Looking to liven up some tuna, I went into the back of the spice cabinet for some Old Bay Seasoning. I wonder if it's still good?
Same great taste for over 50 years
Wow; I guess it's still good then.
This weekend's White Chocolate Raspberry cheesecake came out as I intended.
What I did was combine two recipes - the Cook's Illustrated "Rich and Creamy" cheesecake recipe and the "chocolate raspberry truffle" cheesecake recipe from Houghton's book, with white chips substituted for dark.
The crust is just graham crackers, butter, and sugar.
The chips (Callebaut, courtesy of my gourmet package store), sour cream, and heavy cream get microwaved and stirred. The cream cheese, sugar, eggs, vanilla, and lemon (I used juice instead of zest here), are mixed, and the chocolate mixture is added during the last step, so that everything gets a nice blend.
The raspberry jam is melted. After pouring most of the batter into the pan, the jam is carefully spooned on top (I want it to rest on the surface at this point, not pour through to the bottom), then the rest of the batter added on top (again, trying not to pour through). A quiet stir with a spoon now swirls the raspberry through the batter, leaving pretty lines on top. I wasn't entirely satisfied with the effect, so I scraped a bit more raspberry out of the pot and added one last spiral on top.
Waterbath. Bake. Cool. Refrigerate overnight.
The cake seems to bleed a little bit - jam escaping from the bottom of the pan, I'm not sure why. I've been using a bottom only crust (as opposed to running the crust up the sides of the pan as well), so that may be part of it (although I use the same technique with the pumpkin cheesecake, and have no problems). Maybe it was the pan I used, or I didn't secure it properly.
But I managed to avoid the chocolate sink problem I had last time. Yay me.
As usual, I offered Raspberry Chambord sauce (strawberries, raspberries, OJ, Chambord; blend, chill) for folks to spoon onto the cake.
I am not discouraged, because every wrong attempt discarded is another step forward.
I've not yet found a recipe for white chocolate raspberry that I like ("Mmmph mmmFuh mmph," objects the tasting panel with their mouths full - but I'm a perfectionist). Last nights attempt invoked a standard cheesecake recipe (courtesy of Cooks Illustrated), with the sugar cut way back.
One step that I like, stolen from a previous (and now abandoned) formula, calls for pouring in most of the batter, then adding a layer of raspberry, then the remaining batter, and carefully swirling the batter so that nice raspberry spirals appear.
I tried adding the chocolate with the raspberry - that didn't work. It looked like everything was going fine; the chocolate swirled splendedly and so on. But when I cut into the cake I discovered that the chocolate, being denser than the batter, simply sank. Not what I'm after.
My current theory is that I have to abandon adding the chocolate as a seperate step, and mix it thoroughly into the batter. We'll see how that goes this weekend.
Instead of getting sleep last night, I spent a bunch of hours working on Hermite Polynomials. Yeah, yeah - I could have looked everything up on MathWorld; that's not the point.
It comes about, in exploring the eigenstates of one dimensional harmonic oscillators, that
φn = An Hn(ϵ) exp ( - ϵ ² / 2 )
where
Now, if you apply the lowering operator to this, you should get the next lower state; that is:
a φn = sqrt(n) φn-1
The lowering operator
a = sqrt(1/2) ( ϵ + ∂/∂ϵ )
is straight forward enough, and with a bit of jiggering you can demonstrate that the desired relationship holds, IF you can establish that
( ∂/∂ϵ ) Hn = 2 n Hn-1
So that was the goal - to demonstrate that last equality.
The Hermite polynomials are solutions to the Hermite equation
Hn '' - 2 ϵ Hn ' + 2n Hn = 0
So
H0 = 1
H1 = 2 ϵ
etc. The text I was working from had several entries, so I was able to establish a pattern, defining Hn in terms of Hn-1. This suggests an induction proof; assume the Hermite equation holds for Hn-1, describe Hn in terms of Hn-1, and demonstrate that the Hermite equation must also hold for Hn.
I found the most effective way to achieve this was to rewrite everything in terms of operators.
The Hermite equation can be rewritten
(∂²/∂ϵ² - 2 ϵ ∂/∂ϵ + 2 n ) Hn = 0
[note: n is a number, not an operator]
And we want to demonstrate that this definition of Hn satisfies it
Hn = ( 2 ϵ - ∂/∂ϵ ) Hn-1
So substitute it in
0
= (∂²/∂ϵ² - 2 ϵ ∂/∂ϵ + 2 n ) ( 2 ϵ - ∂/∂ϵ ) Hn-1
and then work carefully through the operators using the product rule. Nothing cancels out, but the terms can be redistributed to produce
= ( 2 ϵ - ∂/∂ϵ ) [ (∂²/∂ϵ² - 2 ϵ ∂/∂ϵ + 2 (n-1) ) Hn-1 ]
The term in the brackets is the Hermite equation for n-1, which is satisfied by Hn-1. Establishing that H0 = 1 satisfies the Hermite equation is trivial, so we can build up the rest by induction.
Having established the correct definition of Hn, proving the relationship of the derivative is straightforward, again following the operator approach
Hn'
= ∂/∂ϵ ( 2 ϵ - ∂/∂ϵ ) Hn-1
= ( 2 + 2 ϵ ∂/∂ϵ - ∂²/∂ϵ² ) Hn-1
That bolded expression matches the Hermite equation, so substitute...
= ( 2 + 2 (n-1) ) Hn-1
= 2 n Hn-1
qed.
The first seven or eight pages of my notes show that, while I was getting a better grip on what Hn was, I wasn't getting closer to proving anything. Once I hit on the idea of working directly upon the operators (without actually applying them to anything), both pieces of the proof just fell into my lap.
Casey and Andy. That should be where the permalink ends up (at the moment it's a redirect to the right place, so there you go).
Actually, in my life this was Amazon, not Netflix....
One of my first dates, lost in the darkness of memory, included the movie Go Fish, showing at Derinderpentent Filmhaus. Not my usual fare, but when a date suggests starting the evening by watching lesbian sex scenes... sweet! Anyway, fond memories, decent movie; when I discovered it was available on DVD, I ordered it.
And for interminable period afterwards, Amazon was certain I was a lesbian. The "recommended for you" list got pretty wild, since I was ordering many physics texts at the time. So tell me, Amazon, what are the other lesbian particle physicists reading this week?
Finally, I got fed up with it, and poked around until I tracked down the "don't use this for recommendations" option, effectively reclassifying myself as a straight particle physicist. Getting warmer.
A week later, I ordered a copy of Antonia and it started all over again....