 17 '15 at 10:07
  • Ignore now the spin part since it undergoes an identity transformation in this particular situation. Question: given the state $|\Psi\rangle$ above, how do you show that the transformed wave function is given by $\psi(\Lambda^{-1} p')$ or by your last but one equation?

    [1] W.-K. Tung (1985). Group Theory in Physics.

     – konstam56 Feb 17 '15 at 10:07
  • In order to answer to this question we have to use rather formal terms, starting from $\langle p'|p\rangle = \delta(p'-p)$. Given this we then have $\langle p'|\Lambda p\rangle=\delta(p'-\Lambda p)=\delta(\Lambda(\Lambda^{-1}p'-p))=\delta(\Lambda^{-1}p'-p)$, since $\det(\Lambda)=1$. Hence the transformation sends $\psi_p(p')$ to $\psi_p(\Lambda^{-1}p')$. – Phoenix87 Feb 17 '15 at 10:30
  • This is correct, but the problem is, as shown above in my question, that we need to calculate the representation of the wave function in the new basis $|p' \rangle$, that is, $\langle p' | \Psi \rangle$. Taking now a fixed $\langle q'| = \langle q | U^{\dagger}(\Lambda)$, we need to evaluate, as above in the question, $\langle q | U^{-1}(\Lambda) | p \rangle$. This gives $\psi(\Lambda q)$, which is wrong since has no inverse $\Lambda$.

    If instead we had $\langle q | U(\Lambda) | p \rangle$, the correct result would follow.

     – konstam56 Feb 17 '15 at 10:58
  • you have to decide if you want an active or a passive transformation. You cannot expect to use the rules of an active transformation to derive the result of a passive transformation. Indeed active differs from passive by an inverse in this case. – Phoenix87 Feb 17 '15 at 11:16
  • Thanks, I guess this is where the mistake in my reasoning lies. Could you possibly elaborate on which is the correct expression for a passive transformation and which is the correct one for an active one? – konstam56 Feb 17 '15 at 12:10
  • See this answer http://physics.stackexchange.com/questions/158398/peskin-and-schroeder-passive-and-active-translation/158448#158448 that hopefully will solve your question. Here, if you apply the transformation on the coordinates you are actually moving the system somewhere else. Conversely if you transform the wave function you are determining how this change at the same point in space-time. – Phoenix87 Feb 17 '15 at 12:21
  • Still trying to connect the dots. In the answer, you say `a passive transformation is a mere change of coordinates'. This seems in line with how passive transformations can be understood in vector spaces: when basis changes $|p\rangle \mapsto |p'\rangle = U|p\rangle$, the coordinates $\psi$ of a vector $|\Psi\rangle$ go in the reverse direction, $\psi' = U^{-1}\psi$. The latter can be easily obtained in the finite dimensional case by calculating $\langle p' | \Psi \rangle$. – konstam56 Feb 18 '15 at 07:17
  • That is, if the basis change is $|p'i\rangle = U|p_i\rangle = \sum_j |p_j\rangle U{ji}$, then $\psi'k = \langle p'_k | \Psi \rangle = \langle p_k| U^{\dagger} | \Psi \rangle = (\sum_j \langle p_j| U^{\dagger}{kj}) (\sum_i \psi_i |p_i\rangle) = \sum_{ji} U^{-1}{kj} \psi_i \langle p_j | p_i \rangle = \sum_i U^{-1}{ki} \psi_i$. However, applying exactly the same reasoning in the case of continuous spaces (in my question above) does not give the inverse transformation $\Lambda^{-1}$ for the wavefunction. – konstam56 Feb 18 '15 at 07:18
  • I guess my question is that since a vector space is a manifold covered by a single coordinate system, then if I change the basis in the vector space and then ask what are the (new) coordinates of the vector $|\Psi\rangle$ in the (new) basis $|p'\rangle$, the answer is precisely: they are $\langle p' | \Psi\rangle$ and this works out perfectly in the finite case. Why does it not in the continuous case? – konstam56 Feb 18 '15 at 07:20