# Confusion on symmetry and basis transformation

Let {$$|a_n\rangle$$} and {$$|b_n\rangle$$} be two basis related by: $$|b_n\rangle = \hat{U}|a_n\rangle \forall n$$.

From my understanding then the unitary operator $$\hat{U}$$ only transforms the basis {$$|a_n\rangle$$} into {$$|b_n\rangle$$} (just like in 2D geometry having a rotation operator which changes the basis $$\hat{x},\hat{y}$$ to $$\hat{r},\hat{\theta}$$).

If there is an operator $$\hat{\Omega}$$, then its representation in basis {$$|b_n\rangle$$}: $$\langle b_n|\Omega|b_m\rangle = \langle a_n| \hat{U}^\dagger\Omega\hat{U}|a_m\rangle$$ $$\Omega \to \hat{U}^\dagger\Omega\hat{U}$$

On the other hand, consider the following unitary transformation: $$|\psi\rangle = \Omega|\phi\rangle$$ $$\hat{U}|\psi\rangle = \hat{U}\Omega\hat{U}^\dagger\hat{U}|\phi\rangle$$ $$\Omega \to \hat{U}\Omega\hat{U}^\dagger$$

1)I am getting very confused by the difference between these, shouldn't the operator $$\Omega$$ transform in the same way?What is the difference between the two things I am doing?

• See Hanting's comment. This is a directional confusion: $U|\psi\rangle$ would be going from $b$ basis to $a$ basis. For consistency with your first part, you would want to substitute $|\psi\rangle \rightarrow U^{\dagger}|\psi\rangle, |\phi\rangle \rightarrow U^{\dagger}|\phi\rangle$, and $\Omega \rightarrow U^{\dagger}\Omega U$ into your $|\psi\rangle = \Omega|\phi\rangle$, and then you have consitency. – dsm Mar 16 '19 at 19:23
• I don't think I am going from $b$ basis to $a$ basis; on contrary I believe that transformation corresponds to $a\to b$. Imagine I have an operator $\Omega$ with a known matrix representation in basis $a$. What I am doing is finding the corresponding representation in basis $b$ i.e. $a \to b$; which transforms as $\Omega \to U^\dagger \Omega U$. Isn't that correct? – Daniel Duque Mar 16 '19 at 20:52
• In the first part yes, but lets look at the components of $\psi$ in the $a$ basis: $\langle a|\psi\rangle = \sum_{b}\langle a|b\rangle \langle b|\psi\rangle$, which if we use subscripts to note the basis is saying that $\psi_a = U\psi_b$. What you are wanting to look at in that second to last equation is $U^{\dagger}|\psi\rangle$. And note that this is all because in the first part you originally established your $U$ as going from $a$ to $b$. – dsm Mar 16 '19 at 21:15

I suppose I'll formally write this up since there seems to still be some confusion. Let's firmly establish that our $$U$$ is a transformation from $$a$$ to $$b$$, that has it's representation in the $$a$$ basis as

$$\langle a_i |U|a_j\rangle = \langle a_i|b_j\rangle$$

Let's look at how the representation of $$|\psi\rangle$$ in the $$a$$ basis transforms when we go to the $$b$$ basis:

$$|\psi\rangle = \sum_{j}|b_j\rangle\langle b_j|\psi\rangle=\sum_{j}\sum_{i}|b_j\rangle\langle b_j|a_i\rangle\langle a_i|\psi\rangle$$

Now pick out the $$k$$'th component of $$b$$

$$\langle b_k|\psi\rangle = \sum_{j}\sum_{i}\delta_{kj}\langle b_j|a_i\rangle\langle a_i|\psi\rangle = \sum_{i}\langle b_k|a_i\rangle\langle a_i|\psi\rangle \\ = \sum_{i}(\langle a_i|b_k\rangle)^{\dagger}\langle a_i|\psi\rangle = \sum_{i}(\langle a_i|U|a_k\rangle)^{\dagger}\langle a_i|\psi\rangle.$$

Letting subscripts denote the basis (i.e. $$|\psi\rangle_a \equiv \langle\vec{a}|\psi\rangle$$, and likewise for $$b$$), we see that this is telling us $$|\psi\rangle_b = U^{\dagger}|\psi\rangle_a$$. Now, from $$U|a_i\rangle=|b_i\rangle$$ we know that $$\Omega_b = U^{\dagger}\Omega_a U$$, so lets check that everything is consistent with your $$|\psi\rangle = \Omega|\phi\rangle$$ when we hit it with $$U^{\dagger}$$. Keeping subscripts to denote the basis for absolute clarity:

$$|\psi\rangle_a = \Omega_a|\phi\rangle_a \to U^{\dagger}|\psi\rangle_a = U^{\dagger}\Omega_a|\phi\rangle_a$$

Looking at each side individually, we have

$$U^{\dagger}|\psi\rangle_a = |\psi\rangle_b \\ U^{\dagger}\Omega_a|\phi\rangle_a =U^{\dagger}\Omega_a U U^{\dagger}|\phi\rangle_a = \Omega_b |\phi\rangle_b,$$

which shows us that everything is nice and consistent:

$$|\psi\rangle_a = \Omega_a|\phi\rangle_a \xrightarrow{U^{\dagger}} |\psi\rangle_b = \Omega_b|\phi\rangle_b$$

• I just want to add something in case someone has the same confusion I had. The operator $\hat{U}$ transforms the basis $|a\rangle$ into the basis $|b\rangle$; a helpfull analogy would be a rotation in 2D; imagine that our new basis is a rotation by an angle $\theta$. If we are interested on how a 2D vector looks like from the perspective of the rotated basis, all we have to do is rotate the vector in the opposite direction that the basis rotated. That is why $|\psi\rangle_b$ is obtained by multiplying $|\psi\rangle_a$ with $\hat{U}^\dagger$. Please correct me if I'm wrong. – Daniel Duque Mar 17 '19 at 19:02
• That is correct. Not to add any confusion, but I will add that although $U$ has been constructed to take $a$ to $b$, it likewise can take $b$ to $a$: $|\psi\rangle_a = U|\psi\rangle_b$ and $\Omega_a = U\Omega_b U^{\dagger}$. This is just from hitting our $|\psi\rangle_b = U^{\dagger}|\psi\rangle_a$ with a $U$ and the left, and sandwiching $\Omega_b = U^{\dagger}\Omega_a U$ between a $U$ on the left and $U^{\dagger}$ on the right. You can then carry over your rotation analogy to see why these also make sense. – dsm Mar 17 '19 at 19:31

In the first example, you're taking $$|b_i\rangle$$ into $$|a_i\rangle$$, while in the second, you're doing the opposite. Explicitly, if $$|\psi\rangle = \sum_n b_n|b_n\rangle$$ then $$\hat U|\psi\rangle = \sum_n b_n\hat U|b_n\rangle.$$

But we don't know anything about $$U|b_n\rangle$$! To go from $$|b_i\rangle$$ into $$|a_i\rangle$$, we what we need is $$\hat U^\dagger$$. Then the calculation will come out as $$\Omega \mapsto \hat U^\dagger \Omega \hat U$$.

• Could you clarify why am I going from $|b\rangle$ into $|a\rangle$ in my first example? The way I see it is: I know the matrix representation of $\Omega$ in basis $|a\rangle$, then I find the matrix representation in $|b\rangle$ which is $U^\dagger \Omega U$ i.e. I went from $|a\rangle$ to $|b\rangle$. Isn't that correct? – Daniel Duque Mar 16 '19 at 20:55
• Well, you can see from the matrix elements that you've gone from $\langle b_n|\Omega|b_m\rangle$ to $\langle a_n| \hat{U}^\dagger\Omega\hat{U}|a_m\rangle$. The left side has $|b_n\rangle$, the right side has $|a_n\rangle$ – cxx Mar 17 '19 at 0:01
• Yes, in the left hand side I have $|b\rangle$ which means that I have obtained it from the rhs $|a\rangle$ i.e. $|a\rangle \to |b\rangle$. Sorry if it is too obvious, but I just can't see it. In the comments on the question someone agrees that I went from $|a\rangle \to |b\rangle$. – Daniel Duque Mar 17 '19 at 0:11

$$\let\Om=\Omega \let\dag=\dagger \def\ket#1{|#1\rangle} \def\bra#1{\langle#1|} \def\braket#1#2{\langle#1|#2\rangle} \def\mxelm#1#2#3{\bra#1\,#2\,\ket#3}$$

As you'll see, I'll make some change of notations I can't explain for brevity. In your first definition you're asking for an operator $$\Om'$$ which in $$a$$-rep has the same matrix elements $$\Om$$ has in $$b$$-rep: $$\mxelm{{a,m}}{\Om'}{{a,n}} = \mxelm{{b,m}}\Om{{b,n}}$$ $$\Om' = U^\dag\,\Om\,U.$$

Afterwards you assume some operator $$\Om$$ sends ket $$\xi$$ into $$\eta$$: $$\ket\eta = \Om\,\ket\xi.$$ Then you define $$\ket{\xi'}$$, $$\ket{\eta'}$$ such that they have in $$b$$-rep the same components $$\ket\xi$$, $$\ket\eta$$ have in $$a$$-rep: $$\braket{b,n}{\xi'} = \braket{a,n}\xi$$ $$\mxelm{{a,n}}{U^\dag}{{\xi'}} = \braket{a,n}\xi$$ $$\ket{\xi'} = U\,\ket\xi \qquad \ket{\eta'} = U\,\ket\eta.$$ Lastly you look for $$\Om'$$ sending $$\ket{\xi'}$$ into $$\ket{\eta'}$$: $$\Om'\,U\,\ket\xi = U\,\ket\eta = U\,\Om\,\ket\xi.$$ If this is to hold for all $$\ket\xi$$ then $$\Om' = U\,\Om\,U^\dag.$$

It should be clear that in your former part you transformed an operator, whereas in the latter you transformed kets under the opposite requirement.